The area of the blue piece isn't just rational just like the pink piece is, it is identical to the pink piece (since the cubic is rotation-symmetric around its inflection point).
According to wikipedia, Pythagoras worked out solution of quadratic equation using geometry...so it's kind of one of the many Pythagoras "theorems" ;-)
Do you think it was an intentional joke or not? Sometimes content creators can put corrections on the screen while keeping the video, with little effort.
Why does the thumbnail show the 2 regions being bounded by the x-axis and not the horizontal tangents? I’m paused at 5:40 after the transformation to g(x).
I wonder, how do the authors come up with such sophisticated problems. Do they stumble upon them during their own scholar routine or carry out special research to find something worth publishing in this kind of math almanacs? :D (I'm from nat. sciences world and was never good with pure math and a corresponding PoV)
The thumbnail doesn’t show the areas bounded by the curve and the tangent line. Rather, it shows the regions bound by the curve and the x-axis between each pair of x-intercepts.
I'm stuck at why not take double integral on the original curve and horizontal lines passing thru y@ t1 and y@t2 Of course there will be two double integrals to evaluate - but all required data is there plus hints of symmetry associated with the wobbly bits about a point of symmetry. I very vaguely recall some symmetries of cubics. Conic sections and all that? Another excellent example of: that is a good place to stop and an excellent place to start easily partitioned into educator and learner duties. Speculation 1a: rotational symmetry of the cubic curve lies precisely at point of intersection with a straight line passing thru t1 and t2 as given in video. Proof: Speculation 1b: there is precisely one distinct rotational symmetry at π and infinitely many rotational symmetries at 2kπ and 2(k+1)π for k in ℤ Specualtion 2a: areas bounded by G(x)= x^3 + ax^2 + bx + c and -G(x) are identical to bounded areas between G(x) and G(-x) Proof Speculation 2b: areas identified in 2a are rational Proof Speculation 3a: rotation by 2qπ for q in ℚ forms a dense partitioning of ℝxℝ with continuity confined to each strand of rotated curve plus, of course, original curve itself. That is if (a,b) lies on a strand then continuity exists along that strand. Or does it? (In effect does rotation by a quotient spoil continuity?) Proof or negation:
I think it has to do with isomorphisms and stuff like that. Every x^2 +ax +b can be morphed from x^2 by transformations. Likewise, and I am not too sure about this, every x^3 + ax^2 + bx + c can be morphed from x^3 OR x^3 + x OR x^3 - x. It seems a basis for solving difficult problems by taking it back to its generator. Also brings in ideas that can be built on such as isomorphisms and that a 1 to 1 is really a 2 to 2 too. It has a pedagogical link in sense that some folks understand from algebraic representation alone whereas some folks like a pictographic insight (hmmmm pictographic incite?) (hmmm or both?) or a geometric approach. Apparently it is something to do with learning theories. Rest assured tho that topics covered in this video are rock solid important and not to be ignored unless you really really really want to. So some easy examples might be multiplication by transforming to addition. That is 1 to 1. Once an answer is reached transforming it back to original form is also 1 to 1. Both together give a 2 to 2 too. These simple notions can be applied to many things such as integration why does substitution work? why do endpoints of integral have to be changed? Once solved why do things need to be changed back to original terms?
There's an easier way to do this. Simplify to a monic-depressed cubic, by scaling it vertically, and shifting it horizontally. It now has the form: y = t^3 - p*t + q I put the negative sign for p, since we're interested in a "roller coaster" cubic with a negative linear term, and so p can be positive. All cubics have half-turn symmetry around the inflection point. By symmetry, the area between the lower horizontal tangent line at its intersection points and the cubic, will equal the area above the cubic to the upper horizontal tangent line. So the area we're interested in, is twice the area between the cubic and the lower horizontal tangent line. Let A be the full area of interest, and A0 as the area we'll calculate to find it. Thus A = 2*A0 When the lower tangent line intersects the cubic, it'll have a distinct real solution with t being negative, and a repeated real solution, when t is positive. Shift the graph vertically, until its positive t solution is a repeated real solution. The area A0, will be the area under this curve from the negative t solution, to the positive t solution. Find the value of q that makes it have a repeated real solution at t2 which is positive. For this to happen, the cubic discriminant is zero. D = (-p/3)^3 + (q/2)^2 D = 0 (q/2)^2 = (p/3)^3 q = +/- 2*sqrt((p/3)^3) We're interested in q being positive, so the inflection point is above the lower tangent line at the x-axis. Use the cubic formula to find the t1, the distinct real solution, using D=0. t1 = cbrt(-q/2 + sqrt(D)) + cbrt(-q/2 - sqrt(D)) t1 = -2*cbrt(q/2) t1 = -2*sqrt(p/3) And the corresponding t2 is half the magnitude and on the opposite side: t2 = sqrt(p/3) Integrate: A0 = integral (t^3 - p*t + q) dt = A0 = 1/4*t^4 - 1/2*p*t^2 + q*t, eval from t1 to t2 After simplifications: A0 = 3/4*p^2 Thus: A = 3/2*p^2 So, as long as our parameter p is a rational number, the area A, will also be a rational number. For rational roots to occur, p/3 needs to be a square number, or ratio of square numbers, so p = 3*r^2, where r is any positive rational number.
