when a quadratic equation has an infinite root.

[@abhinavanand9032]

Great now solve quadratic equation using cubic formula

[@debtanaysarkar9744]

💀💀💀💀💀 we viewers will die if that happens 💀💀💀💀💀

[@vascomanteigas9433]

Maxima goes brrrr.. 😂

[@mizarimomochi4378]

Why stop there? Cubic equation using the quartic formula.

[@jocabulous]

​@@mizarimomochi4378Forbidden math: quartic equation using quintic formula

[@acekingbones]

And then a quartic equation with the quintic formula

[@logician1234]

0:45 "Let's simplify" * erases 0*x^2 *

[@Synthesz1]

because 0*a = 0, then it turns into bx + c = 0, bx = -c, x = -c/b

[@Kokurorokuko]

no shit ​@@Synthesz1

[@Synthesz1]

@@Kokurorokuko just saying bro, no need to randomly say no shit

[@caetanogarelli6657]

@@Synthesz1 They said no shit since what the other person said is obvious.

[@fiNitEarth]

@@caetanogarelli6657no shit

[@martind2520]

Maths teachers: Make sure to always rationalise your denominators! Mathematicians: What if we did _THE EXACT OPPOSITE!!!_

[@nonchalant-turtle]

Dominate your rationalizers

[@__christopher__]

Let's denominate our rationalisations!

[@douglasespindola5185]

Mathematicians always trying to break the math 😂

[@Alan-zf2tt]

Things are easier to see when they are multiples of λ with ones and zeroes but when λ operates on (x, y, z) or even (x: y: z) does that not mean ℝ3 (image? image class?) is now a squished version of ℝ4? I mean (x, y, z) under λ becomes λ(x, y, z) in other words (x, y, z, λ) of course excluding the obvious 𝟎 in each case (each equivalent class? taking x, y, z and λ as equivalent classes if that is a doable and permitted option)

[@tomholroyd7519]

He sells t-shirts that say "Do hard math". That's not because there isn't an easier way. He just likes doing it the hard way.

[@karolakkolo123]

Finally as an engineer I can feel justified in saying that certain linear systems have roots or zeros at infinity

[@tri99er_]

Y=0 is one… It also can be considered to be an infinitely stretched parabola in a way.

[@Fire_Axus]

your feelings can be irrational

[@sebastiangudino9377]

​@@tri99er_ All lines are just stretched polynomials if you squint your eyes

[@Utesfan100]

This factors as (0x+1)(bx+c). Then the roots are clearly -c/b and infinity.

[@backwashjoe7864]

I'm not sure how to solve 0x + 1 = 0, lol. Are you setting it up as lim(a -> 0):[ax + 1 = 0] , to continue Michael's idea? :)

[@paulkohl9267]

Very good. (0x + 1)^n (bx + c), for any allowed n, is also a variety. Wouldn't it be x = -1/0 = - infinity? Or perhaps with projective spaces such directional data is lost? Not limit safe. Noice.

[@cicik57]

0*x+1 must have no solutions?

[@Utesfan100]

@@paulkohl9267 I am picturing a Reimann circle, so +/- infinity are equal.

[@Alan-zf2tt]

@@backwashjoe7864 redefine is as (x - x + 1)(bx-c) ?? Or even (x - x + 1)(bx-c) = x - x

[@HoneyBunny-rg3jx]

We can see that x = -c/b is an unstable solution. And in terms of complex analysis, this instability is because the system has a pole outside the unit circle. This pole is hiding in the form of the root at infinity.

[@HoneyBunny-rg3jx]

@@Simrealism I'll try to summarize it. A Linear Time-Invariant system is any system that takes inputs, provides outputs and satisfies the criteria of linearity and homogeneity. It can be represented by something called an impulse response, H(t). If you provide an impulse input into this system, then its output is characterized by H(t). Suppose, as a result of an impulse, the system outputs c at time index 0 and b at time index 1. This can be succinctly represented as the z-transform H(z) = c + bz^-1. Now consider the system with H(z) = c + bz. What does this tell you? It tells you that as a result of the impulse input, the system outputs c at time index 0 and b at time index -1. Wait, negative time indices? Is that possible? No!!! This is an example of a non-causal system - a system whose past depends on its present. Such systems cannot exist in the real world. But there is such a thing as the inverse of an LTI system. If you take the output of the system and input that into its inverse, the result is the impulse output. If H(z) is the impulse response for a system, then 1/H(z) is the impulse response of the inverse. So for our non-causal system above, its inverse has the impulse response H(z) = 1 / (c + bz). Now let us simply scale this H(z) by b giving H(z) = 1 / (z + c/b). Carry out the polynomial long division by hand and you get (-c/b)z^(-2) as the second term. So, -c/b is the output of this LTI system at time index 2. We are saying that the output of this system is unstable, meaning small changes in the input lead to large changes in the output. How do we know this? Let us define a pole as all values of z for which the denominator becomes 0, so H(z) becomes infinite/undefined. The theory of complex analysis tell us that a function is stable if all poles are inside the unit circle. But from this video, we can see that the denominator has a zero hiding at infinity which becomes a pole of this LTI system making it instable.

