The quadratic formula is essentially just the completing the square method, which is just algebraic manipulation, so it definitely works for any equation. Whether or not completing the square is actually useful in that case, though, is a different story
I took numerical analysis back in my undergrad. One technique to solve non-standard equations, was to find an initial guess, and then somehow find a way to isolate x, and just repeat the process. Similar to Newton but more basic
There are many such numerical methods for root finding. Newton's Method is shown in the vdo. I learnt it as Newton Raphson Method. Other methods include bisection method, false position method etc.
Hey bprp I was scrolling through your older videos and I saw that you made two promises 1. Sing the quadratic formula song! 2. Solve an easy integral in German language! :-) Can you please do them? Maybe just a Instagram reel but please fulfill your promise!
This is really interesting as the quadratic formula can be derived just from algebraic manipulation so obviously it can work with non-constant terms. But I’ve never seen anyone use it that way and never considered doing that before either! The only condition I can think of is a=/=0 but you can just check the a=0 case before doing the quadratic. Very fascinating stuff!
My first reaction. It should produce a valid "value" for x for any a(x),b(x),c(x) as long as a(x)0. After-all it's just completeing the square. However, the solution (as you point out) may not produce anything useful or simpler. But your sin(x) example is pretty cool as you get a different equation that has the same roots - and that is definitlely very cool. The formula is sort of acting like a transformation - well in my mind anyway.
If I was given that equation with sin(x) and cos(x)^2 , I would have just used cos(x)^2 = 1 - sin(x)^2 and then I have a quadratic equation in sin(x). Just use the quadratic formula on that. That seems more straightforward to me than using the formula on x.
If by "that equation" you meant the equation of the second half of the video, you would not get a quadratic equation in sin(x) by replacing cos(x)^2 with 1-sin(x)^2, because you still also have the plane powers of x.
hi so u can actually solve the previous cubic in this way using the quadratic formula the way u showed u get the eq x=(-1+sqrt(25+24x))/x if u rationalize this u get (2x^2+5)^2=25+24x if simplified u get 2x^3+5x-12=0 if u subtract both eq u get x^3=6 as one solution then using x-6 as a divisor u can find the other roots
Just a small doubt on my part. while we derive the quadratic formula, it is usually established that a is not zero(since it wouldn't be a quadratic if it was). However if a, b and c can be functions in x(How does this generalize to all functions? Piecewise functions too?) zero(or any root of a) can be a perfectly legible root for the equation but the denominator for the quadratic formula can become undefined. Eg. (x-2)*x^2 + x - 2. I am really curious about this idea, but I don't think we will be able to derive powerful conclusions without rigorously answering these questions. I highly enjoyed your video nonetheless. Thank You.
In those cases you'd just check those values manually. If f(x) = 0, then just solve for what value of x makes f(x)=0, and check if any of those values of x are solutions to the original equation, and then include them if they are.
An interesting application would be for finding roots for equations of the form ax^3+bx+c/x, doing the trick from this video, then multiplying by x and taking the square root for a final solution.
@@blackpenredpen The actor that claims 1*1=2 He went viral recently on Joe Rogans podcast so it would be cool if you could explain what his errors in reasoning are
For the end, there's an easier way to think about the second root. It actually make sense that the values are opposites as the original function is even. x^2 is even, xsinx is even as it's the product of two odd functions, and -1/4cos^2(x) is even because cosine is even. The sum of even functions is even. f(x) = f(-x), so f(0.335418) = 0 = f(-0.335418)
Probably not, lol...all one needs to do is find a counterexample to show this doesn't work...maybe a(x)=1/x^2, b(x)=1/x, c(x)=1 could do the trick, lol...the function becomes a constant, p(x) = 1+1+1=3...there are no roots...is that also what the quadratic formula would suggest?...I don't know, lol, I didn't plug it all in, but I'm fairly certain this isn't going to work...and if it does in this case, that it "usually" won't...
try x^5+x^3+x=0, where a=x^3, b=x^2, and c=x. Not as impossible as it looks because we can divide out an x, but it's still a quartic. But you can get all 4 complex roots using the trick you showed in this video.
Recently, I was looking at the equation ta^2+xa+y=0, where t is a function of x and y and assuming a is constant. I'm pretty sure, by just taking for granted the applicability of the quadratic formula to solve for a, you can eventually solve for t and y as functions of x. For example, for t(x) I got: t(x)=-2x/a-2/a^2+-[(x/a)^2+(8/a^2)(x/a)+4a^-4]^1/2 and I'm pretty sure that putting in my y(x) and t(x) into the original equation does yield a valid solution. However, tbh I don't understand why my approach was justified, and I'd love to understand it better. Edit: if it wasn't clear, I didn't start assuming t and y were variables of x. I assumed t, x, and y were implicitly related and tried to solve for t in terms of x and y, and then eventually you get another quadratic equation involving x and y and solving for y in terms of x. Plug that y(x) into t(x,y) to get t(x)=t(x,y(x))
Easier example than the ones in the video: x³ - 5x + 4/x = 0. You can take a = x, b = -5, c = 4/x and get x = (5 +- 3)/(2x), and multiplying by x gives the two equations x² = 1 or x² = 4, yielding the four solutions x1,2 = +- 1 and x3,4 = +- 2. (Obviously, you could also get these solutions by multiplying the original equation with x and then solving that bi-quadratic equation by standard methods.)
