Welcome to another Mathologer video. Let

me show you something really amazing, how to solve an equation by playing laser

tag with a turtle. That sounds very strange but just wait and see. Okay

here’s an equation, here’s our pet turtle and here’s my laser tag gun. Here’s the

turtle at his starting point, facing to the right. The leading coefficient of our

polynomial is 1. This tells the turtle to walk one unit

to the right. Now he makes a quarter turn in the counterclockwise direction.

He’s now ready to go up. The second coefficient is 5 which tells the turtle

to advance 5 units. Yes, yes I know it’s a very speedy turtle. Probably doesn’t like

the look of that laser. Anyway another quarter turn counterclockwise. The next

coefficient is 7. Advance 7. One last turn. The last coefficient is three and the

turtle finishes his trip by advancing another three units. Now we place our laser

at the starting point and aim to tag the turtle. But of course a straight shot

wouldn’t be very sporting. So like billiard players of old we will try

to zap the turtle with a bank sho,t like this. BUT, our bouncing rules are special.

We have a weird laser that always bounces off making a right angle, like

this. And again. Missed! But by angling our laser shot slightly differently we can

arrange a hit. There! Great fun, except maybe for the turtle. But what’s the

point? Well, believe it or not, we’ve also found a solution to our polynomial

equation. It turns out that to make the killer shot our laser beam began with a

slope exactly equal to 1. And for the game we are playing here it turns out

this promises us that minus 1 is a solution of our equation. Don’t believe

me? Let’s check. Let’s see. All right five plus three that’s eight minus one

minus seven. That’s zero. Ok let’s allow the turtle back into the game.

That’s a cubic equation down there so hopefully we can find another solution

and it turns out we can. Tilt the laser a little more. Okay keep going. There, a

second hit. Here the initial slope is 3 which means that -3 is a second solution

and yes there’s also a third solution but we’ll worry about that later ok.

Surprised? I sure hope so I definitely was. And with suitable adjustments our

turtle laser method works for any polynomial equation. What are the

adjustments. Well, obviously, a higher degree polynomial will require more

segments in the turtles path but as in our example the turtle always always

makes a counterclockwise quarter turn after each segment. Also the turtle deals

with a negative coefficient in our polynomial by walking backwards instead

of forwards. For a 0 coefficient the turtle doesn’t move but he still makes a

quarter turn for this zero segment. Now one thing that may go wrong is that the

next turtle segment may not be long enough so that the laser beam will miss.

We take care of this by allowing the beam to bounce off the full line

extensions of the turtle segments. Have a look at this. So it misses but

bounces anyway, it bounces and also ends on one of those extensions. It may seem a

little strange to allow this. On the other hand, you’ve already accepted speedy

turtles and weirdly bouncing lasers so it’s definitely a little late to start

objecting now. Anyway trust me for now and just run with it you’ll see it all

ends up being quite natural. So why does this crazy system work? Let me now take

you on a tour of the bizarre mathematical world of turtle tag. This

turtle shooting game is called Lill’s method named after the Austrian engineer

Edward Lill who discovered the method about 150 years ago. It seems that Lill’s

method was quite well known for a while but it is now largely forgotten. I

learned about it from a very nice article by Thomas Hull,

a mathematician who specializes in origami mathematics. We’ll get to the

origami angle soon. As for the turtles I imagine a number of you have already

guessed how they got involved. If not just Google turtle graphics and all will

be clear. Anyway, as far as I can tell it was also Thomas Hull who was the first

employed Turtles to explain Lill’s method. I’ve set up this web page over

there so you can play with Lill’s method. The link is shown at the top and also in

the description of the video. On input of the coefficients of a cubic polynomial

it draws the corresponding turtle path. You can then aim your laser by dragging

the mouse. There, really quite stunning isn’t it? Let’s now have a closer look at

the changing numbers at the top. At the top is the polynomial p(x) whose zeros

we are hunting. We’ve shot the laser at a positive slope of about 0.28. Then

evaluating p(x) at minus that slope we get about 1.4. Close but no banana. Now,

and this is super cool, 1.4 etc. is exactly the distance between the blue

and red points. I’ll give the super super cool proof near the end of this video. So,

in effect, what we’re doing when we shoot at the turtle at some slope is to

evaluate the polynomial at some point x and then we graphically adjust things

until p(x) becomes zero of course p(x) can also take on a negative value

which in the picture would look like this. Okay I think we can all agree that

just on its own solving equations by shooting turtles is really really cool.