[@gametimewitharyan6665]

​@@HoneyBunny-rg3jx Thanks for taking the time to write this info

[@jmcsquared18]

Projective spaces have some of the most beautiful geometry. For instance, two-branch hyperbolas are connected and smooth, if you move them to the Riemann sphere. On that complex projective space, their asymptote corresponds to the hyperbola crossing itself at the point at infinity, which is the north pole.

[@DeanCalhoun]

woah that’s amazing, never heard of that

[@lwmarti]

Indeed. One of the big lessons of algebraic geometry is that things really want to be projective.

[@sergiogiudici6976]

As a Physicist, I fall in love with this! This the approach to projectivity I have always wanted and never dared to ask for. What about conic degenerating in line?

[@Qhartb]

Some neat ideas. I enjoyed in intuitive "sanity check" that as a gets small, the quadratic will have a solution near the linear solution where the quadratic term has been made small enough not to matter, and a solution far from the origin where the quadratic term will inevitably dominate, and how far out we have to go for that to happen runs off to infinity. I feel like this also offers an alternate perspective on the nature of polynomials. Normally we think of them as having finitely many roots, but this lets us think of them as having infinitely many roots, all but finitely many of which are at infinity in some projective space. Somehow that feels natural to me that those roots have somewhere to "come from" instead of just appearing when you add a higher-order term.

[@brosisjk3993]

Doesnt it feel like cheating though, saying its "0x^2) when youre just approaching zero in reality. It should be made clear "0x^2" is undefined. Maybe im just not deep enough into math, but this video left me with a sour taste in my mouth

[@RepChris]

@@brosisjk3993 I mean defining stuff as something, and figuring out the ramifications is one of the big things in advanced mathematics (e.g. square roots of negative numbers are undefined => we define them => suddenly we have the math needed for quantum physics; of course this almost certainly has no where near as much ramifications for further mathematics but defining undefined stuff isnt unusual, and neither is replacing a value with its limit (even if they are distinct, and the usual limit rules wouldn't allow you to)) ; Personally I fell the video was a bit fast and loose with definitions and notation, but it was within acceptable parameters (unlike a certain numberphile video involving -1/12)

[@shruggzdastr8-facedclown]

"Deleted Neighborhoods" (@1:37) would make a great name for a dark-wave industrial band!!

[@Chrisuan]

Einstürzende Neubauten when Deleted Neighborhoods walks into the room: 😳

[@shruggzdastr8-facedclown]

@@Chrisuan: Blixa Bargeld approves this message 👌 😏

[@samueldeandrade8535]

I prefer Madness Squared.

[@davidwright8432]

... and also the fate of some manufacturing neighbourhoods under offshoring!

[@mtaur4113]

We could be more finicky about tracking whether a->0 from above or below, and then depending on the signs of b and c, we could find the signs of the infinite solutions, and this should agree with the visualization of the parabola flattening gradually to the line and the far zero running away on the x-axis.

[@lornacy]

You brought this into perspective for me with this comment - thank you! The math is way above my pay grade.

[@trueriver1950]

15:42 By turning a linear into a quadratic you've introduced a false root at infinity. Interestingly the other solution is the well understood situation to the original linear. The process only makes any kind of sense after you introduce the RP² ideas.

[@Austin1990]

As a approaches 0, the roots go to infinity. Also, the equation becomes a line. A horizontal line, when b is 0, has no roots. But, a line has a root at infinity or negative infinity depending upon the slope, b, being greater than 0 or less than 0. So, as the roots become infinite, you also lose one of the roots.

[@allozovsky]

But that projective "solution" doesn't really make sense. Michael got (x:y:z) = (0:1:0), and we were substituting x ↦ x/z, so it should be x ↤ 0/0, which is indeterminate (i.e. amy value), not "infinity".