Newtons method was another programming execise I did on my Sinclair ZX81 in the early 1980s. I did have to explicitly program in the x function ands its derivative . It was very satisfying. I did need the extended memory module for this though, which was all of 16k.
Newton's Method uses a constant amount of memory. Unless you store all the iterations in an array and do hundreds if not thousands of iterations, you wouldn't need extra memory.
Without the expansion module the ZX81 only had 1 k of memory including the screen display, albeit compressed. The program took a chunk of memory so what was left if any was for the variables etc. arrays seemed to take up a lot of space.
@@jeffeloso Newton's method has trouble when it needs to find a solution that turns out to be a repeated root. I've noticed this happens with goal-seek in Excel, which I believe may be using Newton's method behind the scenes.
It works more generally on any commutative ring of characteristic either 0 or not divisible by 2 when a is invertible. This is a particular case. Also i'm sure people have used the quadratic formula like this, unlike you claim in the discription, i think i have before when solving some differential equation.
Another cool example, where we would use the Lambert W function is b=3exp(x) and c=-4exp(2x), we'd get 25exp(2x) under the square root which is 5exp(x)
i really like this even if i don't understand everything because i still lack some math knowledge that is used in this video... but i would love to learn it and hopefully school will teach it to me soon lol.
For a standard parabola, the vertex's x-coordinate will always be at the average of the two roots, since parabolas have reflectional symmetry. This isn't true for polynomials in general, but it is true for parabolas specifically.
No I wasn't talking about that By the relationship between the roots and the vertex I mean the both the x and y coordinate with the actual quadratic equation
@@carultch no actually you are not understanding it so let me explain it in a simpler way What's the graph of quadratic equation looks like? Yes it's a parabola or a "U" shaped whose RHS and LHS sides are little bit stretched on their sides now if you will join the roots and the vertex you will get "v" shaped or simply a triangle 🔺️ And my question is what is the relationship between all the vertices of the triangle 🔺️
Using the fact that [f(D_x)^2 - 1] y = 0 => [f(D_x) +/- 1] y = 0, you can show that, given ' [b(D_x) / a(D_x)] y ' exists, the quadratic formula also holds for certain sorts of differential operators a, b, c. (Analytic functions evaluated at the d/dx should work.) Notably, this is _not_ useful in solving most DEs, as I am aware of no techniques for solving the sort of non-linear non-sense that is x*y = [ - b(D_x) +/- sqrt( b(D_x)^2 - 4a(D_x)c(D_x) ) / 2a(D_x) ] y Best I can see is that the value of a solution at x=0, after applying the Quadratic-Formula'd Operator, would be 0. It is nifty to be able to isolate the non-constant part of the coefficients of a (very particular) DE, I guess.
The simplified equation is very nice, but since you still need to use a numerical method to solve, you might as well stick with the original. Here the Sage code: sage: f(x)=sin(x)+2*x-1 sage: nm(x)=x-f(x)/diff(f(x),x) sage: x=0.5 sage: x=nm(x);x.n()
Yes it works, you are doing algebraic manipulation In every step (1) to (2), you are actually saying (for every x, if x is a solution to (2), x is a solution to (1)), the "for every x" effectively fixes a,b,c. For example: a(x)x=b(x) to x=b(x)/a(x). You are saying " For every possible x, if x' (one of the possible values of x) makes x'=b(x')/a(x'), then x' will also make a(x')x'=b(x') " Given correct assumptions, you can also say the converse. In the quadratic formula, just make sure a(x)=/=0 and every step is equivalent, you won't lose any roots. This is just rarely useful, isolating x and bundling every other functions with x together with a square root is rarely useful when you want to solve actual answers. However, this is quite useful in finding a numerical answer, if the right side with a(x),b(x),c(x) is convex around the answer, you can iterate to it
Hey @blackpenredpen I have a question for you What does it 'exactly' mean by 'taking log both sides' Let's say I have an equation x-1=3 so the usually it's said that '-1' is transferred to the other side and it's sign changes but in reality we add +1 on both sides making x=4 Now let's say logx(100) =2 So by saying taking log both sides Does it mean that log(logx(100)= log(2) ??? I don't understand this concept can you explain this in a video my teachers never really write this step and jump to the answers 😅
I'm assuming x is the base of the logarithm. If that is the case, x^2 = 100 based on the definition of what a logarithm is. The logarithm is the exponent that the base of the logarithm is raised in order to reach the number in the logarithm. The alternative is to put it into a more familiar term if you wished, which would be changing the base as log_x(100) = ln 100 / ln x = 2. If you are really uncomfortable with domains, keep it rational and make it ln 100 / ln x - 2 = 0, (ln 100 - 2 ln x) / ln x = 0 by finding common denominator. ln 100 - 2 ln x = 0 by using the property of multiplication of 0. ln x = ln 100 / 2. ln x = ln 100^(1/2). Raise e to the power of both sides, which means that x = 10. 10 doesn't satisfy the ln x = 0, so it is a real answer.
