That was arduous. Old René sure loved his quartics 😅

Seems like Descartes was sooo close to discovering differential calculus!

This method is so beautiful and I also appreciate the historical aspect of it. Thank you for sharing it!

Great video. Loved it. Descartes also modified or refined his method to by-pass the circle. It's all in the double root. Late 1630's maybe? For your example: Center is still C(a,0) with P(3,9) slope_R = 9/(3-a) ====> slope_T = (a-3)/9 Equation Of Tangent: y - 9 = (a-3)/9 * (x - 3) ===> y = (a-3)/9 *x + (30-a)/3 Intersection of Curve with Tangent: x^2 = (a-3)/9*x + (30-a)/3 ====> x^2 + (3-a)/9 * x + (a-30)/3 must equal x^2 - 6x +9 for a double root at x=3. Equate co-efficients of x: (3-a)/9 = -6 =====> a = 57 m_t = (57-3)/9 = 6 You can verify the constant term if you want: (57-30)/3 = 9 Easier to force the Tangent to have a double root than the Circle. ======================================================= Here's a cubic with no circle in sight :Find the slope of the tangent to the curve y = 2x^3 + 4x^2 - 5x + 3 at x = 1 ===> y = 4. 1) The Equation of the tangent: y - 4 = m(x - 1) 2) Intersect with the curve: 2x^3 + 4x^2 - 5x + 3 = mx - m + 4 2x^3 + 4x^2 +x[ -5 -m] + [m-1] = 0 and FORCE a double root at x = 1 (necessary condition for tangency) 3) 2x^3 + 4x^2 +x[ -5 -m] + [m-1] = (x-1)^2*(2x - t) 4)Expand the RHS : 2x^3 + x^2 [-t - 4] + x[2t + 2] - [ t ] and equate coeffiicients. 5) (a) 4 = -t - 4 ===> t = -8 (b) -5 - m = 2t + 2 ====> m = 9 ( c) -t = m - 1 not needed, but -8 and 9 verify. You can also adapt this method for finding Stationary Points. Requires polynomial long division though - but no Calculus.

So one last comment on the "circle method". If you try y = x^3 at (2,8) you quickly discover that the y^2 term in the circle equation becomes x^6. That means forcing a double root gets you (x - 2)^2 * [quartic] . No wonder Descartes revised his method. I bet he got a lot of push back on that one. On the other hand, if you intersect y = x^3 with the tangent you get : x^3 - mx + (2m - 8) = 0 and then force a double root at x = 2: (x - 2)^2*(x - t). You very easily get t = -4 and m = 12. One of the giant's shoulders that Newton mentions he stood upon was most certainly Descartes ( not to mention Kepler, Galileo, ...). Pretty fine praise from the tallest giant of them all. Thanks very much Michael for such a great classic solution.

Here is another cool method - no calculus. Given f(x) = 5x^4 - 3x^3 + 2x^2 - 4x + 3. Divide by x-2. to get constant remainder R and quotient Q(x). Evaluate Q(2) and note remaiber R. Now use come calculus to find the equation of the tangent at x=2 and voila! Can you prove this always works?

I can see professor's pain when avoiding himself to use derivates (which wasn't invented yet).

Another cool video might be: Fermat's Method for Finding Tangents.

This is just creepy. I was just thinking about what a version of Taylor series where you approximate the function using circles instead of polynomials, and it reminded me of Descartes's method of finding tangent lines. Now I see a video about it in my feed. Is Michael stalking me?

I really love this video, keep up the good work sir.

Bruh everything can be bruteforced in coordinates U could even just use linear equations and some geometry for this I thought it was some secret method Turns out it was something SIMILAR to what i did in highschool

I think I imagined that mathematicians before the development of calculus weren't even trying to find tangents to curves, but of course that's too important of a mathematical idea to leave behind. So *of course* they were finding tangents. Never even wondered how they did it until now!

