Nice! To prove this in general, the math simplifies considerably if one translates the general cubic such that the first two roots are centred on 0, so that: f(x) = (x+a)(x-a)(x-r) = (x^22 - a^2)(x-r) and f'(x) = 3x^2 - 2rx - a^2. Then f(0) = -a^2r and f'(0) = -a^2, so 0 - a^2r = -a^2(x-0) => -a^2r = -a^2x => r = x and the tangent intersects the x-axis at x=r. Q.E.D.

@starter497

Жыл бұрын

I think the second line is a typo. Should be x^2. Moreover, do you think its generalizes to odd degree polynomial??

@ThAlEdison

Жыл бұрын

I'm not sure it holds if you're a is 0. (i.e. the two roots you're taking the average of to find the tangent are repeated roots).

@ThAlEdison

Жыл бұрын

OK, so if you're saying pick a root r, and take the average of all other roots, does the root of the tangent line at that average value equal r. Then after some working out, it depends on this: let I={1...2n}, Sum_i=I(a_i)=0, and Prod_i=I(a_i)=A=/=0 does that imply that Sum_i=I(A/a_i)=0? if it does, then the answer is yes. I don't know how to go further.

@pietergeerkens6324

Жыл бұрын

@@starter497 Thank you. Now corrected - except I don't have privilege to save an edit to a pinned tweet.

@pietergeerkens6324

Жыл бұрын

@@ThAlEdison By definition of a repeated root, the tangent at that point has slope 0 (since it's either an extremum or an inflection) through a point on the x-axis; hence that tangent line **is** the x-axis, and by definition goes through all roots. I.e. if f(x) = (x-a)^n * g(x) for n>1 then, applying product rule and factoring, f'(x) = (x-a)^(n-1) * ( n*g(x) + (x-a)*g'(x) ) and f'(a) = 0.

I can confirm that I have never heard of this before.

@drpeyam

Жыл бұрын

Whoa hi bprp!!!

@acuriousmind6217

Жыл бұрын

same for me !

@SuperYoonHo

Жыл бұрын

Hello!!! bprp!

@SalsaInMarburg

Жыл бұрын

New for me, too.

@SalsaInMarburg

Жыл бұрын

But this is not a general proof.

Proof of the general case y = (x - a) * (x - b) * (x - c): The tangent at x = (a + b) / 2 is given by y = -(a - b)^2 * (x - c) / 4, which vanishes at x = c.

This result can be generalised to the following: the tangent line at (a+b)/2 of the function (x-a)(x-b)f(x) has the same root as the tangent of f(x) at (a+b)/2 (provided it exists).

Love to see it Dr. Peyam!❤️

Thanks so much Dr Peyam! Glad to have you back again!😍

sir, x³ - 9x² + 23x - 15 divides x - 1 the result will be x² - 8x + 15 not x² - 8x - 15. 1:19

@drpeyam

Жыл бұрын

Ok

@geraldillo

Жыл бұрын

You are right it was a mistake but did write it correct later when he wrote (x-1)(x² - 8x + 15)

@danielleza908

5 ай бұрын

Yeah I'm surprised no one mentions it, I was wondering what he was doing and how did he subtract those two expressions at the end and got 0

Welcome Back!!

nice! feels related to every cubic being symmetric around it's point of inflection

Finally!! A new video missed you so much

Love it, thank you.

I love your energy

awesome, with each video of yours i learn more and more

So glad you’re back. I started getting worried.

welcome back, Dr!

The 1:33 sign is wrong. (I know that later the sign is correct.)

@dabby249

Жыл бұрын

yeah clocked it too

welcome back!!! we were missing your presence!

Wow I was just saying yesterday how I missed your videos and we are blessed with one today. Christmas early🎉

Great. This gives a new perspective om solving cubic equation of states in physics.

Hi Dr. Peyam! I haven't heard this either. But it's very cool!

