I finally took the limit of the quadratic formula

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I finally took the limit of the quadratic formula

Рет қаралды 70,836

2 ай бұрын

Math for fun! Taking the limit of the quadratic formula as a goes to infinity. Enjoy!
Related videos:
I differentiated the quadratic formula: • I differentiated the q...
I integrated the quadratic formula: • I integrated the quadr...
Solve 0x^2+bx+c=0 by ‪@MichaelPennMath‬ • when a quadratic equat...
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Пікірлер: 158

@blackpenredpen 2 ай бұрын

Integral of ln(x) from 1 to "what" =2? kzfaq.info/get/bejne/fbaWn5ucza6riZs.html

@Charity4Chokora 2 ай бұрын

Stay amazing and thank you. Sorry you have to deal with horrible people saying nasty things.

@Charity4Chokora 2 ай бұрын

@@Math-Fact I would but my arms are to short to reach the pencil

@Charity4Chokora 2 ай бұрын

@@Math-Fact 😁 They probably have longer arms outside of tyrannosaurid family. I totally understand why you need to ask for help with the math.

@Maths_Wonderland 2 ай бұрын

*_CHALLENGE_* Calculate the integral of *log₂₇ 9 + log₈ 4* from *−(log₂ √6 + log₂ √⅔)* to *log₅ 50 − log₅ 2* . Please upload this sum’s solution video. I have specially made it for you to solve.

@notfound6969 2 ай бұрын

@@Maths_Wonderland answer is 8 eazy bro

@shimrrashai-rc8fq 2 ай бұрын

This makes geometric sense. As a -> oo, the parabola "tightens up" onto the y-axis. An infinitely "tight" parabola, that also crosses the x-axis (hence c < 0), is like a ray, and when at the y-axis, that means it intersects the x-axis at x = 0.

@programmerpctheory1413 2 ай бұрын

5:33 He's either talking about the dirac delta function or sin 1/x which oscillates infinitely

@jaysonbunnell8097 2 ай бұрын

I also was expecting the dirac delta (or one of its cousins, unit doublet/triplet/etc)

@ppantnt 2 ай бұрын

Now, take limit of a cubic formula.

@frimi8593 2 ай бұрын

I suspect you’d get the exact same result, given the geometric demonstration at the end. This should apply to all n-polynomials taking the limit at the x^n coefficient goes to infinity

@heinrich.hitzinger 2 ай бұрын

@@frimi8593There are no formulae for quintic and higher polynomials. 😔

@frimi8593 2 ай бұрын

@@heinrich.hitzinger you don’t need such formulae. The leading coefficient always determines the steepness of the curve and thus as long as the y intercept is below 0, the x intercepts to the polynomials will always approach (0,0)

@Patrik6920 Ай бұрын

@@frimi8593 ..actually no... consider ax³+bx²+cx+d ... as a and b goes towards 0 or infinity, x intercept x goes towards d, as well as some extremes for 'ax³+bx²+cx+d' its at d/2 as ab goes to infinity, u actually get a very odd triangle(infinite high) connected to a line, actually a very long smoth parabola with so low curvature we can consider it a triangle, sloope goes toward some constant depending on abc -> it approaches -(∞/2)x + d edit: as löong as c in the ' cx ' term does not apprach infinity, in that case x intercept → =0

@frimi8593 Ай бұрын

@@Patrik6920 so I mean I haven’t formally proven it, but I’m quite confident that as a goes towards infinity, all x intercepts approach 0, provided the y intercept is below 0. Consider, as all the other coefficients remain constant, they quickly become irrelevant and we end up looking for the solutions of lim a->inf ax^n-d=0 where n is the degree of the polynomial. With all the other terms removed it becomes much more obvious that x is approaching 0, no?

@ItsPungpond98 2 ай бұрын

If he could marry the quadratic formula, he'd definitely do so lmfao

@hqTheToaster 2 ай бұрын

Nah, he'd probably marry taking the limit of every possible smoothstep.

@iceIceCold666 2 ай бұрын

He'd marry the Lambert w function

@Charity4Chokora 2 ай бұрын

Grow up, childish and hateful.