Meanwhile, in an alternative universe, BPRP wonders if he can use the quadratic formula if a, b and c are not constants, but functions of x. He realises that this could give something nice if the discriminant is a perfect square, so the square root can be eliminated. For good measure, he decides that the "quadratic" will be in sin x, rather than just x. Now for the discriminant Δ: If we let a=1 and b=4x (the coefficient 4 is to avoid fractions when solving for c), then b²-4ac=16x²-4c. If we want Δ=4 (a perfect square), then 16x²-4c=4, 4c=16x²-4, c=4x²-1 So he gets the "quadratic" equation in sin x: sin²x+4x sin x+4x²-1=0, with "solutions" given by the quadratic formula: sin x=(-4x±2)/2=-2x±1 which can be rewritten sin x+2x±1=0 He then gamely uses the Newton-Raphson method to find a numerical solution of each to 5dp. As luck would have it, each of the new equations has only one real solution and the original equation exactly 2 real solutions. As the "quadratic" is even in x, the solutions are opposites. He then notices that the "quadratic" in sin x he found is actually also a "quadratic" in x itself, so he now rewrites the equation in "standard" form, so as to be able to apply the quadratic formula again, but this time to a "quadratic" in x: sin²x+4x sin x+4x²-1=0 4x²+4x sin x+sin²x-1=0 4x²+4x sin x-cos²x=0 x²+x sin x-¼cos²x=0 For the rest of the story, see the video...
A question: I understood that you could integrate over an interval (because that's what you do in your final year), by considering the limits of this interval, but in prep school, you integrate over different domains, so what's the point of integrating over an edge, a surface or something else?
I don't know what you mean with "integrating over an edge". Integrating over a surface is useful for lots of things, e. g. determing the area of the surface, or its center of mass, or the flux of a vector field (e. g. the velocity field of a fluid, or an electric or magnetic field).
@@IlyesBenahmed-vf6gi It's hard to explain in a TH-cam component, because one needs to describe it with formulas... but essentially, you divide the surface into small pieces, multiply the area of each piece with the value of the function there which you want to integrate, and sum up all of these contributions. And then you take the limit where all lengths of these small pieces go to zero.
Thanks for an interesting way to use the quadratic formula. I notice if you are trying to solve a linear equation, a=0, and the formula seems to break down.
It turns out, you can rationalize the numerator of the quadratic formula, and you can show how you can reconcile it for the case of a = 0. Given: a*x^2 + b*x + c = 0 a = 0 x = (-b +/- sqrt(b^2 - 4*a*c))/(2*a) Multiply by (-b -/+ sqrt(b^2 - 4*a*c))/(-b -/+ sqrt(b^2 - 4*a*c)): x = (-b +/- sqrt(b^2 - 4*a*c))* (-b -/+ sqrt(b^2 - 4*a*c))/(2*a* (-b -/+ sqrt(b^2 - 4*a*c))) x = 4*a*c/(2*a* (-b -/+ sqrt(b^2 - 4*a*c))) Cancel the a: x = 4*c/(2*(-b -/+ sqrt(b^2 - 4*a*c))) Now you can directly plug in a=0: x = 4*c/(2*(-b -/+ b)) -b + b = 0, so we cannot use that solution, as it will be degenerate. We need to use -b - b instead. x = 2*c/(-2*b) x = -c/b And this is precisely the x-intercept of: b*x + c = 0
Your question seems to be highly connected to abstract algebra and Galois theory and maybe even algebraic geometry. Yes, a cubic (and quartic) equation exists, but iirc, you cannot write them down without nested roots, so your quadratic formula method wouldn’t get far. The proof for this I believe involves studying ring theory and examining field extensions. Then, once you get to 5th degree polynomials, you can no longer do this; there exist 5th degree polynomials with no roots which can be written with just addition, multiplication, and exponentiation, namely x^5-x-1. This is proven with Galois theory. As for your last example, you could have done exactly that with completing the square, and that’s a common technique used in more advanced studies over all sorts of algebraic rings and fields. My favorite example of such a thing is the derivation of the “Dirac equation” in physics, where you quite literally use the quadratic formula and get 2 solutions - one which corresponds to regular matter, and the other which corresponds to antimatter, which was the first prediction that positrons exist.
Back when I was an undergrad in Chem Engineering (40 years ago), we didn't bother with closed form solutions. I just pile the terms on both sides of the equation, make a program in BASIC on my Sharp handheld - yes, they had computers back then - and let the program keep iterating until both sides of the equation equals each other.
My answer to this question is "That depends." For instance, it would not work with x*x^2 + b*x + c = 0. It would work with x^2 + b*x*y + y^2 = 0 or x^2 * x^2 + b * x * x + c = 0 though.
I mean, it "works" on any equation.. it's just that in most equations the quadratic equation you end up with will be more difficult to solve than the one you started with which makes it kind of impractical (but it's not mathematically incorrect).