But there’s lots and lots more. What I want to do now is show you some of the

super cute features and applications of this method. That includes a clever way

to reinterpret the turtle shooting to get free solutions to closely related

equations and a slick way to solve quadratic equations by simply drawing

circles, and an ingenious way to solve cubic

equations using origami, and some super efficient rapid-fire iterative turtle

shooting, and, finally I’ll also show you a very surprising very beautiful and

apparently new incarnation of the super famous Pascal triangle. And, of course, this is Mathologer and so

along the way I’ll also show you some proofs of how all this works. Quite a

program but we’ll go at a turtle’s pace, that pace. Okay, well anyway I’ll begin

with the simpler ideas and then go on to the ever more challenging.

Feel free to bail or begin to skim when things get too scary. Okay off we go.

Here’s the cubic again and the two ways we found of shooting our turtle to give

the solutions minus 1 and minus 3. Let’s flip the picture vertically. Alright this

flipped diagram translates into solving another closely related equation. What’s

the equation and what are its solutions? Well, our turtle knows it all. So, let’s

just chase it around the path. Okay, the turtle is on its way and again begins by

walking one unit to the right. Then the usual counterclockwise turn. But that

means the turtle walks backwards on the next stage. Therefore there’s got to be a

minus sign in front of the five. Turn again.

Okay the turtle is facing forward again and so we can leave the plus sign in

front of the seven in place. Turn once more. Backwards again and so the

sign has to change again. So we know the equation for the flipped path but what

about its two solutions. Well, because of the flip the starting slopes of our

laser beam are just the negatives of the slopes we had previously and that means

that the solutions to this new equation are plus one and plus three, the negatives of

the original solutions. And, of course, this works for any polynomial equation.

Changing every second sign in the polynomial the solutions to the new

equation are the negatives of the original solutions. Very cute, hmm? 🙂 Anyway here’s a first easy challenge for you: Try to find a turtle-free proof of this

fact. As always you can answer in the comments. Now by

further flipping and rotating this diagram we get more free solutions and

insights into other related equations. Here’s a rotation that gets us to

something pretty surprising. So your second challenge: chase the turtle around

his path and figure out the equation it represents. How are this equation and its

solutions related to the original equation below and what’s the general

principle? Share your thoughts in the comments. So in my last video I tried to convince

you that parabolas and quadratics are much more interesting than the aimless

and tedious school exercises we all seem to have suffered through. Now here’s some

more evidence. A really, really cool way to solve quadratic equations. Okay here’s

a quadratic equation which happens to have solutions minus 1 and minus 2. A

quadratic has three coefficients and so the corresponding turtle path has three

segments and a laser beam solution consists of two segments making a right

angle. But there is a very old and very beautiful theorem about right angles and

this theorem allows us to replace the trial and error approach of swivelling

the laser with drawing a simple circle. Do you know the theorem? No? Well maybe once did? We’ll find out in a second. Start with a circle, draw one of its

diameters and pick a point on the circumference. Then this triangle there

will always be right-angled. Remember that? This beauty is called Thales theorem.

Now watch this. Magic, magic, magic. Very pretty, isn’t it and you can see what it

does, right? It tells us that to solve the quadratic equation using the turtle path

we can just connect the end points find the midpoint and draw the blue circle.

Then the two intersection points tell us where to aim our laser and so also the

solutions to the quadratic. How super pretty is that. Definitely

makes my day when I learn about something like this. How about you? If you draw the turtle path on a

physical piece of paper it is also possible to use paper folding based on

Lill’s method to solve quadratic equations but there’s more. Lill’s method

shows how you can also solve cubic equations with paper folding. This amazing

discovery was made by the mathematician Margarita Piazzola Beloch in 1936.

Let me demonstrate how this paper folding trick works using our cubic

example from before. Here we go. Here’s the distance from the final red point to the

top horizontal line. Draw or fold another horizontal line the same distance above.