This would've simplified so many Exam problems in high school and early college. Glad to see that the legend is posting again 🙌

@drpeyam

Жыл бұрын

Omg hi Dylan!!! Hope all is well 😁

@meettrout419

Жыл бұрын

@@drpeyamWow, it is going well, I'm surprised you remember me still! I have been working with several differential equations and Laplace & Fourier transforms in signal analysis work since the day you first taught me that math behind it.👍 Solving second order ODE's is definitely more applicable than I initially thought it'd be 😂

@drpeyam

Жыл бұрын

How could I ever forget you?? Thanks again for the cake 😁 And I’m so happy to hear that you’re still working with diff eqs 😄

Well... now I have to go fox hunting

welcome back, dr. peyam

Curious on whether this works for a two complex roots such that you can still relate the 'x-coordinates' of two roots to the third. Also wondering what the relationship between the two farthest apart roots are (a and c this case) related to the middle root. Fascinating what properties are hidden in polynomials!

@koenth2359

Жыл бұрын

For the complex case you'd first have to specify how you are going to geometriclly represent a complex function. For the other question: as long as the third root is not equal to the mean of the other two, and the derivative there is nonzero, the same equation holds. Check my derivation.

The sum of the x coordinates of the intersections of a cubic and a line are constant when the line varies. Proof: Intersections are when y=mx+r and y=ax^3+bx^2+cx+d. Putting them together gives ax^3+bx^2+(c-m)x+(d-r)=0. Vieta says the sum of the roots is -b/a, so this is the sum of the x coordinates of the points of intersection. Now we know that if l, m, n are the roots of the original equation, this magic sum is l+m+n (the intersection of the cubic with y=0). But then taking the tangent at (l+m)/2, two of its roots are (l+m)/2, so the third must be n again. But the x-coordinate being n means the y-coordinate is 0, since n is a root of the cubic. So the tangent goes through (n,0), and we’re done.

YES YOU'RE BACK!

Can you do videos about exterior algebra 🤓💞

Nice! Here's the more general case: If f(x)=(x-a)(x-b)(x-c) then by the product rule f'(x)=(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b) Provided f(x) is not 0, note that f'(x)=f(x)(1/(x-a)+1/(x-b)+1/(x-c)) The tangent line at (x0, f(x0)) has y=f(x0)+f'(x0)(x-x0) and therefore, if f'(x0) is nonzero, it will cut the x-axis at x = x0 - f(x0)/f'(x0) Now, provided f(x0) and f'(x0) are both nonzero, then the tangent line cuts the x-axis at x = x0 - f(x0)/f'(x0) = x0 - 1/(1/(x0-a)+1/(x0-b)+1/(x0-c)) Plugging in x0=(a+b)/2 into this indeed gives x=c: x = (a+b)/2 - 1/(1/( (a+b)/2-a)+1/( (a+b)/2-b)+1/( (a+b)/2-c)) = (a+b)/2 - 1/(1/( (b-a)/2)+1/( (a-b)/2)+1/( (a+b-2c)/2)) =1/2 [ (a+b) - 1/(1/(a+b-2c))] =1/2 [ (a+b) - (a+b-2c)] =1/2 [2c] = c

Yea 🥰 Docter Peyam is back. I hope you are my real analysis teacher 🤣😂

Surprising fact

Great ! Never heard of it.

This is not a projectile solution. But that's because this is not a physics problem. The bullet fired from the hilltop traces a hyperbolic trajectory that is tangent to the hilltop but is not the straight tangent line to the root.

Welcome back. This is magic, somehow. How did you figure it out ?

So here’s a follow-up question: what is x for a line tangent to the curve between between 1 3 or x>5?

@drpeyam

Жыл бұрын

Interesting question!

I had this as an exam problem in Advanced Calculus for chapters 5 and 6 of Baby Rudin.

@drpeyam

Жыл бұрын

Fun!!

For when you find a cubic cliff

We in Italy to factor things use a formula from a guy named Ruffini, using the zero of a polinomia

@drpeyam

Жыл бұрын

Oh that’s just synthetic division

@radualexa1356

Жыл бұрын

Thanks!