Overall, I wonder why there was no call to the intermediate value theorem in order to talk about how much solution there is. It would be because he didn't want to bother talking about it simply enough, but I rarely saw it precised in any of his videos sadly
It doesn't have an elementary solution. The first part, x^2/(x^2 + 1), is extremely easy. Add zero in a fancy way, to form a term we can cancel, so we can rewrite it as 1 + 1/(x^2 + 1). This integrates as x + arctan(x) The second part has no elementary solution. Wolfram Alpha's solution is: ((e^2 - 1)*Ci(i - x) + (e^2 - 1)*Ci(x + i) + i*(1 + e^2)*(Si(i - x) + Si(x + i)))/(4*e) Where Ci(x) is the special cosine integral function, and Si(x) is the special sine integral function,, e is Euler's number, and i is the imaginary unit. So the combined solution is: x + arctan(x) + ((e^2 - 1)*Ci(i - x) + (e^2 - 1)*Ci(x + i) + i*(1 + e^2)*(Si(i - x) + Si(x + i)))/(4*e) + C
That one is trivial: -1/(4*x^4) + x + C If you meant 1/(x^5 + 1), that is much more complicated. He has a video on that one, which involves finding all 5 roots of the quintic. One root is trivial to find, which is x = -1. From there, you can use polynomial division to find the quartic and linear factors. You can then use the quartic formula to find the 4 remaining complex roots. Once you have all roots, you can then use partial fractions to integrate. Unpacking the meaning of the partial fraction terms with complex coefficients, is the final challenge.
Unless you're working with a cherry picked equation where it was specifically designed to work, it makes it way, way harder to solve not easier. The equation you get after plugging it into the quadratic formula is almost always more complicated than the one you started with.
I mean.. you probably could.. whether or not it would actually be useful on the other hand is a lot more questionable. It will almost certainly just make the problem more complicated than it was to start with, because you won't have any constants in the equation that you can just calculate so you're just exchanging one equation you don't know how to solve for another different equation that you also don't know how to solve (but is also way messier than the first one).
I created an incredible example based on this: th-cam.com/video/j6ri-2S-hxU/w-d-xo.html
Something weird happens when you plot , y = tan(xy) , in desmos.
Great, keep it going
Yeah
Ignore 👍🏼
this also is consistent with Galois theory since x^4 and above you cannot decompose like this wow
I mean what else are you supposed to think about in the car
agreed
Exactly!
😅 haha yeah
how badly you want to drive into a brick wall at 120 mph
@@MikehMike01if ur going 120 mph u got a speeding problem 💀
The black and red pens at the back are the most underrated thing on the frame though
Not enough credit gets given to the blue pen.
The quadratic formula is essentially just the completing the square method, which is just algebraic manipulation, so it definitely works for any equation. Whether or not completing the square is actually useful in that case, though, is a different story
i’m boutta complete the hypercube
The first thing i thought after seeing the thumbnail is "What!!! is this man going insane?"
This is just about as weird as that time he solved a quadratic equation not in x but in 5 instead (I think it worked, too).
@@IvyANguyen imagine ∫x⁵ d5
@@yoylecake313 Would that be (x^5)/ln(x)?
WEIRD? I've been wondering about this for months, but I've had bigger fish on my lines.
@@IvyANguyen
+c
Somebody stop him, he's getting too powerful
Said the grievous lady
@@dyltan lol
@@Rando2101 yeah sorry I play arcaea too, great game
blud plays arcaea
we can't stop him
I took numerical analysis back in my undergrad. One technique to solve non-standard equations, was to find an initial guess, and then somehow find a way to isolate x, and just repeat the process. Similar to Newton but more basic
Fixed point iteration!
I did something like that in Numerical Methods when I went to the University of Toledo in the late 80s or early 90s.
There are many such numerical methods for root finding. Newton's Method is shown in the vdo. I learnt it as Newton Raphson Method. Other methods include bisection method, false position method etc.
This gives me similar vibes to that equation you did long time ago about a quadratic equation in terms of 5.
Oh yea. I remember that.
@@blackpenredpen,
that was great fun , five as an unknown calculated by the abc-formula containing all kinds of other things .
your vibes are irrational
" a, b, c
Qua-, dra-, tic"
😂
Now plug a quadratic equation into a quadratic equation
real
I did it once
Then plug it in a quadratic eqn😂
@@Ghostwriter_zone Then ask a question about the limits of repeatedly plugging in probably varying quadratics, naturally as one does in the car
@@soupy5890 ooh laa
That's why car was dancing
At this point, bro is waiting for math 2 to release.
Hey bprp I was scrolling through your older videos and I saw that you made two promises
1. Sing the quadratic formula song!
2. Solve an easy integral in German language! :-)
Can you please do them? Maybe just a Instagram reel but please fulfill your promise!
I actually did the first one but I forgot in what video. I need the help from dr peyam for the second one but he never did.
@@blackpenredpen I am gonna wait patiently for the second one and until then I will find the song in your voice! Great work!
@@blackpenredpen Check DorFuchs
This is really interesting as the quadratic formula can be derived just from algebraic manipulation so obviously it can work with non-constant terms. But I’ve never seen anyone use it that way and never considered doing that before either! The only condition I can think of is a=/=0 but you can just check the a=0 case before doing the quadratic. Very fascinating stuff!
After getting the first solution, we can just check the original equation is even, so with that we know x_2 = - x_1.
very nice
My first reaction. It should produce a valid "value" for x for any a(x),b(x),c(x) as long as a(x)0. After-all it's just completeing the square. However, the solution (as you point out) may not produce anything useful or simpler. But your sin(x) example is pretty cool as you get a different equation that has the same roots - and that is definitlely very cool. The formula is sort of acting like a transformation - well in my mind anyway.
If I was given that equation with sin(x) and cos(x)^2 , I would have just used
cos(x)^2 = 1 - sin(x)^2
and then I have a quadratic equation in sin(x). Just use the quadratic formula on that. That seems more straightforward to me than using the formula on x.