There we go. Now do the same for this distance and

draw a new vertical line. Okay, copy the final segment of the laser beam and

notice this copy can now slide snugly between the two horizontal lines, there

just fits. And we can do the same with the first segment of the laser beam over

there. Time to begin folding. Take the paper and

fold along the middle segment of the laser beam. Then because we’ve got a

right angle here this laser beam segment will fall smack on top of its blue copy

and therefore the red point will end up on the green horizontal. Similarly, the

black point will end up on the green vertical. Let’s do this.There, magic 🙂 There

the red and black points end up on the green lines. This means that starting with the path of the turtle we can find solutions to

our equation by folding the red and black points onto the green lines. Okay

let’s do it. And unfold. Then the paper crease pins down the middle segment of a

successful laser path, right? And the rest is auto pilot. Also a super nice

construction, don’t you think? And just in case you’re wondering, here’s our second

solution. Great! And maybe you’ve heard people say

that origami is more powerful than ruler and compass. Have you heard that? I don’t

have time to go into details here but it is exactly the fact that paper folding can

solve cubic equations that shows it’s more powerful than ruler and compass

which can only handle quadratic equations. Okay, one more super nice property before

I tidy up with some proofy details. Here’s our cubic again. Recall that we

found the solutions minus 1 and minus 3 corresponding to beam slopes 1 and 3. Of

course, we’ll come across all possible real solutions by sweeping the laser

through all possible slopes but there’s also another iterative way of finding

new solutions. To begin find one solution as usual. Now forget about the turtle

path for a minute and pretend that the laser beam path is a new turtle path.

Weird, hmm? Swivel your laser to find a solution for our new path. Now what’s

amazing is that this solution to our new equation is also a solution of our

original equation. In this case it’s the minus 3 corresponding to the slope 3. But

why stop now. Let’s do this one more time. Forget about the turtle path again and

make the laser beam into a new turtle path. And here’s a solution. That’s

another minus 1, corresponding to the slope 1 which, as we know, is also a

solution of our original equation. Let’s combine all our solutions into one

diagram. So we get the same solutions as before.

However, the solution minus one appears twice, corresponding to two green angles.

Why is that? Well when we have a close look at the polynomial it becomes

apparent that minus one is a zero of multiplicity two. In fact we can factor

our cubic polynomial like this. There the green solution minus 1 has multiplicity

2. This turns out to work in general. The iterative method picks up the

multiplicities of the zeros whereas the basic Lill’s method does not. Very neat

and also very very mysterious. Why on Earth should turning laser paths into

turtle paths do what it does? I’ll give some details in the proof at the end of

the video. But just quickly, here’s a sketch of what’s happening. After finding

the first green root minus 1 we get rid of one of the green factors down there.

This leaves us with a quadratic and the second turtle path corresponds to this

quadratic. After we find the blue zero of this quadratic, we get rid of the blue

factor, leaving us with the single green factor. The final turtle path corresponds

to this linear equation and it calculates the remaining green zero. Very

pretty stuff. Time to get down to the proofs. I’ll now show you why Lill’s

method and it’s iterated form work. As an incentive to stick it out, at the very

end I’ll show you that very beautiful incarnation of Pascal’s triangle that I

discovered or perhaps rediscovered while preparing this video.

Mathematical seatbelts on? Here we go. I’ll begin with a sketch of a proof of

Lill’s basic method, again focusing on our cubic. So what I want to show is that our

cubic function evaluated at minus this slope is equal to the signed distance

between the red and blue points. In order to do this, we are going to successively

figure out these four distances here. 1, 2 3 and 4. Okay, so that first distance down there is just the first coefficient which is 1,

right? To calculate the other distances notice that the same green angle pops up

three times in the diagram and the slope of the initial laser beam is just the

tangent of the green angle. And then minus that slope is the input x for our

cubic polynomial. Now remember your Sohcahtoa? The length of this yellow

segment is tan of the green angle times the aqua 1 which is minus x. Next, this

blue side of the turtle pass has length 5. Therefore this aqua segment has length

5 minus minus X. Now Sohcahtoa again. The next yellow segment has length tan of green

angle times 5 plus x which is this. The next blue segment of the turtle pass is

length 7 and so the next aqua segment is this long. Fancy, hmm? Repeating this

calculation one more time gives us the distance between the blue and red points.

