Solving EQUATIONS by tagging TURTLES with LASERS

Solving EQUATIONS by tagging TURTLES with LASERS


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.

100 thoughts on “Solving EQUATIONS by tagging TURTLES with LASERS

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

  2. Professor Edmund Weitz erstellte aufgrund dieses Video sein eigenes Video:
    https://www.youtube.com/watch?v=UkozVF8WRj8

  3. 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.

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

  5. 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?

  6. 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..

  7. 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

  8. 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.

  9. 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…

  10. 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.

  11. 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 ??

  12. Where do you get these type of articles from?
    Suggest some journals.
    I am an undergraduate mathematics student.

  13. 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?

  14. 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!

  15. 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.

  16. 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.

  17. 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.

  18. 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.

  19. 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.

  20. 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.

  21. 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.

  22. 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!

  23. @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?

  24. 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

  25. 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?

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

    Dankeschön!

  27. 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 !

  28. 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.

  29. 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

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

  31. 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

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

  33. 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.

  34. 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?

  35. 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.

  36. 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.

  37. 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

  38. 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!

  39. 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

  40. 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

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

    I prefer burning him with a light beam

  42. 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

  43. Legends say this turtle made his journey from LOGO programing language to this video….. Just to become immortal….. But after reaching here they started shooting lasers at him…. 🐢

  44. 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.

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