If by "that equation" you meant the equation of the second half of the video, you would not get a quadratic equation in sin(x) by replacing cos(x)^2 with 1-sin(x)^2, because you still also have the plane powers of x.
@@Apollorion I'm not arguing semantics. I just explained the way to solve it. It works fine, no matter what you call it.
You just can't get enough of abusing the quadratic equation every time you upload :)
😆
hi so u can actually solve the previous cubic in this way
using the quadratic formula the way u showed u get the eq x=(-1+sqrt(25+24x))/x
if u rationalize this u get (2x^2+5)^2=25+24x
if simplified u get 2x^3+5x-12=0
if u subtract both eq u get x^3=6 as one solution
then using x-6 as a divisor u can find the other roots
Just a small doubt on my part. while we derive the quadratic formula, it is usually established that a is not zero(since it wouldn't be a quadratic if it was). However if a, b and c can be functions in x(How does this generalize to all functions? Piecewise functions too?) zero(or any root of a) can be a perfectly legible root for the equation but the denominator for the quadratic formula can become undefined. Eg. (x-2)*x^2 + x - 2. I am really curious about this idea, but I don't think we will be able to derive powerful conclusions without rigorously answering these questions. I highly enjoyed your video nonetheless. Thank You.
In those cases you'd just check those values manually. If f(x) = 0, then just solve for what value of x makes f(x)=0, and check if any of those values of x are solutions to the original equation, and then include them if they are.
An interesting application would be for finding roots for equations of the form ax^3+bx+c/x, doing the trick from this video, then multiplying by x and taking the square root for a final solution.
Bros taking quadratic to whole another level 💀
Hey you should go over that Terrance Howard thing where (sqrt(2))^3/2=sqrt(2) and explain it in your own words.
?
@@blackpenredpen The actor that claims 1*1=2
He went viral recently on Joe Rogans podcast so it would be cool if you could explain what his errors in reasoning are
For the end, there's an easier way to think about the second root. It actually make sense that the values are opposites as the original function is even. x^2 is even, xsinx is even as it's the product of two odd functions, and -1/4cos^2(x) is even because cosine is even. The sum of even functions is even.
f(x) = f(-x), so f(0.335418) = 0 = f(-0.335418)
Probably not, lol...all one needs to do is find a counterexample to show this doesn't work...maybe a(x)=1/x^2, b(x)=1/x, c(x)=1 could do the trick, lol...the function becomes a constant, p(x) = 1+1+1=3...there are no roots...is that also what the quadratic formula would suggest?...I don't know, lol, I didn't plug it all in, but I'm fairly certain this isn't going to work...and if it does in this case, that it "usually" won't...
try x^5+x^3+x=0, where a=x^3, b=x^2, and c=x. Not as impossible as it looks because we can divide out an x, but it's still a quartic. But you can get all 4 complex roots using the trick you showed in this video.
thanks sir
Recently, I was looking at the equation ta^2+xa+y=0, where t is a function of x and y and assuming a is constant. I'm pretty sure, by just taking for granted the applicability of the quadratic formula to solve for a, you can eventually solve for t and y as functions of x. For example, for t(x) I got:
t(x)=-2x/a-2/a^2+-[(x/a)^2+(8/a^2)(x/a)+4a^-4]^1/2
and I'm pretty sure that putting in my y(x) and t(x) into the original equation does yield a valid solution. However, tbh I don't understand why my approach was justified, and I'd love to understand it better.
Edit: if it wasn't clear, I didn't start assuming t and y were variables of x. I assumed t, x, and y were implicitly related and tried to solve for t in terms of x and y, and then eventually you get another quadratic equation involving x and y and solving for y in terms of x. Plug that y(x) into t(x,y) to get t(x)=t(x,y(x))
Easier example than the ones in the video: x³ - 5x + 4/x = 0. You can take a = x, b = -5, c = 4/x and get x = (5 +- 3)/(2x), and multiplying by x gives the two equations x² = 1 or x² = 4, yielding the four solutions x1,2 = +- 1 and x3,4 = +- 2. (Obviously, you could also get these solutions by multiplying the original equation with x and then solving that bi-quadratic equation by standard methods.)
We can now discover warp drive, alcubiere drive, Dyson sphere and time travel. What a discovery
hmmm, recursion might be neat, where a=ax^2+bx+c.
What do you mean?
@@blackpenredpen say (ax^2+bx+c)x^2+bx+c=0 is 1st level of recursion.
Newtons method was another programming execise I did on my Sinclair ZX81 in the early 1980s. I did have to explicitly program in the x function ands its derivative . It was very satisfying. I did need the extended memory module for this though, which was all of 16k.
Newton's Method uses a constant amount of memory. Unless you store all the iterations in an array and do hundreds if not thousands of iterations, you wouldn't need extra memory.
Without the expansion module the ZX81 only had 1 k of memory including the screen display, albeit compressed. The program took a chunk of memory so what was left if any was for the variables etc. arrays seemed to take up a lot of space.
@@jeffeloso Newton's method has trouble when it needs to find a solution that turns out to be a repeated root. I've noticed this happens with goal-seek in Excel, which I believe may be using Newton's method behind the scenes.
It works more generally on any commutative ring of characteristic either 0 or not divisible by 2 when a is invertible. This is a particular case. Also i'm sure people have used the quadratic formula like this, unlike you claim in the discription, i think i have before when solving some differential equation.