There. And this turns out to be our cubic polynomial just written in a very

special form. To check that this really is our beginning cubic we just expand. So

once, twice yep that’s our cubic. And so if we found an x to make this final

aqua distance 0, then we’ve also found a solution to our cubic equation. And

that’s why Lill’s method works. Tada, magic, isn’t it. But it gets even more magical.

this special form of our cubic is called its Horner form. I still remember

learning about the Horner form and a related mathematical miracle from Herrn

Schwenkert the amazing high school teacher who sparked what became my

lifelong obsession with mathematics. That was over 40 years ago back in Germany.

Yep, I’m that old 🙂 And so it was a wonderful surprise to discover that at

its core our turtle tag game is also just “synthetic division”, the Horner form

miracle I was shown by Herrn Schwenkert all those years ago. Let me finish by

telling you about this miracle which then will also explain why iterating

turtles works. In the first instance the Horner form gives a very simple and

efficient way to evaluate a polynomial. Let’s say we are looking at the specific

value x is equal to minus 2 corresponding to an initial slope of +2.

There. Then we evaluate the polynomial from the inside out like this: 1 times

minus 2 plus 5 that’s 3. Minus 2 times 3 plus 7 that’s 1. And, finally, minus 2

times 1 plus 3 which equals 1. So the lengths of our aqua segments are just

the intermediate results of this super efficient way of evaluating the

polynomial. However, and this is the miraculous bit, by calculating these

numbers we’ve also perform what’s called synthetic division, which means that

we’ve also divided our polynomial by x plus 2 in a very sneaky way. How, what, why, where? Well the result of dividing a cubic like this is a quadratic plus a

remainder, right? Now it turns out that, and you should really check this, the

three coefficients of this quadratic polynomial are the first three

intermediate results of our evaluation, those three there, and that the remainder

is the final aqua number, so like that. So, again, we can divide a polynomial by a

linear factor by simply evaluating its horner form. Pretty damn miraculous, isn’t

it? And just in itself a trick worth learning about and committing to memory

for the rest of your lives, don’t you agree? Now for iterating turtles the

important case is where the linear factor corresponds to one of the

solutions of our equation. In this case the remainder vanishes like this… Almost there. The remaining solutions of

our original equation are then the solutions of this quadratic equation,…,

whose total path is this. But, as you can see, and as one easy

bit of trigonometry proves, this quadratic turtle path has exactly the

same shape as our cubic laser path. And that means that we can use the laser

path to solve the quadratic equation. And that is why turtles iterate. Let it

sink in. Got it? Very, very cool, right? And that’s just

about it for today. There were a few more fascinating aspects that I was tempted

to cover like, for example, the generalization of Lill’s method to also

find the complex zeros of our equations, the beautiful characterization of

equations that correspond to closed turtle paths and what happens when we

bend turtle and laser paths at angles different from 90 degrees. Maybe another

video. But for now, if you’re interested then check out some of the references in

the description. To end, let me fulfill my promise and show you an animation of

that beautiful turtle path incarnation of Pascal’s triangle. I haven’t seen it

mentioned anywhere so this may very well be a cute little original discovery.

However, if any of you have seen it before, please let me know. Anyway enjoy

and bye for now.

https://youtu.be/UkozVF8WRj8 Prof. Dr. Weitz from the HAW Hamburg also spoke about this Method after watching this video 🙂

Professor Edmund Weitz erstellte aufgrund dieses Video sein eigenes Video:

https://www.youtube.com/watch?v=UkozVF8WRj8

This is 100% false I used chipmunks and got the same answer…..

What happens if a part of the turtle path intersects another part of the path, and 'blocks' the path i.e. 5x^3 + 3x^2 + x + 10. Where the line for 10 intersects the line for 5x^3.

Why won't we turn turtle clockwise? Then we won't have to change the sign of the slope

This has to be the best of all the mathematics videos with the most elegant property of all time wow… please make a video about complex solutions and what to do when turtle intersects path and maybe even more elegant properties this might have?

Amazing!!!!

What happen if I use other animals beside of turtles?

I think Mathologer has the most click-baity titles. But I don't mind, lol.

The only thing we need now is sharks with frikkin lasers shooting at sea-turtles

Epic. Learned a lot. By the end when the distances were adding up nicely, I was already thinking 'well of course that happens!'

Best part about the solving process is that it's essentially gluing together similar triangles..

If only I knew this method in my 9th grade, when I had to guess the first root using Viet theorems.