@mayankthakur5341 2 ай бұрын

no he would marry the Alpha fish

@lawrencelawsen6824 2 ай бұрын

I saw how he was looking at that equation, it was love eyes lmfao

@magnusmalmborn8665 Ай бұрын

That graph looks a bit like an inverted Dirac-delta, which is a sufficiently cursed concept that you should do a video!

@uwuowo7775 2 ай бұрын

Is there an application for using geometric formulas in calculus?

@soumakirimoto2195 2 ай бұрын

look into analytic geometry

@uwuowo7775 2 ай бұрын

@@soumakirimoto2195 something new to look into! thanks

@daniel_77. 2 ай бұрын

@@soumakirimoto2195 Yes. Analytic geometry is the main bridge to calculus, by combining the cartesian plane, functions and geometry.

@kirtijitchatterjee3564 2 ай бұрын

Aod also provides some numericals based on co ordinate geometry and derivatives

@EigenMaster 2 ай бұрын

Also in Related Rates and Optimization problems

@thabomaleke874 2 ай бұрын

Only Steve can integrate, differentiate, and take limit(s) of the quadratic and not be diagnosed with insanity.

@NarynbekGilman 2 ай бұрын

Is his name really Steve?

@ianfowler9340 2 ай бұрын

Check out Dr. Barker's video on the alternate quadratic formula: x = [ 2c ] / [ -b plus/minus sqrt(b^2 - 4ac) ] As a ----> inf with c0, one from the right and one from the right. kzfaq.info/get/bejne/obWih7aIlcWXaWg.html

@ishaan_lightning6629 2 ай бұрын

We can do this by another approach, so (b^2-4ac)^1/2 we can apply binomial theorem and we will see that we will get the highest power of a as 1/2 now in the denominator we have a^1 so when we will divide by a^(1/2) in numerator and denominator as well, we will get constant in numerator and in denominator we will have (a) ^1/2 so there fore we have constant/infinity thus it tends to zero

@ElderEagle42 2 ай бұрын

Now you only need to take the infinite series of the quadratic formula to complete the saga

@adw1z 2 ай бұрын

u should now do lim a->0 and show that the quadratic root (sometimes!) tends to the linear root in different cases of (a,b,c)

@Sir_Isaac_Newton_ 2 ай бұрын

Did you not watch the video? He said Michael Penn published a video doing exactly that, so he himself won't.

@adw1z 2 ай бұрын

@@Sir_Isaac_Newton_i skipped the first minute until when he actually started

@KarlDeux 2 ай бұрын

And it can't work because there is then a division by zero which makes the formula useless for a=0.

@jelmerterburg3588 2 ай бұрын

"The bottom is going to dominate the top." Well there's a twist :P

@the_nuwarrior 2 ай бұрын

Now try an asymptotic expantion aproach

@TazPessle 2 ай бұрын

@5:50. That graph becomes so hard to draw without losing meaning without context, could you just draw the y=x^2 graph, but have the first mark on the x axis at 1/[infinity] and placed wide.

@paramrituraj1293 2 ай бұрын

@blackpenredpen Hello sir, can you explain how to solve y'' + ky = 0 w/o trial and error? Some solutions i found online use laplace transform but i dont really get it! An explanation would really help of the solution and if required, what laplace transform is! Thank you!

@naitikshukla2020 2 ай бұрын

Bro a request to make videos on functional equations you haven't made videos on it, also i only better understand maths from you only ❤

@KUDIYARASAN- 2 ай бұрын

Excellent Sir

@CrimS0n. 2 ай бұрын

Day 10 of asking if you can do the intergal of 1/(x^i) with π and e as its limits.

@henridelagardere264 2 ай бұрын

"Stay positive, don't be so negative all the time!" While certainly not bad advice, c has to flout it, at least this time. Because today, the motto is, "Keep things real!"

@savitatawade2403 2 ай бұрын

bro what is this

@utkarshjain861 2 ай бұрын

😂😂😂 Upar se nikal gaya kya 😂😂

@tilkkone4257 2 ай бұрын

Stand proud, you are strong

@savitatawade2403 2 ай бұрын

@@utkarshjain861 😭😭😭

@savitatawade2403 2 ай бұрын

@@tilkkone4257 lol💀💀

@utkarshjain861 2 ай бұрын

@@savitatawade2403 😂😂😂

@trueriver1950 2 ай бұрын

Same result if lim a -> -inf by inspection if c also negative. Considering the cases where there's a negative sign under the root, the roots are complex for finite a: but as |a| --> inf the roots tend to 0.i which is the same limit but approached from another direction in the complex plane.