Good old functional analysis. It works point by point so you can define the solution.
Another cool example, where we would use the Lambert W function is b=3exp(x) and c=-4exp(2x), we'd get 25exp(2x) under the square root which is 5exp(x)
I like your Curiosity because im the same way with numbers with different formulas
I completed the square in my mind I knew it's fine but still wanted to watch
Sir, How to know the no. Of rows in DI method? For intergration
Blackpenredpen is back again!!!
People should be glad Police can't check Maths reading like Alcohol reading
i really like this even if i don't understand everything because i still lack some math knowledge that is used in this video... but i would love to learn it and hopefully school will teach it to me soon lol.
Good job now can you think what is the relationship between the roots and the vertex of the parabola ?? TAI BPRP
For a standard parabola, the vertex's x-coordinate will always be at the average of the two roots, since parabolas have reflectional symmetry. This isn't true for polynomials in general, but it is true for parabolas specifically.
No I wasn't talking about that
By the relationship between the roots and the vertex I mean the both the x and y coordinate with the actual quadratic equation
@@Nothingx303 The vertex will occur at the point (-b/(2*a), c - b^2/(4*a) )
Is that what you were asking?
@@carultch no actually you are not understanding it so let me explain it in a simpler way
What's the graph of quadratic equation looks like?
Yes it's a parabola or a "U" shaped whose RHS and LHS sides are little bit stretched on their sides now if you will join the roots and the vertex you will get "v" shaped or simply a triangle 🔺️
And my question is what is the relationship between all the vertices of the triangle 🔺️
@@carultch well I forgot to say one thing that is with the quadratic equation
can that be done with matrices?, at least the nxn ones?
I knew i was not the only one who thought about this stuff on weird times!!! I also wonder upon this while trying to sleep..
Using the fact that
[f(D_x)^2 - 1] y = 0
=>
[f(D_x) +/- 1] y = 0,
you can show that, given ' [b(D_x) / a(D_x)] y ' exists, the quadratic formula also holds for certain sorts of differential operators a, b, c. (Analytic functions evaluated at the d/dx should work.)
Notably, this is _not_ useful in solving most DEs, as I am aware of no techniques for solving the sort of non-linear non-sense that is
x*y = [ - b(D_x) +/- sqrt( b(D_x)^2 - 4a(D_x)c(D_x) ) / 2a(D_x) ] y
Best I can see is that the value of a solution at x=0, after applying the Quadratic-Formula'd Operator, would be 0. It is nifty to be able to isolate the non-constant part of the coefficients of a (very particular) DE, I guess.
The simplified equation is very nice, but since you still need to use a numerical method to solve, you might as well stick with the original. Here the Sage code:
sage: f(x)=sin(x)+2*x-1
sage: nm(x)=x-f(x)/diff(f(x),x)
sage: x=0.5
sage: x=nm(x);x.n()
UR CHANNEL is the BEST PROF)
Could someone link me the vide about Newton thing? I dont see it in the description
Just added. Thanks.
Newton's method (introduction & example)
th-cam.com/video/iVOsU4tnouk/w-d-xo.html
Why didnt you solve with the newton's method to solve the original equation directly?
Yes it works, you are doing algebraic manipulation
In every step (1) to (2), you are actually saying (for every x, if x is a solution to (2), x is a solution to (1)), the "for every x" effectively fixes a,b,c.
For example: a(x)x=b(x) to x=b(x)/a(x). You are saying " For every possible x, if x' (one of the possible values of x) makes x'=b(x')/a(x'), then x' will also make a(x')x'=b(x') "
Given correct assumptions, you can also say the converse. In the quadratic formula, just make sure a(x)=/=0 and every step is equivalent, you won't lose any roots.
This is just rarely useful, isolating x and bundling every other functions with x together with a square root is rarely useful when you want to solve actual answers. However, this is quite useful in finding a numerical answer, if the right side with a(x),b(x),c(x) is convex around the answer, you can iterate to it
Bro boutta add a chapter in the curriculum
Even calculus is afraid of this man , somebody please stop him !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Nice thought!!!
This channel is just epic
Where did you buy the euler on the wall?
It's his product
Ouuhh, where would i find it? I can only find hoodies? @@prince_bk
This is very cool!
Hey @blackpenredpen I have a question for you
What does it 'exactly' mean by 'taking log both sides'
Let's say I have an equation
x-1=3 so the usually it's said that '-1' is transferred to the other side and it's sign changes but in reality we add +1 on both sides making x=4
Now let's say logx(100) =2
So by saying taking log both sides
Does it mean that log(logx(100)= log(2) ??? I don't understand this concept can you explain this in a video my teachers never really write this step and jump to the answers 😅
I'm assuming x is the base of the logarithm. If that is the case, x^2 = 100 based on the definition of what a logarithm is. The logarithm is the exponent that the base of the logarithm is raised in order to reach the number in the logarithm. The alternative is to put it into a more familiar term if you wished, which would be changing the base as log_x(100) = ln 100 / ln x = 2. If you are really uncomfortable with domains, keep it rational and make it ln 100 / ln x - 2 = 0, (ln 100 - 2 ln x) / ln x = 0 by finding common denominator. ln 100 - 2 ln x = 0 by using the property of multiplication of 0. ln x = ln 100 / 2. ln x = ln 100^(1/2). Raise e to the power of both sides, which means that x = 10. 10 doesn't satisfy the ln x = 0, so it is a real answer.
