A minimum problem that is too much for calculus to handle. 

I don't think you have to make the problem harder first. Otherwise a very nice solution!

Beautiful construction

What thebheck..hiw does this make any sense..why in God's name would you introduce a second variable y and why would youbdo thatm.i dont see how that is necessary or logical sorry? Why is math so needlessly fucking stupid and convoluted sometimes??

@@leif1075 sometimes you do things because you can, and see what happens when you do.

@@Daniel-ef6gg I see..thanks for answering..can you share what made you think of this though I'm curious?

@@sunnyarora7177 That's not true at all, because the derivative of any sum is equal to the sum of the derivatives of each term.

​@sunnyarora7177 no. Completely false. The derivative is a linear operator

With this method you proved that there is an extremum at x=1, but you did not prove that it is a minimum, nor did you prove that there was no other deeper minimum.

@thierrypauwels true, but whatever. I'm not trying to do proofs lol. Trying to have some fun

You've shown the function has a stationary point at x=1, but why is this where the global minimum is achieved?

Update: a) it turns out it only works this simple if 2) & 3) are swapped or else the coefficients of the power series don't line up properly with the exponents of x powers & don't give simple analytical results. b) also need to check that the difference of infinite power series converges for the rewriting of the finite degree polynomial to be valid.

I tried pursuing this approach to then get the minimum but you end up with a somewhat complex polynomial that you need to find the minimum of so I'm not sure it helps that much.

You could factor an x^2023 out first, that gets you something of the form d/dx(geometric sum of 1/x)

In order to make it look like the 'derivative' of a geometric series, the easiest way is to introduce an extra variable and represent it as a derivative with respect to that extra variable, as in the solution I posted

I also removed the x^2022 term as well. Mind you, it got gnarly enough that I stopped.

Why should that work, given that a geometric series doesn't converge outside (-1,1)?

@@Alex_Deam it's a finite geometric series

​@@abblabaabblaba823Well now I feel dumb lmao

@@Alex_Deam i thought of this same solution and abandoned it thinking "well, this doesnt converge tho" haha

​​@@Alex_Deamhy would you think to divide by x at all though?? Still seems random to me..why nko take the derivative and set equal to zero to find the minimum? Or something else?

where algebraic nightmare is equal to $(algebraic nightmare) $(algebraic nightmare)

Great point. It shouldn't be too repetitious of a calcultion. Michael really likes induction, though, which is fine

Can you lay out a sketch to do so? I thought the same but it was really hard to factorise the resulting sum after subtracting n+1

@thesecondderivative8967 1. write (x-1)^2 as x^2-2x+1 2. write the sum it's multipliying in sigma notation 3. Multiply the two by distributing, i.e. (x-1)^2*sum=x^2*sum-2x*sum+sum 4. Reindex the sums so that they have the same powers of x 5. combine like terms 6. Add in the terms that are outside that product 7. Check that you get the original function

@@Keithfert490 Thanks. It's been bugging me since

You almost wrote why, the x^2024 term is negligible for x->0 and thus f_2N/2N = (x+1+o(x^2023))/(x+1)^2 ~= 1/(x+1) = 1+x+o(x)

What area of mathematics is this? I'm still a student but I haven't heard of most of this stuff. Is there something I can look up to learn this? it seems super interesting.

