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

[@iabervon]

Polynomial division by x²-2x+1 is pretty neat in this case.

[@Daniel-ef6gg]

Don't rule out calculus so quickly. But if you want to use calculus, you need to make things harder before they get easier. Let f(x,y) = x^2022 - 2 x^2021 y + 3 x^2020 y^2 - ... + 2023 y^2022. Let g(x,y) = - x^2023 + x^2022 y - x^2021 y^2 + x^2020 y^3 - ... + y^2023 Partial d g(x,y)/dy = f(x,y), and f(x,1) = f(x) Note that g(x,y) * (x+y) = y^2024 - x^2024 In order to find the minimum, we need to find where d(f(x,1))/dx = 0. But we can find f(x,y) in a simpler form using the quotient rule, f(x,y) = [2024 y^2023 * (x+y) -y^2024 + x^2024]/(x+y)^2 Now set y =1 and differentiate with another quotient rule to get f(x) = (x^2024 + 2024x + 2023)/(x^2 + 2x +1) and df(x)/dx = (x+1)^(-3) * [2022 x^2024 + 2024 x^2023 - 2024 x - 2022] Note: f(-a) > f(a) for all a>0. Thus, we only need to find solutions where x>=0. Also, note that the numerator polynomial only has a single sign change: thus, there is only a single solution with x>0. Also note that f(1/x) is a solution only if f(x) is a solution. Thus, the only solution is when x = 1. Also for completeness, we should check x=0, but it's easy to check that f(0) = 2023. f(1) = (1^2024 +2024*1 + 2023)/(1^2+2*1+1) = (2024*2)/(2*2) = 1012. Thus the minimum is 1012.

[@kareolaussen819]

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

[@OuroborosVengeance]

Beautiful construction

[@leif1075]

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??

[@Daniel-ef6gg]

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

[@leif1075]

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

[@Keithfert490]

If you write f(x) as sum_{k=0}^N (-1)^(k+1)kx^(N-k), then take its derivative with the power rule, you can show that f'(1) is (-1)^N times itself by reindexing the sum by taking k -> N-k. Therefore, when N is odd, f'(1)=0. Then, just find f(1)=(N+1)/2

[@megauser8512]

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

[@Keithfert490]

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

[@thierrypauwels]

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.

[@Keithfert490]

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

[@ChronoQuote]

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

[@FrankHarwald]

My first thought for a solution I came up with: 1) remember (1+x)^(-1) = 1 - x + x^2 - x^3 + x^4 - x^5 +-... 2) figure out d/dx -(1+x)^(-1) = 1 - 2x + 3x^2 - 4x^3 + 5x^4 -+... = (1+x)^(-2) 3) try to rewrite this polynomial of finite degree as the difference of two infinite power series starting at different indices scaled by appropriate amount of powers of x 4) that difference of two infinite power series is equal to the difference of two rational functions 5) I'm pretty sure simple calculus is powerful enough to handle the minimum of the difference of two fractional functions

[@FrankHarwald]

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.

[@OllieClarke8787]

Cool method. I found it natural to multiply f(x) by (x+1)^2 to get rid of most of the terms from the sum. This idea isn't just a little trick - for those interested look up generating functions. You get (x+1)^2*f(x) = x^2024 + 2024x + 2023 and differentiating looks kinda bad at first but still doable.

[@stefanotorelli3688]

Good presentation of proof "by induction", very good example!

[@GaneshKumar-hy8pp]

Just divide f(x) with x and add that to f(x) you will get a geometric progression with common ratio -x . Which is easily solvable

[@Alex_Deam]

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

[@abblabaabblaba823]

@@Alex_Deam it's a finite geometric series

[@Alex_Deam]

​@@abblabaabblaba823Well now I feel dumb lmao

[@ametamaa]

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

[@leif1075]

​​@@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?

[@kareolaussen819]

By differentiating the finite sum S_n(u) = 1 + u + ... u^(2n+1) = [1-u^(2n+2)]/(1-u), setting u=-1/x, and multiplying with x^(2n), one finds that the series in question equals g_n(x) = x^(2n) S'_n(-1/x) = [x^(2n+2) + (2n+2)x + (2n+1)]/(x+1)^2 = [x^(2n+1) -1](x+1)^2 + (2n+2)/(x+1), and g'_n(x) = {2n[x^(2n+2) -1] + (2n+2)x[x^(2n)-1]}/(x+1)^3. For positive x one can easily see that the numerator of g'_n is monotoneously growing with x, and is zero at x=1. Hence, in this interval g_n has a minimum at x=1, with g_n(1) = n+1. For negative x we can see directly from its definition that g_n >= (2n+1) and is growing with |x|, so the minimum at x=1 is unique. Hence, the problem is quite natural and straightforward to handle by calulus; although it requires some fiddling to write g'_n in a convenient form. Added: I note that @Daniel-eff6gg has solved the problem by essentially the same method.

[@ANTONIOMARTINEZ-zz4sp]

I love this kind of content. Great video Prof. Penn. Thank you for uploading.

[@mathunt1130]

Straight off it looks like the derivative of a geometric series...

[@TheEternalVortex42]

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.

[@minamagdy4126]

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

[@Daniel-ef6gg]

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

[@minamagdy4126]

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

[@Nikolas_Davis]

Why apply induction here? Why not just expand the factored form of f_2n(x) and verify it's equal to the definition?

[@Keithfert490]

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

[@thesecondderivative8967]

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

[@Keithfert490]

@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

[@thesecondderivative8967]

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

[@cxpKSip]

Factorization would be a pain, too... $2023=x^2022-$(algebraic nightmare).