[@allozovsky]

If we take only two variables x and z (explicitly excluding the "free" variable y), then the "solution at infinity" (x:y:z) = (0:1:0) collapses into (x:z) = (0:0), and it is not a valid projective solution, since x and z are not allowed to be both zeros.

[@Austin1990]

@@allozovsky The issue is that he is taking limits, but he did not explain what limits are. And, he didn't discuss how the limits are different when approaching from different directions. Per my last comment, the solution is nuanced, and those are largely lost without making those detailed distinctions. It''s not that he is wrong. Maybe he expected us to have all bad precalculus or calculus I. Or, maybe he just didn't want to mess with all that extra detail, which would have taken time.

[@allozovsky]

@@Austin1990 The limit solution makes perfect sense, but the projective one seems not to be comprehensible.

[@pierreabbat6157]

A few years ago, I was investigating the inverse of the arithmetic-geometric mean step, i.e. given a and g, find x and y whose arithmetic mean is a and geometric mean is g. I had to compute it in any of four ways to maintain precision. Similarly, if you find the roots of ax²+bx+c with the quadratic formula as usually written, and ac is tiny compared to b², you should rearrange the formula so that you don't lose precision.

[@rmandra]

Thanks!

[@TymexComputing]

As a computer engineer i started watching this video thinking to myself: IT IS A TRAP! 🦑🦈

[@filippodifranco8225]

I'm an engineer, not a mathematician, and I studied these things more than 35 years ago so forgive me if I write something wrong. This remind me a lot something similar (or was it the same thing in diguise?) used a lot in Analiytic Geometry, one of my favorite math classes at university. There was something called improper line in R2 or improper plane in R3, whose equation was t=0. In practice the coordinate system was added by one variable, t, that was always 1 except for the points (or lines) at infinity were it was 0. For the R2 case this gives the points at the infinity for a conic while for the R3 case gives the conic at the infinity for a quadric. This helps to classify the conic or the quadric. For example a parabola have two real coincident points at the infinity and the improper line is tangent in this point at the parabola. Moreover the point at the infinity gives also the direction of the axis of the parabola. Very fascinating, only real and complex analysis classes were more fascinating to me, was just a pity that this was only half class, the other half was the one I hated more than all: Linear Algebra...

[@depiction3435]

That's really cool. Thx for the comment.

[@mrhatman675]

How can you like analytic geometry and hate linear algebra at the same time

[@filippodifranco8225]

​@@mrhatman675 Very simple, I like what I can see. I looked at an equation and I was able to imagine the conic or the quadric, I looked at one isomorfphism or a base change matrix and I saw nothing. The only thing I needed from linear algebra at the time was how to calcuate the determinant of a matrix but I already knew it fron high school. I know it can be useful to change the reference system, rotate it, traslate it and so on but this was rarely needed in the exam exercises.

[@mrhatman675]

@@filippodifranco8225 I am mentioning this because analytic geometry was a subject in my previous semester and didn't t really love it the same can be said about linear algebra (although I liked it more than analytic geometry for sure and I thought people who like geometry also like linear algebra cause of how it analises vectors)

[@iabervon]

​@@mrhatman675The abstract aspects of linear algebra appeal to the same sorts of thinking that analytic geometry does, but the calculations can be hard to cope with and a blocker for enjoying the subject. The linear algebra course on MathMajor goes through a lot of interesting stuff and then kind of dies out when Michael has to explain matrix multiplication, and I think that's a similar attitude.

[@enpeacemusic192]

Please do more projective geometry in the future! I don't know much about it but it already looks super cool

[@larspos8264]

There is a cool way to interpret this limit as the derivative of \sqrt{b^2-4cx}/2 at 0

[@steka68]

Nice, but I wondered why you introduced the projective plane instead of just the projective line?!

[@harleyspeedthrust4013]

very cool. i was reading Levi-Civita's Absolute Differential Calculus and he mentioned projective geometry. I had never heard of it, so I looked it up and read through some wikipedia articles. I didn't expect to see it again when I clicked on this video!

[@uglychamaeleon]

Cool! In my last paper, we had to expand the roots of a polynomial equation as Puiseaux series in the neighborhood of leading coefficient = 0.