@tonypapas9854 2 ай бұрын

1:#3 "When keeping it real goes wrong". I wonder if anyone will get this reference?

@NullExceptionch 2 ай бұрын

Can you try finding the exact solution of this? “(e^x)+(3x^3)-(1/x)=0”, Thanks so much for your time at looking this.

@lumina_ 2 ай бұрын

you've integrated, differentiated, and now taken the limit of the quadratic formula!!

@armanavagyan1876 2 ай бұрын

Outstanding 👍

@nekothecat 2 ай бұрын

Can you try letting a approaches to zero? I found it difficult to do so and result was quite strange when I enter in on wolfram alpha, it was not just bx+c=0 Later I found that the first step of getting the quadratic formula is diving a for all terms, that's why the result is strange

@carultch 2 ай бұрын

You can rationalize the numerator, to derive this variant on the quadratic formula: x = 2*c/[-b +/- sqrt(b^2 - 4*a*c)] Now you can let a = 0 directly: x = 2*c/[-b +/- sqrt(b^2)] x = 2*c/(-b +/- b) Of the two solutions, only -b-b is legitimate. The other solution would have us dividing by zero. This gives us: x = -c/b And this is consistent with the solution to the linear equation, b*x + c = 0.

@trueriver1950 2 ай бұрын

See the Michael Penn video on that - see link in the description. He has a very different style than BPRP but I find the maths just as interesting from both channels. I'm really glad that they've started to reference each others work 😊

@gamemakingkirb667 2 ай бұрын

The graph you sketched should have a vertex that lies on x = 0!! I was going crazy for a bit after looking at your graph and imagining the change in A lol

@tgx3529 2 ай бұрын

nx^2+bx+c=f_n(x), there Is not uniform convergention on by Moore Osgood theorem. There Is the problem in zero with changes limits.

@SumanYadav-wr3cn 2 ай бұрын

Sir please provide lectures on sieve theory

@mekaindo 2 ай бұрын

b: i'm only constant after all, i'm only constant after alo don't put the blame on me

@mathmachine4266 2 ай бұрын

-c/b and -b/a

@muhammadhussainsarhandi9928 2 ай бұрын

Great video! Now please take the limit of the Thin Lens Formula. 1/f = 1/p + 1/q where f = Focal Length of the lens, p = Object distance from the lens, and q = Image distance from the lens, take the limit of this formula as f approaches 0,

@cffex3858 2 ай бұрын

As f approaches zero, either p or q approaches zero or both of them approach zero.

@krishanunath2599 2 ай бұрын

Can this result be obtained from root locus

@tobybartels8426 2 ай бұрын

Kind of a casual diss of a→0, considering that there are _two_ expressions in the quadratic formula and only _one_ of them approaches the solution to bx+c=0. (But I suppose that the video by Michael Penn covers that.)

@CharlesStancampiano 29 күн бұрын

Why if a approaches zero does the quadratic formula not give the solution x= -c/b?

@jamiewalker329 2 ай бұрын

The f(x) = ax^2 + bx + c can be thought of as a family of functions parametrised by a. There is no uniform limit of this family of functions. There is a point-wise limit existing at x = 0, which is just a singular point (0,c), but that's all I'm afraid, a point wise limit doesn't exist at any other point. Really you need to explore the surface g(x, a) = ax^2 + bx + c where b and c are fixed. If you consider g(1/sqrt(a), a) = 1 + b/sqrt(a) + c which tends to a finite value as a tends to infinity...

@dominicellis1867 Ай бұрын

So the limit as a approaches infinity of a parabola is a delta function?

@BB13580 2 ай бұрын

We love content like this.

@bokser733 2 ай бұрын

Hi blackpenredpen, since you do a lot of calculus, what do you think of John Gabriel's New Calculus? Is it rubbish or is there some truth in it?