In the American long division I learned, we only bring down as many values each time as we got rid of in the previous subtraction, so that the thing being subtracted aligns with the thing it's subtracted from, and we write each term of the result above the lowest term in that subtraction, so the x^n term in the result is above the x^n term in the dividend. I like to omit the powers of x and just work with coefficients in columns, but that's not standard in America for polynomials (of course, it is standard for decimals to omit the powers of 10).

0:52 typo It should have been x^3 - x^2 But its great to see your videos!!

Great !! Maths are always surprising !!!!

On a test in my AP Calc BC class this was an extra credit question on a test lol 😂

But this only shows it is true for that one polynomial. Would be cool to show it for all cases where a cubic has three roots. If indeed that is true. And if it is true for the average of the two smaller roots, then wouldn't the average of the two larger roots have a tangent that goes through the smaller root?

@birdbeakbeardneck3617

Жыл бұрын

in fact any couple of roots gives rise to a tangent through the third

@drpeyam

Жыл бұрын

See comments

@tutordave

Жыл бұрын

@@birdbeakbeardneck3617 Does that only work for cubics? What about higher degree polynomials?

@birdbeakbeardneck3617

Жыл бұрын

@@tutordave idk but probably not (if you mean tangent at (S-xi)/n goes through (xi,0) with S the sum of roots and xi one root and n the degree of the polynomial) ill think about it

How'd you go about proving this statement for every polynomial of degree 3? I struggle to imagine I need to use the explicit formulae for roots of 3rd degree polynomials

@kowalski6589

Жыл бұрын

Another comment pointed out how to prove it. Centering two roots around the origin simplifies the proof.

@utilizator1701

Жыл бұрын

There is a lot of nasty calculations. I give up at calculating the first coefficient of a 3 degree polynomial derivate at the average value between a and b. Maybe I will try to use symbolic calculation from Matlab.

@dalehall7138

Жыл бұрын

Try this: let p(x) = (x-a)(x-b)(x-c), since the leading coefficient is irrelevant, and let a,b be any two of the roots. The derivative p'(x) = (x - b)(x - c) + (x - a)(x - c) + (x - a)(x - b) Let q = (a+b)/2, the mean of the two chosen roots. Then, we get these (I've cleared out denominators because I can): 8 p(q) = (b - a)(a - b)(a + b - 2c) = -(a - b)^2 (a + b - 2c) 4 p'(q) = (a - b)(a + b - 2c) + (b - a)(a + b - 2c) + (b - a)(a - b) = -(a - b)^2 The equation for the line tangent to the curve y = p(x) at x = q is then y = T(x) = p'(q) (x - q) + p(q), and since I don't like denominators when I can avoid them: 8 T(x) = -2 (a - b)^2 (x - q) - (a - b)^2 (a + b - 2c) = - (a - b)^2 (2 (x - q) + (a + b - 2c)) Now, find out where T(x) lands when x = c: 8 T(c) = - (a - b)^2 (2 (c - (a + b)/2) - (a + b - 2c)) = - (a - b)^2 (2c - a - b + a + b - 2c) = 0 We find that the tangent line to y = p(x) at x = (a+b)/2 passes through the x-axis at x = c. Further, we never used the identity of any root, so this result holds when a,b are any two of the real roots of the cubic p(x), and c is the third real root.

@utilizator1701

Жыл бұрын

Ok, I written a Matlab program to do the calculation. Here is where the tangent of the polynomial (at the average value) is intersecting the x axis. ((- a^4 + 2*a^3*b + 2*c*a^3 - 6*c*a^2*b - 2*a*b^3 + 6*c*a*b^2 + 12*c*a + b^4 - 2*c*b^3 + 12*c*b)^2*(- a^4 + 2*a^3*b + 2*c*a^3 - 6*c*a^2*b + 6*a^2 - 2*a*b^3 + 6*c*a*b^2 + 12*a*b + b^4 - 2*c*b^3 + 6*b^2))/(864*(a + b)^3) Note that Matlab does not fully simplify expressions, so you may need to manually wrote the expression and reduce it yourself. (Also maybe I am misinterpreting the average value with the average value of the function and not of the roots.)