[@chri-k]

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

[@VaradMahashabde]

Multiplication by x^2-2x+1 can be visualized as taking the symmetric discrete second difference operator on the sequence 1,0,2,0,3,0,... which gave us the original sequence of 1,-2,3,-4,.... . The similarity can be seen using the convolution perspective on polynomial multiplication. Incidentally, since the orignal polynomial has almost linear coefficients, we can apply a variation of the second difference operator by multiplying by (x+1)^2 = x^2+2x+1 which cancels everything down to (x+1)^2 f_2022(x) = x^2024 + 2024x + 2023. Calculus should be easier at this point. I guess one could also get a smaller form like this using the Arithmetico-Geometric series formula. Also, f_2N / 2N really approximates x+1 in the region -1 to 1 and I dont know why

[@bonzinip]

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)

[@ichigo_nyanko]

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.

[@VaradMahashabde]

@@ichigo_nyanko Mathologer did a video using finite difference operators

[@samosavaglio2141]

Note that f(0)=2023, then assume X to ne nonzero and define g(x)=f(1/x)x^2022. g(x) has the f coefficients reversed, check for yourself, so g(x) can be easily seen to be the derivative of a much nicer polynomial. Can you conclude from here?

[@JamesLewis2]

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)

[@Songvbm]

6:07 So when we writing f2, f4, f6 in the format :- (x-1)^2 × (a polynomial) + constant , then do the lower bounds of (x-1)^2 and of the other polynomial get multiplied?

[@johannbauer2863]

A thing I did was leaving the exact polynomial out and only claimed: f_{2n}(x) = (x - 1)^2 P(x) + n + 1, where P(x) is a polynomial always greater or equals to zero. I wonder if you can use that to generalize it somehow. (probably not, because I still used the definition of f)

[@burk314]

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.

[@JamesLewis2]

If you reverse that recursion, then f_0 is the constant polynomial 1, and I think f_(−2) is the zero polynomial; both fit the pattern of what their global minimum values are and at least one point at which each minimum value is attained.

[@Nellak2011]

This is like the tabulation programming technique, but using algebra instead of some table we are building up.

[@bruhliver]

Where does this problem come from?

[@rifqi2006]

It is similiar to indonesian mo 2009 problem 6

[@wyboo2019]

at 10:36 you say the factored part, (x-1)^2(x^4+2x^2+3) is "zero at zero" but isn't it (0-1)^2(0^4+2*0^2+3)=1*3=3 at 0? unless i'm misunderstanding

[@thichhochoi766]

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

[@luisaleman9512]

He meant to say is zero at one

[@goodplacetostop2973]

16:58

[@bruhliver]

How are you so fast every time?

[@kpaasial]

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

[@ezequielangelucci1263]

​@@kpaasialits not a bot lol

[@kpaasial]

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

[@stephenbeck7222]

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

[@mab9316]

Isn't the minimum the leftover in the polynomial n+2=1011+2=1013 ???

[@martincohen8991]

I got the min for 2n is n+1. Also, f_{2n+2}(x)x^2f_{2n}(x)- (2n+2)x+ (2n+3) so f_{2n+2}'(x)=x^2f_{2n}'(x)+2xf_{2n}(x)- (2n+2) so f_{2n+2}'(1)=0.

[@psycdalex]

Ahead of the curve

[@benbooth2783]

Wow, I actually understand the question.

[@philp4684]

The Spice must flow!

[@bb5a]

Might have to get that tee shirt myself!

[@tokajileo5928]

I envy your math talent, really great videos, I wish I had them when I was a kid, many many years ago but then there was no internet nor I understood english. However and do not take this the wrong way do you ever smile? i saw many many of your videos but not even a millisecond of smile happened. this is rigid math presentation.

[@matematicacommarcospaulo]

17 min waiting Calculus to get into the mix and it did not come 😅

[@travishayes6037]

Are you wearing pajamas only? Or boxers?

[@USS_Relativity]

even powers positive odds negative so 2022 / 2 = 1011 its a odd number so we take only this part -k (x^1011).other parts cancel each other so f((x) = -1012 x^1011 + 2023 -> x=1 and -1012 + 2023 = 1011 its interesting i found 1011 not 1012 😞its just too close 😜

[@padraiggluck2980]

Very nice. ⭐️

[@chimetimepaprika]

Extremely valuable

[@damascus21]

What happened to stephanie being goofy in the description? 🥺

[@souvikmondal5191]

Marvelous❤

[@jaikumar848]

Hello sir ! Could you please make video on Z- transform and bode plot ? How it is useful for mathematician ?

[@tomholroyd7519]

Reminds me that proofs are like programs, and sometimes mathematicians are software engineers. Building practical general purpose tools for example

[@realGBx64]

nice pants (or underwear?!)

[@minwithoutintroduction]

تحليل رائع و ذكي.

[@ecoidea100]

Nice 🙂

[@matematicacommarcospaulo]

What happened to the intro?

[@artsmith1347]

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.

[@matematicacommarcospaulo]

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

[@mapleandsteel]

Apply netwon's method, lol

[@lobsterfork]

Just sucks because I have experience with this from school, but where the hell did f4 come from? You just pull it out of thin air. You would hate me as a student. I operate under no assumptions or very few assumptions. This might as well have been in another language when you just expect (expecting is an assumption) me to make blind assumptions.

[@mab9316]

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

[@PetraKann]

What is the physical or natural significance of this polynomial? Why would you want to determine the minimum? You should apologise

[@harambesson1098]

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

[@mrhatman675]

You shouldd get lost

[@fahrenheit2101]

Apologise? Lmao

[@psycdalex]

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

[@adrycough]

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.