[@ramieskola7845]

I'm an electronic engineer and we have a practical circuit solution to 'push' an inconvenient positive root to infinity... and beyond. Then it emerges from the negative infinite and becomes beneficial if we push it close enough... We use it all the time.

[@abraham5276]

Stop using i as j , mr.electrical engineer

[@ゾカリクゾ]

no ​@@abraham5276

[@jon9103]

​@@abraham5276it's the other way around.

[@carultch]

@@abraham5276 Ok: J'm an electronjic engjneer and we have a practjcal cjrcujt solutjon to 'push' an jnconvenjent posjtjve root to jnfjnjty... and beyond. Then jt emerges from the negatjve jnfinite and becomes benefjcjal jf we push jt close enough... We use jt all the tjme.

[@Alan-zf2tt]

Bravo! It makes for a conundrum: A good place to stop and an excellent place to start!

[@cdamerius2895]

We can use a variation of the quadratic formula x_1,2 = -2c/(b+-sqrt(b^2-4*a*c)) to solve this. That would also give a solution for a=0 without the need to calculate a limit.

[@joshcal7370]

Now do it with a sample equation, plug in infinity, and show that b*(infinity)+c=0. Maybe something like 4x+2. I want to see how multiplying 4 times infinity, then adding 2 will equal 0.

[@allozovsky]

But that doesn't really seem to be infinity, since we were replacing *x* with *x/z,* so the point *(x:y:z) = (0:1:0)* must correspond to *x = 0/0,* which is indeterminate, not infinity.

[@douglasmiller1233]

The quadratic formula applies only to actual quadratic equations, in which the second-degree term has a non-zero coefficient. This is not a quadratic equation at all, but rather, a linear equation with only one solution (x = -c/b).

[@allozovsky]

1:30 > _and of course, if we're taking the limit as _*_a_*_ approaches zero, that means _*_a_*_ is _*_never_*_ zero, because limits are all about _*_deleted_*_ neighborhoods if you will_

[@user-zz3sn8ky7z]

He's taking the limit, the "actual" equation he's solving isn't 0x^2 + bx + c = 0, but instead ax^2 + bx + c = 0 where a is arbitrarily close to 0. He never actually reaches the linear a=0 case by definition of a limit

[@david_porthouse]

For roots known to be real, if b>0 then calculate s = - b - sqrt (b*b - 4 a c). If b

[@brunogrieco5146]

Curious that you arrived at Projective Geometry through an algebraic way. IMHO the interesting part about it is the concept of negating Euclid's 5th postulate arriving at the Projective plane, where a point is a line and a line is a plane, and they all cross at the single point at the origin. Which is rather beautiful. And I had to study it for Computer Graphics until Quaternions were resurrected from oblivion.

[@BullaMax]

In mathematics, a quadratic equation is an equation that can be rearranged in standard form as where x represents an unknown value, and a, b, and c represent known numbers, where a ≠ 0. (If a = 0 and b ≠ 0 then the equation is linear, not quadratic.)

[@PawelS_77]

So it's like interpreting a line crossing the x axis as an infinitely stretched parabola that crosses the x axis again at infinity.

[@allozovsky]

You can even plot both the quadratic and the linear functions on the *closed* interval *[−∞; +∞]* (by transforming the coordinate grid appropriately to squeeze the whole range into the *[−2; 2]×[−2; 2]* square) and notice how the parabola (with *a → 0)* follows the linear function almost over the whole range and then rapidly falls down at the very end of the plot (near infinity point) to give a second root.

[@numbers93]

What I find fascinating is that in RP2, the "infinite" root is unique, whereas the solution "-c/b" is actually an infinite collection of roots of the form (-c/b: y: 1) since y can be anything.

[@kristianwichmann9996]

More in-depth material on projective geometry would be great

[@allozovsky]

*Part II (projective plane)* 13:14 > solutions to *bx + c = 0* in *RP²* But isn't it more like *y = bx + c* ∧ *y = 0,* since *y(x) = bx + c* is a function that we are looking for zeros of, so our *y* is in fact present and equal to zero. 13:37 > we'll homogenize this by replacing *x* with *x/z* So we actually got *y/z = bx/z + c* ∧ *y/z = 0* => *y = bx + cz* ∧ *y = 0.* And if we replace *x* with *x/z,* is *z* allowed to be equal to zero, since we then "multiply through" by *z.* 14:09 > and observe that here *y* is really allowed to be anything we want because it's not represented inside of the equation Well, but is it really not represented there? It makes very little sense to me. When we solve *x² = 4* for *x,* it means that we solve *x² = y* ∧ *y = 4* for *(x; y),* doesn't it? Otherwise we would not really need a (projective) *plane,* since a *line* would suffice. 14:57 > if *λ* is equal to zero, then observe that that means that *y* is not allowed to be zero But *y* *has* to be zero, otherwise it has no relation to the equation whatsoever. 15:08 > scale this down to the point *(0 : 1 : 0)* And which value of *x* in the original equation does this "solution" correspond to? We were replacing *x* with *x/z,* so it must be *x = 0/0,* which is indeterminate, not infinity.