Here's one you'll like especially the ending. Enjoy

Alternative Math | Short Film https://www.youtube.com/watch?v=Zh3Yz3PiXZw&list=WL&index=237&t=0s

Pure coincidence: we had origami and solving the cubic at our Montreal Math Circle just two weeks ago! Adding to the origami part of the "bouncing" laser: the crease line from taking each point to the line is equivalent with constructing a tangent to a parabola with the focus and directrix being the point and line, respectively. Practically, the slope of the common tangent to two parabolas is the solution of a cubic equation. Very nicely explained in R. Geretschlager's Geometric origami book. Maybe a future video? Since this topic is so rich, it would be nice to see more videos about. Thank you for your videos and the links, they are a rich source of wonder.

A modified carpenter's square (as mentioned in the book Practical mathematics in a commercial metropolis, page 110), as instrument for practical solution, is also related to the turtle…the solution going back to Ancient Greece…

Did you know that this can be modified to find derivatives of functions? Here is an app that allows you to find the value of f(x), f'(x) and f''(x).

https://www.geogebra.org/classic/fgdwvhwx

It's a work in progress, but the basic idea is that when you repeatedly divide by the same factor when doing synthetic division, some of the reduced polynomial coefficients lead to derivatives evaluated at that value.

I don't understand. How can the path be equal to 1 (1min56), when this is the hypotenuse of a square triangle that has one of its other sides equal to 1 ??

Sub to Ruth Almgill

Can you explain where Terrance Howard's view on math and physics is wrong? Please?

This is so beautiful! Thank you!!!

And how about complex numbers? 😉

i love this chanel

Can you please make a video about surreal numbers? (And pasudo numbers)

Cool

You look 45 at most

Where do you get these type of articles from?

Suggest some journals.

I am an undergraduate mathematics student.

I remember reading somewhere that it is, in general, impossible to solve polynomials of degree <=5 (that is, there is no 'quintic equation' or higher), and that indeed it had been proved that no such equation should exist. However, at around 4:10 you show an application of Lill's method to a quintic equation and it seems that this method should be able to provide (at least the real) solutions of polynomials regardless of degree. Why does this not work? Is there a geometric intuition for the cases in which it does and doesn't?

But but the law of reflection… your laser is impossible in the real world

Hello, you might not see this comment, but I would like to see a video about Bernoulli numbers. I have been working with partial and infinite summations for a while now and they keep popping up mysteriously! I don't know why!

Brilliant!

This would be a good lesson into intro to quadratics and trigonometry as a visualising tool. This could also help understand the math behind some artist techniques.

Is there another way to solve equations? Also, on an unrelated note, does anyone know where I can find turtles? No particular reason, just asking for a friend.

Hello Prof. Polster, I'm a Mexican mathematics super-fan since childhood. Even if I end-up becoming a Plastic Surgeon, I still enjoy math a lot and love your videos. Today I have a question (and video suggestion) for you… I've been struggling to plot the power tower function f(x)=x^x and I have seen a lot of partial answers around, but not one that satisfies me…. so the positive integers and positive real roots of real powers are pretty easy (which is basically what everybody tackles), but I assume, for example that 0.2^0.2 has five answers right? just taking in account the complex plane… still easy.

First question: would 3.2^3.2 still have 5 possible answers as 3.2^3.2=(3.2^3)(3.2^0.2)? I'm struggling as I don-t know if positive non-integer powers of real numbers higher than 1 also have more than one answer to plot on the complex plane.

Next step… going to negatives and using the complex plane is not difficult, only the previous problem applies too… and plotting numbers such as -0.1^-0.1 would end up in 10 possible answers right?

Lastly… what if we input complex numbers in the f(x)=x^x function as well? what would the result be? I'm assuming we would need a 4D graphic to plot that function as f(x) would also be a complex number… how would that graph look? this problem has daunted me for a couple of weeks now and I thought you might enjoy the challenge and help completing the power tower function for us.

Very cool, but this is not laser tag, because lasers reflection corner is not always 90°

Please consider using dark mode. the white background is too bright.

Very nice work . . .!