Meanwhile, in an alternative universe, BPRP wonders if he can use the quadratic formula if a, b and c are not constants, but functions of x.
He realises that this could give something nice if the discriminant is a perfect square, so the square root can be eliminated.
For good measure, he decides that the "quadratic" will be in sin x, rather than just x.
Now for the discriminant Δ:
If we let a=1 and b=4x (the coefficient 4 is to avoid fractions when solving for c), then b²-4ac=16x²-4c.
If we want Δ=4 (a perfect square), then 16x²-4c=4, 4c=16x²-4,
c=4x²-1
So he gets the "quadratic" equation in sin x:
sin²x+4x sin x+4x²-1=0,
with "solutions" given by the quadratic formula:
sin x=(-4x±2)/2=-2x±1
which can be rewritten
sin x+2x±1=0
He then gamely uses the Newton-Raphson method to find a numerical solution of each to 5dp.
As luck would have it, each of the new equations has only one real solution and the original equation exactly 2 real solutions.
As the "quadratic" is even in x, the solutions are opposites.
He then notices that the "quadratic" in sin x he found is actually also a "quadratic" in x itself, so he now rewrites the equation in "standard" form, so as to be able to apply the quadratic formula again, but this time to a "quadratic" in x:
sin²x+4x sin x+4x²-1=0
4x²+4x sin x+sin²x-1=0
4x²+4x sin x-cos²x=0
x²+x sin x-¼cos²x=0
For the rest of the story, see the video...
Espectacular... gracias.
Where is the beard? It makes you look wise.
ax² = qua
bx = dra
c = tic
0 = equation
Thanks for the vedio
How about: a=x, b=2x, c=x. Following your approach we get to x=-1. But we somehow lose the alternative x=0 solution.
pretty cool!
A question: I understood that you could integrate over an interval (because that's what you do in your final year), by considering the limits of this interval, but in prep school, you integrate over different domains, so what's the point of integrating over an edge, a surface or something else?
I don't know what you mean with "integrating over an edge".
Integrating over a surface is useful for lots of things, e. g. determing the area of the surface, or its center of mass, or the flux of a vector field (e. g. the velocity field of a fluid, or an electric or magnetic field).
And what is the definition of an integral over a surface?
@@IlyesBenahmed-vf6gi It's hard to explain in a TH-cam component, because one needs to describe it with formulas... but essentially, you divide the surface into small pieces, multiply the area of each piece with the value of the function there which you want to integrate, and sum up all of these contributions. And then you take the limit where all lengths of these small pieces go to zero.
@@IlyesBenahmed-vf6gi Professor Dave Explains has a video on surface integrals in his Calculus playlist.
Oh, thank you.
I’m at the beginning of the video. It seems that it should be possible for this to work although I’ve never even considered it before.
Thanks for an interesting way to use the quadratic formula. I notice if you are trying to solve a linear equation, a=0, and the formula seems to break down.
It turns out, you can rationalize the numerator of the quadratic formula, and you can show how you can reconcile it for the case of a = 0.
Given:
a*x^2 + b*x + c = 0
a = 0
x = (-b +/- sqrt(b^2 - 4*a*c))/(2*a)
Multiply by (-b -/+ sqrt(b^2 - 4*a*c))/(-b -/+ sqrt(b^2 - 4*a*c)):
x = (-b +/- sqrt(b^2 - 4*a*c))* (-b -/+ sqrt(b^2 - 4*a*c))/(2*a* (-b -/+ sqrt(b^2 - 4*a*c)))
x = 4*a*c/(2*a* (-b -/+ sqrt(b^2 - 4*a*c)))
Cancel the a:
x = 4*c/(2*(-b -/+ sqrt(b^2 - 4*a*c)))
Now you can directly plug in a=0:
x = 4*c/(2*(-b -/+ b))
-b + b = 0, so we cannot use that solution, as it will be degenerate. We need to use -b - b instead.
x = 2*c/(-2*b)
x = -c/b
And this is precisely the x-intercept of:
b*x + c = 0
How fantastic man
Heck yes
Your question seems to be highly connected to abstract algebra and Galois theory and maybe even algebraic geometry. Yes, a cubic (and quartic) equation exists, but iirc, you cannot write them down without nested roots, so your quadratic formula method wouldn’t get far. The proof for this I believe involves studying ring theory and examining field extensions. Then, once you get to 5th degree polynomials, you can no longer do this; there exist 5th degree polynomials with no roots which can be written with just addition, multiplication, and exponentiation, namely x^5-x-1. This is proven with Galois theory. As for your last example, you could have done exactly that with completing the square, and that’s a common technique used in more advanced studies over all sorts of algebraic rings and fields. My favorite example of such a thing is the derivation of the “Dirac equation” in physics, where you quite literally use the quadratic formula and get 2 solutions - one which corresponds to regular matter, and the other which corresponds to antimatter, which was the first prediction that positrons exist.
Back when I was an undergrad in Chem Engineering (40 years ago), we didn't bother with closed form solutions. I just pile the terms on both sides of the equation, make a program in BASIC on my Sharp handheld - yes, they had computers back then - and let the program keep iterating until both sides of the equation equals each other.