[@robobuilder1335]

I want to note, but dividing the original equation after removing the 0x^2 by b and multiplying it by 1 essentially gives the same result as the factor. Because you can then add 0x to the 1 and it wont change anything

[@AlbertTheGamer-gk7sn]

This actually has a marvelous connection: The quadratic equation y=x^2+2x+1 is a parabola. However, when you make a closer to 0, the parabola "zooms in" until a reaches 0, where it becomes its tangent line, proving the concept of the derivative. Same thing with conic sections (r=1/(a+cos(θ))): If you stretch an ellipse until its eccentricity gets closer to 1, it will eventually look like a parabola. You can go further and make it a hyperbola, where it breaks the infinity barrier and pops out on the other end. Finally, you can do the same thing with a bell curve (1/(x^2+a)), where you can stretch it to infinity, and breaking the infinity barrier makes it pop out on the other side, forming a bottleneck curve.

[@tomholroyd7519]

I love it when you ask questions that seem ridiculous at first but turn out to have important answers. No such thing as a bad question, no sir.

[@PC_Simo]

0:00 That’s just a linear equation, though. You’re also gonna have a problem with dividing the damn thing with 2*0 = 0. 😅

[@dmytryk7887]

Another way of looking at it (it has been a while si some of this could be wrong). Use stereographic projection. The x axis becomes a great circle, C, through the north and south poles. (The north pole being the point at infinity). Any line becomes a circle, D, also through the north pole. It is the intersection of the plane through the north pole and the line with the sphere. The two circles typically intersect at two points, the north pole being a guaranteed point. Circles that intersect at only the north pole will correspond to lines parallel to the x axis.

[@a.lollipop]

i was kinda understanding, then you got to projective geometry and i was instantly completely lost lmao

[@allozovsky]

Same here. Still don't get how we obtained a solution "at infinity" (and whether it really is a solution at infinity), since *(x:y:z) = (0:1:0)* presumably corresponds to *(x/z : y/z) = (0/0 :1/0),* as we were dividing by *z* to get this solution. Or maybe there's really smth crucial about the solution I don't quite understand yet.

[@user-zz3sn8ky7z]

@@allozovsky We don't, review again the original equivalence relation : "two points (x:y:z) and (r:s:t) are equivalent is and only if there exist a λ so that λx=r, λy=s and λz=t" This, as pointed out, implies that almost any arbitrary point (x:y:z) is equivalent to (x/z : y/z : 1) by setting λ=z However this implication doesn't hold for (x : y : 0), because no matter what λ you set, 0*λ =/= 1, which is required to get the (x/z : y/z : 1) form, meaning the two are not equivalent. And if you review the video closely you'll see that the author did include a small z =/= 0 note ;)

[@KipIngram]

This all kind of makes sense. Fundamental theorem of algebra - a second order poly has to have two roots. And b*x + c = 0 darn sure can't have a second FINITE root anywhere. So infinity is the only place it could be, right?

[@nou2844]

When you homogenize the equation, why can we presume that the x from the second equation is the same as from the first? Wouldn't the solution instead be x over z, which would be 0/0, which is undetermined, and (-c/b)/1? That would be more logical, since the initial equation should only have 1 solution. For context, I do not know a single thing about projectional geometry.

[@cyberagua]

Same thoughts 👍

[@cyberagua]

The second part is unrelated to the topic.

[@NHGMitchell]

At 7:10, 2b or not 2b? That is the question

[@allozovsky]

Ha-ha-ha!!!! 😂 That's just marvelous! Very Shakespearean indeed! Michael (in a mysterious voice): _Now it may not seem like anything interesting is going to happen, but something _*_will_*_ happen!_

[@ironbutterfly3701]

Isn’t RP^1 enough for this?