@MyOneFiftiethOfADollar it's probably about the same length as the video tbh. Just uses some stuff that most people don't know. Obviously in text form you'd expect it to be kinda long, same with the transcript of this video no doubt
@@Mmmm1ch43l I meant that we could have argued right from the start that we could shift the graph of the function f, which has rational coefficients, so that one of the points with horizontal tangent of the shifted graph lies on the origin. Then at the end, one would only have to argue why alpha^4 is a rational number. No need to actually calculate the value of alpha, it would have been enough to show that it is a square root of a rational number.
@@bjornfeuerbacher5514 so you do the same steps as in this video except in a different order? seems like that would be less understandable, no? and showing that alpha is the square root of a rational number pretty much just amounts to calculating what it is, doesn't really get much easier than that
The great discoveries normally are made putting together tons of tiny bits of information like this one. A tiny metal sphere is not that useful by itself, but bikes and cars (for example) would not work (at least not for long) without a lot of them.
@@maxhagenauer24 picture on the right is a translation of picture on the left..so g(x) and f(x) have same high order monomial coefficient : 1 for x^3. Visually : curves have the same shape and orientation. Check it for two elementary translations parallel to x-axis and y-axis or using change of variables X = x - t1 and Y = y - f(t1). Another way is to check that f'''(0) = g""(0) (using chain rule, derivative of composite function g).
Yes. You can construct a general case of a cubic with one real distinct root (at x = r1), and one repeated real root (at x= r2). To keep it simple, let it be a monic, depressed cubic. I.e. coefficient on x^3 is 1, and inflection point is at the y-axis. All we are doing is shifting it horizontally until its inflection point is at the y-axis, vertically until its lower stationary point is at the x-axis, and scaling it until its monic. So it will be the same shape as any possible original cubics with real coefficients, that is the "roller coaster cubic" with two stationary points. A property of depressed cubics with the real/distinct and the real/repeated root, is the following: r1 = -2*r2 The distinct root is twice as far from the origin as the repeated root, and of opposite sign. This is because the roots must all add up to zero, for a depressed cubic. Let r be the repeated root. Thus, our cubic becomes: y = (x + 2*r)*(x - r)^2 Expand: y = x^3 - 3*r^2*x + 2*r^3 Integrate from x = -2*r to x = +r Integral: 1/4*x^4 - 3/2*r^2*x^2 + 2*r^3*x + C Evaluate from -2*r to +r: [1/4*(r)^4 - 3/2*r^2*(r)^2 + 2*r^3*(r)] - [1/4*(-2*r)^4 - 3/2*r^2*(-2*r)^2 + 2*r^3*(-2*r)] = Which simplifies to: 27/4*r^4 The area Michael is interested in, will be twice this area, which is: 27/2*r^2 So as long as r is a rational number, or at least has a rational fourth power, the area between the two horizontal tangent lines and the cubic curve, will be a rational number. In order for the repeated root r to be a rational number, the linear term of the depressed cubic will be a multiple of -3*r^2, where r is any rational number.
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The area of the blue piece isn't just rational just like the pink piece is, it is identical to the pink piece (since the cubic is rotation-symmetric around its inflection point).
Nice. I think this propeety is used when deriving the metastable region of a van-der-waltz vireal gas in physics.
11:44 and that’s a good
…. place to stop 😂
... place to cut... not.
...place to continue a little bit! 🥺
It is a really good
2:37 Pythagorean theorem?
what the brain associates with "quadratic formula" apparently ;-)
According to wikipedia, Pythagoras worked out solution of quadratic equation using geometry...so it's kind of one of the many Pythagoras "theorems" ;-)
That's what I thought when Michael said that!
Do you think it was an intentional joke or not? Sometimes content creators can put corrections on the screen while keeping the video, with little effort.
Maybe he was thinking and talking too much about newly found proofs for the Pythagorean Theorem these days? 😉😂
3:58 is a good place to sneeze
By Pythagorean theorem we know the roots of quadratic polynomial?
He just misspoke. You can see he uses the quadratic formula.
3:58 salud!
and that's a good ..... (place to stop)
Why does the thumbnail show the 2 regions being bounded by the x-axis and not the horizontal tangents? I’m paused at 5:40 after the transformation to g(x).