Is it possible to get rid of any particular molecule on a global scale… with a system complex enough to exchange information through a field that includes every point in and around the earth?… Using scalar dimensional relativity and inducing quantum energy exchange, if we could understand the harmonic balance between our relative dimensions and the quantum dimensions- the ones that make the quarks and protons and atoms and such- we could vacuum up all the radioactive particles right back into quantum energy space. differentials at every relative dimensional boundary. i'm thinking of a synthetic boundary around target acting as a valve and waiting for the exterior environment to push or pull, then open the valve at and for the desired time with the controlled quantum filter running. Dibs on the whole new power company system if there really is anything to this and that EmDrive.

Square root using 0x^3+1x^2+0x+-2=0 to find square root of 2

Oh so that is where “Turtle Graphics” comes from. I remember my computer teacher talking about the Turtle Graphics library in Turbo Pascal back in 1991. I didn’t know it was that old of an expression, like 18-something.

This is very interesting.I've been involved and delving into math academically, professionally, and recreationally for most of my 62 years on this planet and I still haven't seen this method until now.

Lol I couldn’t stop laughing. A turtle and lasers to solve this problem — this is mad; this is genius!

Can you do a video on what some call 'Eulers blunder S = lim S'? I've seen some arguments claim that you cannot distribute over an infinite series, and that the infinite series S is indeterminate, but the limit of S is determinate.

Where can I get that shirt?

So cool!

This is probably the most beautiful method to solve anything in maths i have ever seen! please make the second video about the complex roots and closed turtle paths and different angles about which you talked about i'd really like to know what happens!

Burkard, lookout for flammable maths! He is a dangerous person in the math community! Beware!

Very cool sir!

There was more beauty in this video than all the 🌹

Mind = Blown

Amazing video, as usual

WTF

@Mathologer Could you point us to a reference to what happens when you reflect the turtle by the angle of incidence, instead of 90 degrees every time?

Nobody:

Mathologer:Solving EQUATIONS by shooting TURTLES with LASERS

Nice, if only he wasn't so cringy haha

The description of pascals turtle was just amzing

no…your videos are making me interested in maths! that cannot be stop! rofl awesome way to present the content, love your vids! keep it up! cheers

Have a look at turtletoy.net

I love YouTube for directing me here but I hate YouTube that it took this long. I've subscribed to many math channels over the years and it never once brought up this channel, until now. Come on man! What gives?

This is actually a really interesting idea.

You are GREAT!

Just amazing, really amazing. An absolutely different vision of a known topic. Thanks for making such an interesting video.

Dankeschön!

your videos are delightfully engaging entertaining and instructive.

however – at times i find it hard to understand what you are saying.

closed captions clarified your use of the word "aqua" spoken at 18:44 & printed at 18:44.

however – closed captions was (apparently) equally confused as i am about

the word "Sohcahtoa" spoken at 18:30 & printed at 18:30.

also spoken at 18:52 & printed at 18:44.

can anybody help ?

thx.

ah -got it – SOH.CAH.TOA

SOH (SINE = O/H)

CAH (COSINE = A/H)

TOA (TANGENT = O/A)

i don't remember being taught this mnemonic when i suffered through high school trig.

ANYWAY – GREAT VIDEO !

First my sincere thanks for I have learn a lot from many of your videos and used some of them for teaching. After some investigation about – x = tan (angle), I found a different explanation for not taking negative sign. Just take x as positive, then we have ax, b – ax, then x(b-ax), then c – x(b-ax) and x[c – x(b-ax)] = d to obtain a cubic equation of the form ax^3 – bx^2+ cx – d = 0. The root of this equation can be used to deduce the roots of original ax^3 + bx^2+ cx + d = 0 because a(-x)^3 – b(-x)^2 + c(-x) – d = 0 tells us to take the opposite sign. This is also true for polynomials of higher degree. Hence there is no need to bother about negative coefficient , the rotation of the turtle and the backward movement of the turtle. For example the roots of x^2-3x +2=0 can be obtained from x^2+3x+2= 0 because (-x)^2 + +3(-x) +2=0 tell us to take the opposite sign obtain from x^2 + 3x + 2 = 0. To conclude its choosing a diagram that solve an equation whose roots can be used to find the roots of the desired equation. You may wish to try (x-1)(x+2)(x+3) =0 where there is a need to draw 1 unit left then 4 unit up then 1 unit right and 6 unit up (not down for -6).We choose the diagram and not follow the sign of the coefficient. In this example we need to reflect on the extended line. Hence much of your explanation in the video are still applicable.

I shall make an attempt on complex roots.