@@ichigo_nyanko Mathologer did a video using finite difference operators

If G is the antiderivative of the continuous extension of g at which G(0)=−1, then G(x)=Σ(−(−x)^k,k,0,2023), and for x≠−1, G(x)=(x^2024−1)/(x+1), while G(−1)=−2024; additionally, G(x)=(x−1)Σ(x^(2k),k,0,1011). I'm not so sure about how this helps analyze f, because g′(x) looks a bit complicated: Σ(−(k+2)(k+1)(−x)^k,k,0,2021). In terms of f, g′(x)=2022f(1/x)x^2021−f′(1/x)x^2020=2022g(x)/x−f′(1/x)x^2020. Substituting in, to eliminate explicit mention of g, Σ(−(k+2)(k+1)(−x)^k,k,0,2021)=2022Σ(−(k+1)(−x)^k,k,0,2021)+2022/x−f′(1/x)x^2020, from which f′(1/x)x^2020=Σ((k−2020)(k+1)(−x)^k,k,0,2021)+2022/x, from which f′(1/x)=Σ((k−2020)(k+1)(−x)^(k−2020),k,0,2021)+2022x−^2021. With the substitution k=2021−j, j now ranges from 0 to 2021, and f′(1/x)=Σ((1−j)(2022−j)(−x)^(1−j),j,0,2021)+2022x−^2021, from which f′(x)=Σ((j−1)(j−2022)(--x)^(j−1),j,0,2021)+2022x^2021. --- I think I did something wrong there; a more direct approach, starting with f(x)=Σ((2023−k)(−x)^k,k,0,2022), results in f′(x)=Σ(−(2023−k)k*(−x)^(k−1),k,0,2022), which after noticing the k=0 term is zero and re-indexing with j=k−1, becomes f′(x)=Σ((j+1)(j−2022)(−x)^j,j,0,2021). You could get lucky while testing out the possible rational zeros and noticing that f′(−x) has no sign-changes and f′(x) has 2021 sign-changes, which means all real zeroes are positive and there is an odd number of them; then trying out 1 and going in pairs, f′(1)=Σ((2k+1)(2k−2022)−(2k+2)(2k−2021),k,0,1010) =Σ(−4042k−2022+4038k+4042,k,0,1010) =Σ(2020−4k,k,0,1010) =2*1010*1011−4Σ(k,k,0,1010) =2*1010*1011−4*½*1010*1011=0, so 1 is a critical point for f. To finish it off, f″(x)=Σ((k+2)(k+1)(2021−k)(−x)^k,k,0,2020), and again taking the terms in pairs, f″(1)=Σ((2j+2)(2j+1)(2021−2j)−(2j+3)(2j+2)(2020−2j),j,0,1009)+2022*2021 =Σ((2j+2)((2j+3)(2j−2020)−(2j+1)(2j−2021)),j,0,1009)+2022*2021 =Σ((2j+2)(−4034j−6060+4040j+2021),j,0,1009)+2022*2021 =Σ((2j+2)(6j−4039),j,0,1009)+2022*2021 =Σ(12j^2−8066j−8078,j,0,1009)+2022*2021 =12Σ(j^2,j,0,1009)−8066Σ(j,j,0,1009)−8078*1010+2022*2021 =12Σ(j^2,j,0,1009)−4033*1009*1010−8078*1010+2022*2021 =2*1009*1010*2019−4033*1009*1010−8078*1010+2022*2021 =5*1009*1010−8078*1010+2022*2021 =−3033*1010+2022*2021 =−1011*3030+1011*4042 =1011*1012>0, so 1 is a local minimum for f. (yeah, more insightful than just punching all that into a calculator)

It is similiar to indonesian mo 2009 problem 6

That can work. If you claim f_{2n}(x) = (x - 1)^2 P_n(x) + n + 1 with P_n(x) always nonegative and calculate f_{2n+2})(x) by the identity f_{2n+2}(x) = x^2 f_{2n}(x) - (2n+2)x+2n+3, you can simplify to f_{2n+2}(x) = (x-1)^2 [x^2P_n(x)+n+1] +n+2. Since P_n(x) is a always nonnegative polynomial, P_(n+1)=x^2P_n(x)+n+1 is also an always nonnegative polynomial. Since the base case of f_2(x) = (x-1)^2 + 2 where P_1(x) = 1 clearly follows your claim, then by induction all f_(2n)(x) can be written in that form without even worrying about what the form of P_n is.

It is Michael. He ate the words. He meant x-1= zero

He meant to say is zero at one

How are you so fast every time?

It's a bot reacting to certain words or phrases in the video.

​@@kpaasialits not a bot lol

Even a cursory search on google would reveal to you that such bots exist and are used actively by people for this very purpose.

@@kpaasialbut the user has commented before that he does it manually. And often adds personal comments beyond just the stop time.

Might have to get that tee shirt myself!

Do you mean the gentle intro that was used until a few months ago? I liked, it, too. But no intro is better than a jarring one.

@@artsmith1347 i would say a few weeks ago.. I did not t it was jarring 😂

Since f(x) power is 2022 (even), then we take care only for smaller even forms (2, 4, 6, ... 2n)

Math has absolutely nothing to do physical or natural significance, although it can. You should apologize first

You shouldd get lost

Apologise? Lmao

I feel like this problem and the result/possible extended structure inferable from it really stimmed my brain. Based math W

Welcome to discrete mathematics, where all the problems are abstracted away into arbitrary meaninglessness. I think the point is not the problem itself, but the process. I think a lot of mathematics seems dull and pointless until you reach a point where you realize you actually need it to solve a problem. I believe that is how many branches of mathematics start out; they start from abstract nonsense, then, one day, you figure out that it's the most important thing ever to solving some problem you face and then you thank the math gods that some guy 100+ years ago was psychotic enough to devote half their life to do all the, potentially useless, mathematical groundwork for you, and they probably did it for pure interest, or, for the sake of knowledge itself. Beautiful.