Great now solve quadratic equation using cubic formula

@debtanaysarkar9744

Ай бұрын

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

@vascomanteigas9433

Ай бұрын

Maxima goes brrrr.. 😂

@mizarimomochi4378

Ай бұрын

Why stop there? Cubic equation using the quartic formula.

@jocabulous

Ай бұрын

​@@mizarimomochi4378Forbidden math: quartic equation using quintic formula

@acekingbones

Ай бұрын

And then a quartic equation with the quintic formula

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

@Synthesz1

Ай бұрын

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

@Kokurorokuko

Ай бұрын

no shit ​@@Synthesz1

@Synthesz1

Ай бұрын

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

@caetanogarelii6657

Ай бұрын

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

@fiNitEarth

Ай бұрын

@@caetanogarelii6657no shit

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

@vyvlad

Ай бұрын

Dominate your rationalizers

@__christopher__

Ай бұрын

Let's denominate our rationalisations!

@douglasespindola5185

Ай бұрын

Mathematicians always trying to break the math 😂

@Alan-zf2tt

Ай бұрын

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

Ай бұрын

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.

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

@tri99er_

27 күн бұрын

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

@Fire_Axus

25 күн бұрын

your feelings can be irrational

@sebastiangudino9377

2 күн бұрын

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

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

@backwashjoe7864

Ай бұрын

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

Ай бұрын

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

Ай бұрын

0*x+1 must have no solutions?

@Utesfan100

Ай бұрын

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

@Alan-zf2tt

Ай бұрын

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

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

29 күн бұрын

@@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

Сағат бұрын

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

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

Ай бұрын

woah that’s amazing, never heard of that

@lwmarti

Ай бұрын

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

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

18 күн бұрын

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

17 күн бұрын

@@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)

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?

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

@Chrisuan

Ай бұрын

Einstürzende Neubauten when Deleted Neighborhoods walks into the room: 😳

@shruggzdastr8-facedclown

Ай бұрын

@@Chrisuan: Blixa Bargeld approves this message 👌 😏

@samueldeandrade8535

Ай бұрын

I prefer Madness Squared.

@davidwright8432

Ай бұрын

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

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

Ай бұрын

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

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.

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

Ай бұрын

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

28 күн бұрын

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

28 күн бұрын

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

28 күн бұрын

@@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

28 күн бұрын

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

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

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

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

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!

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

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

Ай бұрын

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.

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.

More in-depth material on projective geometry would be great

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

Ай бұрын

That's really cool. Thx for the comment.

@mrhatman675

Ай бұрын

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

@filippodifranco8225

Ай бұрын

​@@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

Ай бұрын

@@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

Ай бұрын

​@@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.

as always impressive videos, thank you.

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.

I love the projective geometry videos 🔥

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.

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.

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

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.)

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.

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

this is such an interesting video. subscribed

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

That's kinda crazy. Love it.

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?

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

28 күн бұрын

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

27 күн бұрын

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

Correct me if I'm wrong but I'm thinking about this from the lens of the limit as a --> infinity, so, graphically/visually, as the curve gets wider and wider (like using slider on desmos). Eventually you get a line when a = 0. But as we approach 0, there's still two zeros, one approaching -c/b and one really rally far away on the x-axis approaching either positive or negative infinity depending on the sign of b. Is this another way to look at it or is my reasoning flawed. I'm sorry I'm a physics student and not a mathematician and not trained in mathematical proofs!!!

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

Ай бұрын

Stop using i as j , mr.electrical engineer

@zokalyx

Ай бұрын

no ​@@abraham5276

@jon9103

29 күн бұрын

​@@abraham5276it's the other way around.

Thanks!

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

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.

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.

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

Very interesting! I had never seen this.

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.

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?

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

Its like using a conformal transformation z=1/x

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.

Anyone for adding +0x³?

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?

Isn’t RP^1 enough for this?

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

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

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

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

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

@allozovsky

Ай бұрын

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

27 күн бұрын

@@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 ;)

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

Ай бұрын

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

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

28 күн бұрын

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.

*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.

Remind me of the catalan generating function

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.