Your videos are always informative and I learn much. You and Hannah Fry are my two favorite Mathematicians. I must be honest though. Hannah is prettier. lol

Mind blowing mathematics! Amazing and interesting thank you mathlogger

Note to myself: watch it two more times to understand it

I'm trying to figure out how much of this was inspired by Robot Turtles lol and now i'm stuck watching this

This is an analog computer!

Dear Mathologer; your mind is brilliant and your videos very educational.

I did not know that you could do clickbate with mathematics, I guess this topic will never stop to surprise me!! Amazing video and thank you mathologer for making this sorts of things as quadratic equation fun or way funnier than how the educational system does it. And Hurra for the discovery, this pascal’s turtle is quit astonishing

Thank you for the video! All of you friends are super awesome! Oh, moments in this video are sad.

Today in my school, I showed this method and my friends went crazy. ''How you can solve cubic equations using turtles and lasers!?''

Oh my gosh!, you're a boss, respect to you Grand Master.

I'm sorry, I thought this was women's lasers. When are we playing with the lasers again?

Is it also related to Alex Eskin's work. https://youtu.be/66A1EdFKQTg

I'll have to watch it again a few times…

The this geometric solution is really interesting…. but It seems like this solution is a lot more work to do than other traditional method unless you can have it modeled. Probably why it fell out of fashion and it isnt taught to students much anymore.

Can we go the other way and use the turtle to generate a cubic function from a quadratic? As in, using only the 'turtle maths' to continue building on pascal's turtle?

A great video. The humour is a nice touch but confirms one's belief that mathematicians are at root members of the monster group.

I wonder about using the geometric algebra notation of David Hestenes. (multivectors, directed arcs and rotors) since two reflections are equal to a rotation and two vectors are orthogonal when the inner product is zero.

This was actually very cool.

Next you’ll tell me you can do calculus with some variation on this,

There were lasers in the 18xx's?

Yikes! You had the Horner form in high school? (Gymnasium?) thx. Was that an advanced class?

P, P' in R[X], P and P' are the same polynomials exept the pair degrre therms are inversed. We have P'(-x) = – P(x), so the two plynomials have the same 0.

Me: Okay, I'm not all that interested in the proof for this one, i'll just ski-

Mathologer: As an incentive to stick it out, I'll show you that very beautiful incarnation of Pascal's triangle…

Me, knowing I could just skip to the end: …fine

Oh, i remember the Horner Schema taught to me and how easy it is to use. And I will stick to that, not to shooting poor turtles with lasers!

Why does the turtle turn counter clockwise? If it turned clockwise you wouldn’t need to take the negative slope

Something just occurred to me. If you substitute i into any polynomial and consider each term in the polynomial as a vector in the complex plane, and then connect each successive vector tip to tail, the resulting diagram is equivalent to Lill's method. The only difference is that the leading term vector could point in any of the four directions instead of always pointing to the right initially. For example, x³ + 2x² + 4x +3 yields ‐i – 2 + 4i +3

That's absolutely amazing

l'article de Hill : http://www.numdam.org/article/NAM_1867_2_6__359_0.pdf

I just found this channel today but my god i love it

I SHOT TURTLES WITH LASERS TO DO MATH (GONE WRONG) (100% REAL) (NOT CLICKBAIT)

This is like my dream come true…since the very day I watch Mathloger, I was waiting for this video… I knew there was a relation

The true marvel is that this Edward Lill invented the concept of lasers 100 years before the first laser. 😀

Best clickbait for mathematicians ever.

Why did you allegorize this geometrical process as playing laser tags with a turtle?

I prefer burning him with a light beam

That’s because for n=2k

(-a)^n=((-a)^2)^k=(a^2)^k=a^n

Therefore

(-a)^(n+1)=((-a )^n)(-a)=(a^n)(-a)=-a^(n+1)

So ((-a)^(n+1))-(-a)^n=-a^(n+1)-a^n=-(a^(n+1)+a^n)

Q.E.D

Legends say this

turtlemade his journey from LOGO programing language to this video….. Just to become immortal….. But after reaching here they started shooting lasers at him…. 🐢If you enter a value of x with the opposite sign, the terms with even powers (including 0) will not change so for the cancellations to still happen to get 0, we need to change the signs of the coefficients with even powers of x.

Hi synthetic division, didn’t expect to see you here.