Isn't the hypotenuse the longest side of a right triangle? If so, O-P couldn't be the hypotenuse. Did I hear that wrong?

Question: What is the symbol Michael uses for "at" at 7:49. It looks like a 3 with a dot above it. I've been doing mathematical proofs for over 40 years, and I am not familiar with it.

The quadratic does not need to be irreducible. Try y = x^3-9x^2+24x-3 at x=3. So cool. What is happening at x = 3??

You can tell this is a historical method of solving a problem because it requires knowing half a dozen highly specific Polynomial Facts

I like this.

You sound liked "Dr Peyam" ! 🤣

Descartes should have discovered the derivatives instead 🤭

Here are my thoughts on why I think the quadratic factor, (x^2+bx+c), in (x-3)^ (x^2+bx+c) is automatically irreducible. In general a parabola and a circle can have as many as 4 intersection points. But here you are forcing the circle to have its center on x-axis and the parabola to open up or down. Such a circle and such a parabola can intersect in at most 2 points. Also, you are forcing the circle to be tangent to the parabola - hence the double root. So if the quadratic (x ^2 + bx + c) on the RHS : (x-3)^2*(x^2 + bx + c) was reducible then it would produce 2 more real roots ====> 3 intersection points between the circle (center on the x - axis) and the parabola (opening up). This cannot happen. Therefore x^2 + bx + c = 0 will have no real roots. If the parabola opens right or left then just place the center on the y-axis. Thoughts?

OP2 + PQ2 = OQ2 !!!!

can you explain how thats the irreducible form of a quadratic?

Why don't you use the Ruffini theorem?

3:00 shouldn't it be OP^2 + PQ^2 = OQ^2? OP is not the hypotenuse!

@3:38 why does this imply P=Q? Think of an isosceles triangle ABC where BC lay on the same line and AB=AC yet B does not equal to C

What!

1,4,9 = 3,5,4, while 19 = (a,0 & 3,9) , 1/6= (0,54) gremlins

Noice

The proof that the radius is perpendicular to the tangent is flawed. In particular, OP = OQ does not (by itself) imply P = Q. (They could be sides of an Isosceles triangle, so you have to bring in the fact that the base angles would then both be 90°, meaning that the triangle was degenerate.) A simpler argument is to go back to the inequality OQ^2 + PQ^2 ≤ OQ^2, hence PQ^2 ≤ 0, from which it follows that P = Q since PQ^2 ≥ 0.

Going through all that for such a simple problem makes me really appreciate differential calculus. Imagine if we could only calculate slopes at points, and every time we had to do all of that!

For the radius being perpendicular to the circle I think a proof by contradiction would have been easier to follow. Suppose OP is not perpendicular to the tangent. Let Q ≠ P be such that OQ is perpendicular to the tangent. That makes OQ the shortest distance from O to the tangent and therefore OQ < OP. But OP is the hypotenuse of the right triangle OPQ and therefor OP > OQ. By contradiction then the assumption that OP was not perpendicular to the tangent must have been wrong.

The fact the y-co-ord of the tangent circle is zero comes from a symmetry argument?

Can you explain why a symmetry argument is insufficient?

I think he made a mistake. OP^2+PQ^2=OQ^2.

Now generalise it .... see you next year.

OQ^2+QP^2 = OP^2 REALLY? A small mishap. You are actually very good and relaxing.

Euclid, Descartes, Newton and Einstein: 0D (indivisible, subatomic, non-natural): not fundamental 1D, 2D, 3D and 4D (divisible, atomic, natural): fundamental Leibniz: 0D (indivisible, subatomic, non-natural): fundamental 1D, 2D, 3D and 4D (divisible, atomic, natural): not fundamental Can someone explain to me how stuff that can be divided further is fundamental and stuff that cannot be divided further is not fundamental? In geometry any new dimension has to contain within it all previous dimensions. Kinda seems like Euclid, Descartes, Newton and Einstein sucked at geometry.