[@JobBouwman]

When a --> 0, you can use this alternative formula: 2c/(- b +- sqrt(D)) It is derived by substituting z = 1/x

[@davidgillies620]

I like to think of this as a space with three basis vectors, all of them lines, with one each along the axes and one "at infinity". Then lines in the plane become three dimensional objects, which is cool. It's useful from a geometric algebra perspective.

[@tomholroyd7519]

"deleted neighborhoods"

[@zerochan2915]

this video casually jumps from middle school math to something I completely don't understand

[@dubsed]

Anyone for adding +0x³?

[@jneal4154]

I love the projective geometry videos 🔥

[@dougr.2398]

Intercept over slope is one solution and the other is the point at infinity

[@YamiSuzume]

Do you relally say "limit" in english for "lim"?

[@spdking01]

en.wikipedia.org/wiki/Limit_(mathematics)

[@YamiSuzume]

@@spdking01 funny. Most other languages I know use Limes, the latin term

[@hhhsp951]

Uh. Yeah. It's kinda the proper grammatical use-case for English.

[@edwardmacnab354]

this is why Gauss destroyed his notes. He knew that in that way , no one could not marvel at his elegance . As for math and me , I gave up when I had to confront the non dimensionality of points meaning that lines are not made up of points

[@davidbizzozero3458]

Neat video. I didn't know about a projective geometry interpretation but I have seen a similar idea of (0x^2 + bx + c = 0) in singular perturbation theory (where the leading term is taken as epsilon x^2) and then examining the same behavior.

[@DrYankeeDoodle]

Are we allowed to multiply by 0/0 (in the second case) at 5:34? Is this possible bc we're working under limit sign there?

[@galoomba5559]

Yeah, that's the reason we're taking the limit

[@miguelmendoza3403]

Idk if I'm missing something but 8:00 absolute value shouldn't always have an positive output for any negative and positive input? So, there shouldn't be two paths but one because absolute value should always return a positive number. Absolute value is an even function.

[@christophertate3990]

Or use 2c/-b+-sqrt(b^2-4ac), which when a=0 simplifies to -b/c

[@samhess78]

I am glad you finally found the solution. According to the fundamental theoreme of algebra there have to be 2 solutions for this kind of linear equations. How many did you find?

[@mate.maticamente]

And 0x³ + bx² + cx + d = 0 ????

[@JohnLee-dp8ey]

Then by this logic, every nth degree polynomial has a root at infinity for a n+1 degree polynomial with coefficient of x^n+1 of 0

[@the_nuwarrior]

Its like using a conformal transformation z=1/x

[@adrified9352]

Why not L’hopitals at 3:05?

[@9adam4]

Middle school algebra student: "Look at what they need to do to mimic a fraction of our power!"

[@patrickmclean36]

Seems to be missing the graphical point of view: the parabola degenerates to a vertical line x=-c/b

[@allozovsky]

But why vertical? I was plotting both the quadratic and the linear functions on the closed interval [−∞; +∞] (by transforming the coordinate grid appropriately to squeeze the whole range into the [−2; 2]×[−2; 2] square) and the parabola follows the linear function almost over the whole range and then rapidly falls down at the very end of the plot (near infinity point) to give a second root.

[@ProactiveYellow]

I feel we are having a bit of trouble here: we have said that it'll be type 0/0 but that only works in the case where |b|=b because, otherwise, we have -b/a→-b/0 which is undefined. I know it was to make more sense of things, but I do think it's important to recognize this hidden case.

[@allozovsky]

That's why it is "when a quadratic equation has an _infinite_ root".

[@allozovsky]

And also we are in a "deleted neighborhood" 1:30

[@wilderuhl3450]

Cool. Now do it with Ferraris method.

[@troxexlot18]

You mean Cardano's

[@Karan_k1888]

​@@troxexlot18 both are different!!!

[@jefffuhr2393]

Understood everything **prior** to @0:30 into the video. After that... not so much. Question: if "y" is free, wouldn't the solution be describing a plane? Never mind! 🤯

[@allozovsky]

Same thoughts about *y.* It's rather *bx + c = y, y = 0,* so *y* is obviously not free.

[@juniorcyans2988]

I learned a trick to get rid of the zero in the denominator, which might be useful in solving my physics problems.

[@tomholroyd7519]

I really appreciate the energy here: tie one hand behind your back. You can still do this. You could even make it harder ... um ...

[@supersonictumbleweed]

This reminds me of my lost joy for mathematics. Thank you very much!