Proposition 1 in Book 1 of Euclid is to find the midpoint between two given points, but the point @infinity is equality valid as the point between the two given points....there were always two solutions, they've been lying to us for 2000 years!! 😡😠

@trueriver1950

Ай бұрын

Three points, surely? One at +inf and one at -inf

@allozovsky

28 күн бұрын

​ @randomuser-xc2wr How are we supposed to calculate the coordinate of the second midpoint? Say, for *a = 0* and *b = 2* the first midpoint is *(a + b)/2 = 1,* and what is the formula for the second midpoint?

@allozovsky

28 күн бұрын

Sure I did. I was trying to compute the other midpoint and just couldn't. I need a formula to gat a result, not just some notion or concept of coordinates.

@allozovsky

28 күн бұрын

I understand what a projectively extended real line is, but I couldn't find the other midpoint on it, I thought maybe you could provide some reasonable approach.

@tsawy6

22 күн бұрын

​@allozovsky working without coordinates we might start with the definition that the midpoint is equidistant from both points. Yes 1 is 2 from 3 and also -1, but the point at infinity is infinitly far from 3 and also -1. For any pair of points one midpoint lies at infinity

"deleted neighborhoods"

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.

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.

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

why is dividing by zero not undefined but instead infinity?

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

23 күн бұрын

Same thoughts 👍

@cyberagua

23 күн бұрын

The second part is unrelated to the topic.

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

Ай бұрын

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.

And 0x³ + bx² + cx + d = 0 ????

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

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

Ай бұрын

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

@allozovsky

Ай бұрын

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

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?😅

16:44

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.

A slight elision on your part in the indeterminant limit of the quadratic. If b < 0 then one has -b +- sqrt(b^2 - 4ac) and this is not zero if a = 0 but +-2b and in this case one has the solution b/a which is your "solution at infinity".

I had been thinking about what happens to the quadratic formula when a = 0. Cool up till ~ 8:50. Am not a maths major myself, so everything starting from ~ 8:52 and beyond is all Greek to me.

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

Ай бұрын

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

@ke9tv

Ай бұрын

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

@drdca8263

Ай бұрын

@@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.

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?

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

Can you do the same thing for second order differential equation ?

@allozovsky

28 күн бұрын

Solve the first order as the second order? Hmm, that's an interesting idea 🤔

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

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!

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

@allozovsky

Ай бұрын

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.

Do we actually need the y Dimension? Could this be done in one Dimension lower?

@steka68

18 күн бұрын

Yes, it could (and perhaps should). I wondered why he unnecessarily introduced the projective plane RP^2 instead of the projective line RP^1.

Ah, projective geometry! Fun! :D

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. 😅

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

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)

Why not L’hopitals at 3:05?

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

@allozovsky

29 күн бұрын

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|.*

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

@steka68

18 күн бұрын

Your math teachers were boring. :)

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

haven’t watched the video yet, but i bet it’s gonna be using L’Hopital’s Rule to solve for a limit of a rational function as a approaches 0 yeah i knew it wait nvm… but just as a general question, could you not use l’hopital to solve this? just let a approach zero and solve for both cases… in one case (subtracting the discriminant) the limit adds to -2b/0 so we discard that solution, but in the second case (adding the discriminant) the limit approaches 0/0 so we can use l’hopital’s rule in the case where we add the discriminant. doing this should yield the same solution that you end up getting using the reciprocal (ik that’s not the name of the method u used, but the name is just slipping my mind rn) method.

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

@allozovsky

Ай бұрын

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

Ай бұрын

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.

Cool! Can you invert this process to deduce the quadratic equation? Can you extend to deduce equations for higher power polynomials? 🙂 Probably not, but it might be interesting to see where it fails.

I love that you're bringing in projective geometry and infinity into your recent videos. I truly believe that the Riemann Sphere and wheel algebra can teach us so much and can generalize so much that we hadn't realized before, so thanks for going down this route when no other math educator on KZread will.

@allozovsky

Ай бұрын

But did we really get a solution at infinity in the second part of the video? How can we verify it?

It's obvious that when a=0 the quadratic is no longer a quadratic and becomes a 1st degree, so x=-(b/c). Probably the same can be applied to any degree but I am not a mathematician.

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

@allozovsky

Ай бұрын

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

@DMSG1981

Ай бұрын

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