0:00 Here's the story
Of a lovely lady
Who was bringing up three very lovely girls
a, b and c
My answer to this question is "That depends." For instance, it would not work with x*x^2 + b*x + c = 0. It would work with x^2 + b*x*y + y^2 = 0 or x^2 * x^2 + b * x * x + c = 0 though.
I mean, it "works" on any equation.. it's just that in most equations the quadratic equation you end up with will be more difficult to solve than the one you started with which makes it kind of impractical (but it's not mathematically incorrect).
Man… I want to Taylor expand the RHS so bad. Maybe to three terms, keeping the x^2 term… apply the quadratic formula again… 😅
Overall, I wonder why there was no call to the intermediate value theorem in order to talk about how much solution there is. It would be because he didn't want to bother talking about it simply enough, but I rarely saw it precised in any of his videos sadly
The quadratic equation has been used before in numerous ways...like finding the inverses of the hyperbolic trig functions
The x’th root of the i’th root of the pi’th root of the i’th root of i = the (√(6x) + √(-x))th root of e^x . Solve for x
I used it on constant numbers once and it worked lol 😂
Sri Dharacharya formula
I’m glad to have woken up early and saw your new post 😊
"I was thinking about the quadratic formula. As one does."
I really liked ✌🏼QUA ✌🏼 DRA ✌🏼 TIC equaion
it should work because the process is nothing but just factoring terms.
Is this already proven to work? Would be interesting to see a rigorous prove that this applies for all functions of c
Interesting
Degree 4 polynomial has radical solution formula. Use a(x) = mx + c. Now any degree 5 polynomials has radical solution formula. QED.
Lol!
I need you help: Integral of ((x^2 + sin x)/(x^2 +1))dx
It doesn't have an elementary solution. The first part, x^2/(x^2 + 1), is extremely easy. Add zero in a fancy way, to form a term we can cancel, so we can rewrite it as 1 + 1/(x^2 + 1). This integrates as x + arctan(x)
The second part has no elementary solution. Wolfram Alpha's solution is:
((e^2 - 1)*Ci(i - x) + (e^2 - 1)*Ci(x + i) + i*(1 + e^2)*(Si(i - x) + Si(x + i)))/(4*e)
Where Ci(x) is the special cosine integral function, and Si(x) is the special sine integral function,, e is Euler's number, and i is the imaginary unit.
So the combined solution is:
x + arctan(x) + ((e^2 - 1)*Ci(i - x) + (e^2 - 1)*Ci(x + i) + i*(1 + e^2)*(Si(i - x) + Si(x + i)))/(4*e) + C
I guess its better to treat them as constants
Please try to integrate 1/x^5+1
That one is trivial:
-1/(4*x^4) + x + C
If you meant 1/(x^5 + 1), that is much more complicated. He has a video on that one, which involves finding all 5 roots of the quintic. One root is trivial to find, which is x = -1. From there, you can use polynomial division to find the quartic and linear factors. You can then use the quartic formula to find the 4 remaining complex roots. Once you have all roots, you can then use partial fractions to integrate. Unpacking the meaning of the partial fraction terms with complex coefficients, is the final challenge.
Is there not a 'Lambert-sine' kind of function for analytically solving equations that involve x and sin(x)?
You could try expressing the sine by complex exponentials, perhaps then it somehow becomes possible to use the usual Lambert function?
Are you excepting challenge problems?
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as long as the image of the functions are still in a field, there is no reason why the quadratic formula wouldn't work.
means, solving cubic or quartic is easier now
Unless you're working with a cherry picked equation where it was specifically designed to work, it makes it way, way harder to solve not easier. The equation you get after plugging it into the quadratic formula is almost always more complicated than the one you started with.
Okay but if you need to do newtons, you could have just done it for the original functuon
in this case it's not actually that difficult to get 2x = -sinx ± 1 without using quadratic formula
x^2 + xsinx - 1/4cos^2x = 0
4x^2 + 4xsinx - cos^2x = 0
4x^2 + 4xsinx - 1 + sin^2x = 0
(sinx)^2 + 2(2x)(sinx) + (2x)^2 - 1 = 0
(sinx + 2x)^2 - 1 = 0
(sinx + 2x - 1)(sinx + 2x + 1) = 0
sinx + 2x = ±1
2x = -sinx ± 1
definitely much faster to do it with the quadratic formula though!
Just like the quadratic formula, you're completing the square.
@@Apollorion it's not completing the square it's just factorising
@@tobysuren Ok, solving A^2-1=0 not as A^2=1 and thus A=±1 but as (A-1)(A+1)=0 and thus A=±1 isn't really 'completing the square'.
@@Apollorion I never said it was?
Now prove that we can use this on the quartic formula for quintic equation.
I mean.. you probably could.. whether or not it would actually be useful on the other hand is a lot more questionable. It will almost certainly just make the problem more complicated than it was to start with, because you won't have any constants in the equation that you can just calculate so you're just exchanging one equation you don't know how to solve for another different equation that you also don't know how to solve (but is also way messier than the first one).
Great
Out of millions thoughts he only thought about the quadratic formula
I mean he operated this formula multiple times already 😂
Nobody talks about this because it's absolutely obvious.
❤
Bruh
Finally, a worthy opponent.