[@NotEvenJokingg]

16:44

[@santherstat]

this is such an interesting video. subscribed

[@subramaniankannapadi1524]

Definition of a quadratic equation is ax2+bx+c =0 where a is NOT equal to zero.

[@allozovsky]

1:30 > and of course if we're taking the limit as *_a_* approaches zero, that means *_a_* is never zero because limits are all about _deleted_ neighborhoods if you will

[@user-zz3sn8ky7z]

He's taking the limit, the "actual" equation he's solving isn't 0x^2 + bx + c = 0, but instead ax^2 + bx + c = 0 where a is arbitrarily close to 0. He never actually reaches the linear a=0 case by definition of a limit

[@MsMumuku]

14:00 The lambda is not the same one that cannot be 0, otherwise the following would not be an equivalence relation, right? Isn't that where the hat box is usually opened?😅

[@bikramadityadas7470]

The definition of a quadratic equation strictly mentioned that the coefficient of the x² term can never be zero. If it is zero, then it's not a quadratic equation but a linear one.

[@tonyutt]

That's why he specifically takes the limit of a as a approaches 0.

[@slowfreq]

1:04 lim a ->

[@Lklibertad]

as always impressive videos, thank you.

[@thehalfbloodprince-nm8uk]

Basic rule taught my math teachers An Equation is quadratic only if coefficient of x2 is non zero

[@steka68]

Your math teachers were boring. :)

[@depiction3435]

That's kinda crazy. Love it.

[@ke9tv]

You don't need all those limits. Let y=1/x, substitute into the quadradic to give a/y² + b/y + c = 0; Multiply both sides by y², apply the quadratic formula to get y=[-b ± sqrt(b² - 4ac))/2a or x=2c/[-b ∓ sqrt(b² - 4ac)]. That hands you the two solutions that you wrote on the lower left corner of the board. SInce the roots of a quadratic can be complex, I'd project them onto the Riemann sphere rather than the projective line, but the rest of your development makes some sense.

[@drdca8263]

But the Riemann sphere *is* the (complex) projective line? :P

[@ke9tv]

@@drdca8263 Well, yeah, so why not go to the general case right away rather than confine yourself to the reals?

[@drdca8263]

@@ke9tv going to be honest, I haven’t watched the full video. I made guesses about the latter portion of it based on what’s in the comments. I had assumed/imagined that the way he introduced the projective line didn’t specify that the variables be real valued, and that just taking what he said and assuming the variables were complex valued (rather than an unspecified field), would result in interpreting what he said as describing CP^1 but from your response I guess what he said probably referred to “the number line” or “the real numbers” or something. My mistake.

[@laranjajefessor]

se eu entendi, o final é ver a equação quadrática como ym vetor 3 e fz a projeção para 2. Pela projeção, por ser uma projeção, existe pontos ondo em 3 seria normal mas no plano acaba "distorcendo" e aparecendo como infinito. vi isso também em um video de relatividade do science asylum

[@milind006]

Watching this at 3:10 AM because I can’t fall asleep, and I have never been so lost. I don’t know if that made me sleepy or more awake than ever!

[@brunogrieco5146]

An actual question: Do you have any video about 1+2+... = -1/12 ? Have you checked Terrence Tao's take on it? If so, please post link.

[@ghlscitel6714]

Did you omit the double product? u²+2uv+v²

[@Grassmpl]

Remind me of the catalan generating function

[@leif1075]

He meant x/y colon 1 colon zero at 12:25 not just x colon 1 right?

[@sebastiangudino9377]

I don't know enough math to understand the second half of the video. But i learned OpenGL a few years back, there you work in 4D as a projective space which gets proyected to 3D to then be rendered in 2D. So i guess thanks OpenGL for discretely teaching me projective geometry?

[@lucahermann3040]

Wait, it's not called the midnight formula outside of Germany? (As in: if your math teacher calls you in the middle of the night, you have to be able to recite this formula)

[@philippemathieu5028]

Funny, I did this exercise exactly a year ago. Studying the cubic and quartic case is still on my to-do-list. Now, I am wondering if there is an interpretation in terms of projective geometry of the fact that, starting from degree 5, there is no solution in radicals. If someone has an answer to this or an idea, it is very welcome.

