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Zvi is a Hebrew Israeli name pronounced "tze vee"

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

Didn’t know we could solve it that way. You used Tonelli théorème but you didn’t say the name. Great work tho

Still no reply Ignore V good 👍🏼

Bro went from high school to college in 9hrs

Wait, but if a^2 = 1/4 then there should be 4 answers, no? Two real answer and two complex answers

It's really simple!

Can this result be obtained from root locus

Thanks l am from Ethiopia 🇪🇹

Looking forward to solve the integral of " e^(x³) dx "

3:45 shouldn’t it be x^2 - x + 1?

I presume quadratic formula lawyers still did not appear just to sue you

5:32:53 the fact that he still cared about his handwriting at that point...

Sir please make videos on sieve theory

Why cant you just make e^x e^e^x to make it solvable

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. DISCLAIMER : This is a repost of my comment on the previous video posted, but I am also putting this here because I believe it would be more likely that you read it here.

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.

Keralites were shocked at the intro sponsor, until we saw the logo😂

Delta = Epsilon/2 is also correct right?

I want to know why do you keep holding the Pokémon ball ?

I have a question Can all irrational numbers be found for their rational approximation values? Example e use Taylor series you can find rational approximation values of e Are number super irational can be exist ?

The 3rd blackboard was the biggest plot twist imo

For the 186 case, notice that after subtracting one the RHS divides the prime, so 185=5*37 must divide it. Neither 5 nor 37 works.

For the 186 case, notice that after subtracting one the RHS divides the prime, so it must be a divisor of 185=5*37. Neither of these work.

is 0.9 the height for the triangles? cause the domain never goes to 1

A. It is an indeterminate form B. Square root of a number is always +ve C. ≠1 since it is tending to 1- D. The answer is 16 as if it was (3×4) then it would be 1 but (4) stands for (1×4) which indicates 16 as the answer So I would say there is no scope for any debate in any of these

Imagine being asked the Primorial of a Googol and you don't know the prime numbers.

Just put x= omega(w) and boom.

Very neat and direct. For those interested in a simple way, using polar coords, ( and simple to understand) I suggest you refer to Prime Newtons' video- it is nice for beginners.😊

x^4-(x^2+2x+1) =x^4-(x+1)^2 =(x^2-(x+1))(x^2+(x+1)) =(x^2-x-1)(x^2+x+1) Perfect square trinomial and difference of squares. I wonder which quartic equations are solvable via your method.

before watching: convert to difference of two squares (x^2)^2 - (x+1)^2 then two quadratic factors [x^2 - x - 1][x^2 + x + 1] = 0 . Real roots are phi, Complex roots (-1+- iRoot3)/2

when you think of exponentiation as dimensions the 0th dimension either has something or nothing

太酷啦

Let me make something simple for you. People who are asking why is lim of Cos h is 1 and lim of Sine h is 1 ??? It' because when we have the h approaching 0(h is same as theta btw). At that time hypotenuse becomes closer and closer and closer to adjacent side. So Cosine h would be 1. And Sine h would be 1 also.

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.

Can't we just substitute x^2 as t and then solve ?

Yeah 👍

UTTPs with childs be like

One way I like to derive the quartic equation is to split the reduced quartic into a difference of squares of the form (x^2+ax+b)^2 = c(x+d)^2. In order to solve for the coefficients in this equation, you need to solve a cubic 'determinant' equation. Of course in this case, the coefficients are trivial.

hyperreal numbers are a joke to you?

that first equation was so cool

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

in the first case we can divide by 6 if we also multiply the 4 in mod by 2

Ok, but what does it mean when it adds 2pi, its a different value after all

When solving for the Z values in the Lagrange resolvent shouldn’t the 2 at the bottom be a -2? Since a=-1 or am I just being stupid

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.

You can do dots to immediately factor it. You have x²+2x+1 - x⁴ → (x+1)² - x⁴ → ( x²+x+1)(-x²+x+1)

Where did you buy the euler on the wall?

Hmm...I don't know what that would mean, lol, to abuse the quadratic formula, but maybe one can rewrite it as x⁴- (x²+2x+1)=0 => x⁴- (x+1)²=0, so, the difference of two squares, that would be (x²-x-1)(x²+x+1)=0, etc...just set each factor equal to 0, regular quadratics, etc...hopefully I didn't make some silly mistake...