Typical for a mathematician: When firing projectiles, physical stuff like gravity can be ignore - the trajectory will be a straigt line, not a parabola. :D

@Zero-ov6xs

Жыл бұрын

the trajectory of a fallen object is actually not a parabola but the hyperbolic cos function

@bjornfeuerbacher5514

Жыл бұрын

@@Zero-ov6xs Pardon?! The hyperbolic cos function describes a hanging chain. The trajectory of a falling object _is_ a parabola.

@muttleycrew

Жыл бұрын

@@Zero-ov6xs Björn is exactly right. More generally: in the Newtonian approximation, all motions in which gravity is the only force acting must always be one of the conic sections and there is a proof for that which should not be too hard to find. The important result here is: objects solely under the influence of gravity must trace out curved paths described by the form of ax²+bxy+cy²+dx+ey+f = 0 for some values of a,b,c,d,e,f. This general form of second order equation recovers parabolic, circular, elliptical and hyperbolic curves but not cosh(x) = 1 + 1/2x² + 1/24 x^4 + ..., which requires a polynomial series in x, for which infinitely many terms of order higher than 2 are needed.

What, no synthetic division? That takes even less space!

He is not killing any foxes with that strategy. Ok , maybe pure mathematics foxes living in perfect mathematics universes with no gravity. Even so it’s a nice video and a fun fact.

Applications to missiles? Awesome. I am interested in programming simulations for anti ballistic missiles. I wonder if this will help.

@General12th

Жыл бұрын

Friendly reminder that the missile knows where it is because it knows where it isn't. :)

But the projectile has a parabolic movement?

American long division 😃 My parents didn't want me confused when they taught long division in school, so they didn't teach me how to lay it out. I invented my totally different layout, with each subtraction in a box at the bottom.

"For the french one you always run out of space" XD XD I'm even more mad because it's true ROFL

This hypothetical hunter's "projectile" appears to be immune to gravity, relative air density and humidity, and windage. ;^}

that for all curve in general ??

@drpeyam

Жыл бұрын

Of degree 3

@drpeyam

Жыл бұрын

I state the main result at the end

I thought it would be true even if we have a=b

@drpeyam

Жыл бұрын

No that would be a problem 😂 Also hi Ricky!!! Hope things are going well 😁

@liehanjiang3640

Жыл бұрын

@@drpeyam Aha! Welcome back to KZread for us!!!

General fact: f(x)=x³+ax²+bx+c, then f(x)=kx+m gives x³+ax²+(b-k)x+(c-m)=0, so the sum of the roots is (always) -a. Because tangents are lines though double intersections, it easily follows. More general fact: For every non-degenerate cubic (wikipedia has an article about those things) there is a function f, such that for points A, B and C on the cubic, f(A)+f(B)+f(C) is equal to a certain constant if and only of A, B and C are colinear (which means there is a line through A, B and C. Cool, right? Even better, A, B, C, D, E and F are on the same conic section if and only if f(A)+...+f(F) is equal to twice that constant! I just proved it myself, after coming up with the easiest proof for Pascal's theorem.

@drpeyam

Жыл бұрын

I like that!

@caspermadlener4191

Жыл бұрын

@@drpeyam Thank you :)

Is that true for each polynomial?

@kowalski6589

Жыл бұрын

For any cubic.

@IoT_

Жыл бұрын

@@kowalski6589 for any cubic with the roots a

@loicetienne7570

Жыл бұрын

​@@IoT_ I believe that the condition a < b < c is not necessary.

So if you are at the top - almost - of the hill you can aim a cruise missile parallel to the ground and kill the fox and the rest of the area. Useful to know. When I saw 1,2,3 and 5 I thought Fib. numbers were going to be used. What would a cubic look like with roots of 2,5, and 8 where 3 is the tangent and hits x=8? Can that even occur?