@francorota8638 Ай бұрын

I think Gabriel is just trying to make an easier version of calculus. He doesn't hate derivatives and integrals, he only hates the idea of limits. And to his defense, a lot of people struggle defining a limit during their college years.

@TheBlueboyRuhan 2 ай бұрын

Hi blackpenredpen! Could you do a video on non-linear regression? Talking about using the jacobian and hessian will be cool!

@cdkw2 2 ай бұрын

I have only recently been watching Michel Penn and now you reference him lol. Nice coincidence.

@victorjiang5827 2 ай бұрын

The thumbnail is fire 🔥🔥🔥😂

@faekalkeinennachname.werwa9703 2 ай бұрын

B = Inf: (sqrt(a) times x times sqrt(c)) times 2 = inf times x; but x irrelevantly smaler than a and or c ? What Happens if you put in inf for b? Is this a stupid question?

@Nikioko 2 ай бұрын

And now the limes of the pq formula.

@ruben2019 2 ай бұрын

Can you do the Spanish math college entrance exam of 2024, it’s rising a lot of controversy because of the difficulty?

@Dau1928 2 ай бұрын

if we put infinity in the formula thet directly gets zero. is that wrong? (im in class 10 but have knowledge of some things of class 11-12)

@Defaulter_4 2 ай бұрын

Well well, you are taking a limit of a given function so you are supposed to deal with limit and then put the value. You'll learn in class 11 limits and derivatives, hope this helps

@deepanjansen2147 2 ай бұрын

hey i have a challenge for u : find the range of the function , f(x)=(3x-x^3)/(1-3x^2)

@samueldeandrade8535 2 ай бұрын

All real numbers.

@downrightcyw 2 ай бұрын

If we take lim a➛0, quadratic equation will become linear and it holds, i.e. x=-c/b, very funny.

@Hailfire08 2 ай бұрын

Bprp always keeping things real 😤✊

@singhavinash76 2 ай бұрын

Is it even possible to find a general solution for x for the equation a^(x ln(x))+bx+c=0, Wolfram Alpha doesn't really give an answer.

@acid8000 2 ай бұрын

Goat

@siddharthahluwalia6595 2 ай бұрын

Cant you use l'hopital's rule?

@dd_hd2 2 ай бұрын

Please make a video on Fourier transform.

@lawrencelawsen6824 2 ай бұрын

This guy is so good

@Asma.Park. 2 ай бұрын

جمييل جداا❤

@romanbykov5922 2 ай бұрын

what about videos on elliptic curves?

@jorgeverde4674 2 ай бұрын

Interesting that the limit is 0 for a formula that is designed to give the zeroes of a quadratic function.

@notafraidofchange Ай бұрын

1:15 why not use complex numbers? Surely inf*i isn’t too bad to work with. 😁

@Brid727 2 ай бұрын

now raise the quadratic formula to the quadratic formula power

@user-tj2fr3zv4j 2 ай бұрын

plz also take some good questios from ventor and 3d geometry plz

@archangecamilien1879 2 ай бұрын

Should be 0, I think, and perhaps makes sense, since if the first coefficient, the square's coefficient is made larger and larger and the others are kept constant, perhaps the only thing that could make the equation 0 would be 0, lol...I mean, should be zero because the power of the numerator is 1/2, and the power of the denominator is 1...

@user-fp4vk9wj8g 2 ай бұрын

What if you make “a” approaching a complex or a irrational number 🤔

@oscarfoley511 2 ай бұрын

Tou dont even need a limit, just ser a to the number and it will work!

@franckdebruijn3530 2 ай бұрын

Can we be sure that for x=0, y = c, when a goes infinite?

@CptGallant 2 ай бұрын

We know that for a general parabola, the vertex is at x= -b/2a. Plug that back into the general function, and after simplifying we find the y value of the vertex is always at -(b^2)/4a+c Take the limit as a goes to infinity, and we get y=c

@leonardobarrera2816 2 ай бұрын

Manx where do we use 0^+ ????

@paultijink4235 2 ай бұрын

Dirac delta function?!