[@jameshart2622]

I've studied an abstract algebra textbook that ends at Galois theory and the proof of the non-solution of 5th+ order polynomials in radicals, and my takeaway is: no. The tools used to prove that result are subtle, complicated, and almost perfectly algebraic rather than topological. Solutions found using radicals and ones not available using radicals are both dense over the entire complex plane and deeply intricate. Their relationship is not well understood using topology, which is all about considering "neighborhoods" of related points, assuming that closeness implies some kind of common properties. For "radical" solutions and "non-radical" solutions, this is patently false. Projective geometry, on the other hand, is almost entirely a topological construct, used to fill conceptual gaps in the complex plane (or other spaces), namely the points at infinity. For non-infinite points it almost trivially reduces back to just the complex plane. I doubt it has anything to say about Galois theory.

[@eyesontheball6481]

How did you know that the quadratic formula needed to be manipulated to derive both roots (as opposed to directly calculating the ratio of limits)? Could both roots be solved for using the original expression by somehow incorporating lim b as a-> 0 after expressing b in terms of x, a and c?

[@DMSG1981]

So, why don't you divide the y by minus lambda b in the end?

[@allozovsky]

But *_y_* is _any_ number, so even if we divide it is still _any_ number.

[@DMSG1981]

@@allozovsky Right, that's what I was missing. Thank you.

[@ultrametric9317]

Very interesting! I had never seen this.

[@CTJ2619]

if a=0 then you run into problems dividing by 0 in the quadratic equation

[@allozovsky]

We are in a "deleted neighborhood" 1:30

[@angeldude101]

Good thing calculus exists to let us get away with dividing by zero.

[@RexxSchneider]

Not if you use the alternative quadratic formula x = 2c / (-b ∓ √(b^2 - 4ac)). That gives the same solutions to ax^2 + bx + c as the more usual formula, and also immediately gives x = -c/b when a = 0.

[@pankajsinha4280]

We may put x=1/t to obtain t=0 and t=-b/c

[@allozovsky]

But does *t = 0* turn *b + c·t = 0* into a true equality? It doesn't differ from *b·x + c = 0* by its form/appearance - both are linear, so, following the logic in the video, *b + c·t = 0* also has to have a root "at infinity".

[@allozovsky]

Oh, I guess I got it: you were talking about turning *0·x² + b·x + c = 0* (by subst *x = 1/t)* into *0/t² + b/t + c = 0* and then into *0 + b·t + c·t² = 0* which then factors as *t·(b + c·t) = 0* Well, I don't know how legitimate it is - we are multiplying by *t = 0* then.

[@Mike-cx4ov]

When the coefficient of x^2 is Zero, it ain't a quadratic any more and the famous formula doesn't apply. X = -b/c

[@haraldmilz8533]

A common way of solving the equation ax² + bx + c = 0 is to divide the equation by a and use the so called pq formula where p = b/a and q is c/a. Then x(1,2) = p/2 +/- sqrt(p²/4 - c). And then you see immediately what kind of nonsense the whole movie is. But them I'm only a dumb engineer.

[@bernardoxbm]

I don't agree with your terms. It's not an "infinite root" because there are two infinities, positive and negative. However, a more appropriate term would be a solution at the horizon.

[@numbers93]

but in projective geometry, there is only one true infinity

[@numbers93]

wait I confused it with stereographic projection 😅 In the sense of projective geometry, there would be an uncountably infinite number of infinities, corresponding to the infinite number of points around a circle. So his title is very suitable: there are an infinite number of roots, all at “infinity”

[@NotYourAverageNothing]

Why is using absolute value at the end "legal", when you implicitly assume otherwise from +- in the quadratic formula?

[@allozovsky]

But the absolute value pops up when we are evaluating the square root of *b* squared: *√(b²) = |b|* - and the ± cases are treated separately, each with its own copy of *|b|.*

[@guidosalescalvano9862]

Is it me or do you not even have to let it approximate the limit....

[@at1with0]

Except when a=0, it is not a quadratic equation. Why not solve it with the quartic equation?

[@user-zz3sn8ky7z]

He's taking the limit, the "actual" equation he's solving isn't 0x^2 + bx + c = 0, but instead ax^2 + bx + c = 0 where a is arbitrarily close to 0. He never actually reaches the linear a=0 case by definition of a limit. The quartic would get too messy for most people to follow i think, that's the only reason, you could if you wanted to, as long as you understand that the solutions you get aren't necessarily real

[@brunogrieco5146]

BTW, Does that mean that if you solve ax + b for a=0, on the projective line, all numbers become infinite? LOL.