@drpeyam

Жыл бұрын

No here the tangent would be at (2+5)/2 = 3.5

Please do not eradicate the fox

@bobh6728

Жыл бұрын

No problem. The fox will not be eradicated unless the projectile can tunnel through the ground from 3 to 5. A whale might be in danger!!

Might want to redo that long division. There are two errors.

@drpeyam

Жыл бұрын

?

@dalenesbitt

Жыл бұрын

@@drpeyam 0:52 it should be x^3-x^2 (instead of x^3-x). 1:17 it should be +15 (instead of -15).

@drpeyam

Жыл бұрын

Ok, I mean the result is still correct

@dalenesbitt

Жыл бұрын

@@drpeyam yes, sorry for any confusion. I meant anybody watching might get confused about how dividing polynomials works. "Redo" as in refilm.

@drpeyam

Жыл бұрын

You want me to refilm the whole video just because of irrelevant errors? 😱

Proof of general statement?

@drpeyam

Жыл бұрын

See comment

Cute calculus, but why attack the foxes? They help control the population of rodents and other small mammals and birds.

Hay ầy. Phép chia đa thức. Nếu nhẩm được nghiêm.

just a simple logic: the summation of x coordinates are always the same when it is about cubic function and a line 1+3+5=2+2+5

@utilizator1701

Жыл бұрын

Why to add the value 2 two times?

عالی بود استاد ❤🖤

@drpeyam

Жыл бұрын

Mersi

Wonderful❗ But when we know the average of the two of three real solutions, we can easily get another solution. It's very intriguing but not so practical.

@drpeyam

Жыл бұрын

Are you sure? If you know a, b, and a+b/2, I don’t think you can get c

@vacuumcarexpo

Жыл бұрын

@@drpeyam Thanks for your reply. Let f(x)=px^3+qx^2+rx+s=p(x-a)(x-b)(x-c), then the solutions of f(x)=0 is a,b and c. a+b+c=-q/p. If we know t=(a+b)/2, then c=-q/p-2t. This is what I mean. Am I wrong?

Şu adamın videolarını nasıl Türkçe altyazı izliyim yok lan

that's really cool

You screwed up the polynomial long division. The answer (quotient) should be x^2 - 8x + 15. One characteristic of a great mathematician is their attention to detail. Why would you post such an embarrassing mistake?

@drpeyam

Жыл бұрын

Why would you write such an embarrassing comment?

@belovedone151

Жыл бұрын

@@drpeyam It's frustrating to find a mistake in a math book since you don't really know if it's a mistake or not. A lot of time is wasted trying to determine if it's really a mistake. You have the power to save a lot of people from confusion, frustration, and wasted time by simply editing your video to alleviate your mistake.

@drpeyam

Жыл бұрын

It’s not that simple to edit the video

The MATHS is great. However the video is spoilt by your unnecessary, flippant reference shooting innocent animals - when you could just as easily used the word TARGET instead of FOX. I hope that you do not hunt animals for fun and, as a subscriber, I beg you not to continue offending your animal loving subscribers.

@drpeyam

Жыл бұрын

Isn’t targeting the same as shooting?

@davidbrown1614

Жыл бұрын

@@drpeyam Used as a noun, as in the context of your video, a TARGET is something that is intentionally shot at - which is the FOX in your narrative. At a shooting range, a person usually shoots at an inanimate TARGET, which the intended object to be hit in order to score points. Hence do I conclude that the noun TARGET could (and I think should) replace the noun FOX. In other words the fox was specified as a potential living target - which I found to be offensive, when you could simply have referred to an unspecified TARGET.. As a verb, targeting is to shoot with intended precision with the intention of hitting a particular object (the target) - as opposed to just randomly shooting at anything.

drive.google.com/file/d/14Fa1mQ2ziYjQJqPUyPRON6NDxl_WR56z/view also at x=4, the tangent passes through (1,0) a root