I wonder, how do the authors come up with such sophisticated problems. Do they stumble upon them during their own scholar routine or carry out special research to find something worth publishing in this kind of math almanacs? :D
(I'm from nat. sciences world and was never good with pure math and a corresponding PoV)
The thumbnail doesn’t show the areas bounded by the curve and the tangent line. Rather, it shows the regions bound by the curve and the x-axis between each pair of x-intercepts.
Nicely worked solution
I'm stuck at why not take double integral on the original curve and horizontal lines passing thru y@ t1 and y@t2
Of course there will be two double integrals to evaluate - but all required data is there plus hints of symmetry associated with the wobbly bits about a point of symmetry.
I very vaguely recall some symmetries of cubics. Conic sections and all that?
Another excellent example of: that is a good place to stop and an excellent place to start easily partitioned into educator and learner duties.
Speculation 1a: rotational symmetry of the cubic curve lies precisely at point of intersection with a straight line passing thru t1 and t2 as given in video.
Proof:
Speculation 1b: there is precisely one distinct rotational symmetry at π and infinitely many rotational symmetries at 2kπ and 2(k+1)π for k in ℤ
Specualtion 2a: areas bounded by G(x)= x^3 + ax^2 + bx + c and -G(x) are identical to bounded areas between G(x) and G(-x)
Proof
Speculation 2b: areas identified in 2a are rational
Proof
Speculation 3a: rotation by 2qπ for q in ℚ forms a dense partitioning of ℝxℝ with continuity confined to each strand of rotated curve plus, of course, original curve itself.
That is if (a,b) lies on a strand then continuity exists along that strand. Or does it? (In effect does rotation by a quotient spoil continuity?)
Proof or negation:
Just wondering 🎉🎉🎉very very thankful to you sir❤
I dont understand why the original curve pass by (0,0) ??? 0 is not a solution.
It is if c = 0. But, yes, the drawing does not reflect the general case (i.e. when c is non-zero).
I think it has to do with isomorphisms and stuff like that. Every x^2 +ax +b can be morphed from x^2 by transformations.
Likewise, and I am not too sure about this, every x^3 + ax^2 + bx + c can be morphed from x^3 OR x^3 + x OR x^3 - x.
It seems a basis for solving difficult problems by taking it back to its generator.
Also brings in ideas that can be built on such as isomorphisms and that a 1 to 1 is really a 2 to 2 too.
It has a pedagogical link in sense that some folks understand from algebraic representation alone whereas some folks like a pictographic insight (hmmmm pictographic incite?) (hmmm or both?) or a geometric approach. Apparently it is something to do with learning theories.
Rest assured tho that topics covered in this video are rock solid important and not to be ignored unless you really really really want to.
So some easy examples might be multiplication by transforming to addition. That is 1 to 1.
Once an answer is reached transforming it back to original form is also 1 to 1. Both together give a 2 to 2 too.
These simple notions can be applied to many things such as integration
why does substitution work? why do endpoints of integral have to be changed? Once solved why do things need to be changed back to original terms?
There's an easier way to do this.
Simplify to a monic-depressed cubic, by scaling it vertically, and shifting it horizontally. It now has the form:
y = t^3 - p*t + q
I put the negative sign for p, since we're interested in a "roller coaster" cubic with a negative linear term, and so p can be positive. All cubics have half-turn symmetry around the inflection point. By symmetry, the area between the lower horizontal tangent line at its intersection points and the cubic, will equal the area above the cubic to the upper horizontal tangent line. So the area we're interested in, is twice the area between the cubic and the lower horizontal tangent line. Let A be the full area of interest, and A0 as the area we'll calculate to find it. Thus A = 2*A0
When the lower tangent line intersects the cubic, it'll have a distinct real solution with t being negative, and a repeated real solution, when t is positive. Shift the graph vertically, until its positive t solution is a repeated real solution. The area A0, will be the area under this curve from the negative t solution, to the positive t solution.
Find the value of q that makes it have a repeated real solution at t2 which is positive. For this to happen, the cubic discriminant is zero.
D = (-p/3)^3 + (q/2)^2
D = 0
(q/2)^2 = (p/3)^3
q = +/- 2*sqrt((p/3)^3)
We're interested in q being positive, so the inflection point is above the lower tangent line at the x-axis.
Use the cubic formula to find the t1, the distinct real solution, using D=0.
t1 = cbrt(-q/2 + sqrt(D)) + cbrt(-q/2 - sqrt(D))
t1 = -2*cbrt(q/2)
t1 = -2*sqrt(p/3)
And the corresponding t2 is half the magnitude and on the opposite side:
t2 = sqrt(p/3)
Integrate:
A0 = integral (t^3 - p*t + q) dt =
A0 = 1/4*t^4 - 1/2*p*t^2 + q*t, eval from t1 to t2
After simplifications:
A0 = 3/4*p^2
Thus:
A = 3/2*p^2
So, as long as our parameter p is a rational number, the area A, will also be a rational number. For rational roots to occur, p/3 needs to be a square number, or ratio of square numbers, so p = 3*r^2, where r is any positive rational number.