@SeekingTheLoveThatGodMeans7648 2 ай бұрын

becomes ax^2 = 0 or x^2=0 i.e. x=0

@nuclearrambo3167 2 ай бұрын

what are you after

@rudyj8948 2 ай бұрын

Wake up babe new delta function just dropped 🗣️🗣️

@Samir-zb3xk 2 ай бұрын

Bro invented dirac delta function part 2

@da3l_525 2 ай бұрын

where can i buy this T-shirt

@da3l_525 2 ай бұрын

nvm

@mortada865 2 ай бұрын

Now take the limit of the quartic formula

@ANIKETPODDER-yz8wy 2 ай бұрын

Bro once come to INDIA we study all these in class 11 for the world's 2nd most toughest exam 😂

@Algebraictivist 2 ай бұрын

Cool

@eagleraj3126 2 ай бұрын

Love from India

@ChessBros-ic5tz 2 ай бұрын

What about if you take the limit of the phytagoras

@m3nny125 2 ай бұрын

as in sqrt(a^2+b^2)? if you take it as a (or b) approaches 0, itd just be 0, and if you take it as a (or b) approaches infinity itd just be infinity And if you take a multivariable limit where both of them approach zero, itd still just be 0, and if you take a multivariable limit where both of them approach infinity, itd be infinity again so he probably wouldnt cover it because no matter how you do it its very boring and straightfoward

@ChessBros-ic5tz 2 ай бұрын

I mean a^2+b^2-c^2 like when you take the c^2 to the other side

@ChessBros-ic5tz 2 ай бұрын

Also what happens if you equate phytagoras in the form(a^2+b^2-c^2) to = the quadratic formula

@m3nny125 2 ай бұрын

@@ChessBros-ic5tz In that case youd be taking the same limit of 2 equations with different values which simply isnt possible the only way you can do it is like the one i showed

@Karcrab_mcshouty 2 ай бұрын

I've come up with a cool geometry problem, I don't know if you would be interested

@carultch 2 ай бұрын

Of course he's interested.

@user-zi7cj6lp2q 2 ай бұрын

Isn’t sqrt(4a^2)=|2a| ?

@supratikchandra3387 2 ай бұрын

a---> ∞ so no need of the modulus

@sandeepkanekal 2 ай бұрын

Yes, you are right. But it has been already assumed that x is tending to POSITIVE infinity. So, the modulus opens with a positive sign, thus, |2a| = 2a

@adamkucera9094 2 ай бұрын

Singular perturbation. Use method of strained variables.

@ronaldking1054 2 ай бұрын

This should be the same as the limit as p -> 0 and q -> 0-. Your manipulation just turned it into the p, q formula.

@danandrews299 2 ай бұрын

hey

@guy_with_infinite_power 2 ай бұрын

@bunthornpgoneshop6603 2 ай бұрын

Can you prove that why (-)(+)=- (-)(-)=+

@Grassmpl 2 ай бұрын

Limit DNE. The domain of the quadratic formula in a is bounded above.

@kerkuesidijes1006 2 ай бұрын

Distance between line and point: kzfaq.info/get/bejne/pNx8o91kqtGUd5s.htmlsi=50J8Vb_TRSCXUXRG

@marshallmanz123 2 ай бұрын

Divide numerator and denominator by a. Numerator goes to zero and denominator is two. The limit is zero

@jpcurley25 2 ай бұрын

Big fan but a boring conclusion considering what a high school algebra student should know about what a does to a quadratic

@amangine 2 ай бұрын

kzfaq.info/get/bejne/m7SigJWjq7PRgaM.html : "And keep things real" - always with the gangsta talk

@Otakumaaa 2 ай бұрын

First 😂

@adiaphoros6842 2 ай бұрын

Ew... 🤢

@craigncube7057 2 ай бұрын

nerd

@FredMaHead 2 ай бұрын

what a waste of a video

@nightytime 2 ай бұрын

I mean, a lot of his videos aren't necessarily "useful." That doesn't mean it isn't interesting or worth watching, at least, it doesn't given the right target audience.

@forcelifeforce 2 ай бұрын

1) Write a sentence. 2) It is a terrible thing to waste your terrible mind for not appreciating the video.

@soroushdarvish9281 2 ай бұрын

In all of this man’s videos I get from his energy that he is flexing math skills rather than actually teaching to the audience. Although the content is still great. I just wish he wouldn’t be an arrogant ass