I have to be honest, this doesn't seem easier than the solution in the video (it also requires knowing the cubic formula etc)
Not as elementary as the method in the video, though
The length of your comment even with significant important omissions refutes your “easier” way to do it claim.
@MyOneFiftiethOfADollar it's probably about the same length as the video tbh.
Just uses some stuff that most people don't know.
Obviously in text form you'd expect it to be kinda long, same with the transcript of this video no doubt
And that’s really good…
It seems the answer does not depend on C parameter, so condition for it to be rational is irrelevant
Wlog, we could have started with the shifted graph. Would have made the calculations a bit easier and shorter.
Sometimes it's nice to show why the WLOG is actually a WLOG, but you're right
not quite. notice that g doesn't have rational coefficients
@@Mmmm1ch43l That's actually a very good point that I didn't think about!
@@Mmmm1ch43l I meant that we could have argued right from the start that we could shift the graph of the function f, which has rational coefficients, so that one of the points with horizontal tangent of the shifted graph lies on the origin.
Then at the end, one would only have to argue why alpha^4 is a rational number. No need to actually calculate the value of alpha, it would have been enough to show that it is a square root of a rational number.
@@bjornfeuerbacher5514 so you do the same steps as in this video except in a different order? seems like that would be less understandable, no?
and showing that alpha is the square root of a rational number pretty much just amounts to calculating what it is, doesn't really get much easier than that
What can we build with this information?
The great discoveries normally are made putting together tons of tiny bits of information like this one. A tiny metal sphere is not that useful by itself, but bikes and cars (for example) would not work (at least not for long) without a lot of them.
@@aldineisampaio The emetic answer you give when you don't have an answer.
that wasn't a good place to stop at the end lol
I had misread the title as "pubic curves".
The video editors are having fun. I approve!
6:52 why does that constant have to be 3?
Since the leading term of g(x) is x^3, it implies that the leading term of g'(x) should be 3x^2
@vaidhymama How do you know the leading term is x^3 and not some other coefficient in front of x^3?
@@maxhagenauer24 Because g(x) = (x+t1)^3+... = x^3 + terms with lower powers of x.
@@maxhagenauer24 picture on the right is a translation of picture on the left..so g(x) and f(x) have same high order monomial coefficient : 1 for x^3. Visually : curves have the same shape and orientation. Check it for two elementary translations parallel to x-axis and y-axis or using change of variables X = x - t1 and Y = y - f(t1). Another way is to check that f'''(0) = g""(0) (using chain rule, derivative of composite function g).
@benno365 Not necessarily because we don't know if that 1st coefficient is 1.
This is such a cute problem
The thumbnail is so wrong
2:38 ?!?!?!
It's not even April
Can a high school student understand this who is taking calculus?
Yes. You can construct a general case of a cubic with one real distinct root (at x = r1), and one repeated real root (at x= r2). To keep it simple, let it be a monic, depressed cubic. I.e. coefficient on x^3 is 1, and inflection point is at the y-axis. All we are doing is shifting it horizontally until its inflection point is at the y-axis, vertically until its lower stationary point is at the x-axis, and scaling it until its monic. So it will be the same shape as any possible original cubics with real coefficients, that is the "roller coaster cubic" with two stationary points.
A property of depressed cubics with the real/distinct and the real/repeated root, is the following:
r1 = -2*r2
The distinct root is twice as far from the origin as the repeated root, and of opposite sign. This is because the roots must all add up to zero, for a depressed cubic. Let r be the repeated root.
Thus, our cubic becomes:
y = (x + 2*r)*(x - r)^2
Expand:
y = x^3 - 3*r^2*x + 2*r^3
Integrate from x = -2*r to x = +r
Integral:
1/4*x^4 - 3/2*r^2*x^2 + 2*r^3*x + C
Evaluate from -2*r to +r:
[1/4*(r)^4 - 3/2*r^2*(r)^2 + 2*r^3*(r)] - [1/4*(-2*r)^4 - 3/2*r^2*(-2*r)^2 + 2*r^3*(-2*r)] =
Which simplifies to:
27/4*r^4
The area Michael is interested in, will be twice this area, which is:
27/2*r^2
So as long as r is a rational number, or at least has a rational fourth power, the area between the two horizontal tangent lines and the cubic curve, will be a rational number.
In order for the repeated root r to be a rational number, the linear term of the depressed cubic will be a multiple of -3*r^2, where r is any rational number.