x^2 + cos x = 2024

[@wehpudicabok6598]

Me: This problem looks impossible! How could he possibly solve it in just 5 minutes??? "Note that x^2 is an even function --" Me: Oh, it's 0. Duh! Possibly the biggest gap I've ever seen between how hard it looked and how hard it ended up being.

[@aljawad]

A great titillating question! I visualized the question as a parabola (x^2) intersecting with a constant line (y = 2024) at two points that mirror one another around the y-axis, hence the two solutions would hover around +/- sqrt(2024), and their sum = 0!

[@sunhouse8616]

when you're doing math you're allowed to be a pedant. it's not 0! rather 0

[@aljawad]

@@sunhouse8616 how about: “ … their sum = 0 !” 😃

[@Tritibellum]

math people will be going crazy with that exclamation sign at the end i swear we need to invent a new exclamation sign to type something like "the answer is 5! it's 5!!!!! trust me!!!!!"

[@anushkamishra5865]

​​@@TritibellumYea because 0! (or 0 factorial) = 1... but can you tell why 0! equals 1 ??

[@dageustice]

​@@anushkamishra5865 f(x)=x!, x>0 converges to 0 as x approaches 0. The binomial formula (a+b)^n gives us 1 only if 0! when n=0. Factorial is used in combinatorics. 0! represents the amount of ways you can sort an empty box of a deck of cards: 1 way. More conceptually, had there been 0 ways to sort an empty box, empty boxes couldn't exist!

[@jimmy_4]

so i did it this way. x^2 +cos x = 2024 x^2 = 2024- cosx so if we draw the graphs of both LHS and RHS it will intersect symmetrically because both the graphs are symmetrical. Hence the sum will be zero! edit: there is another method i could think of: since range of cos x lies between -1 and 1 so here value of x2 would lie between 2023 and 2025. hence x would have two solutions lying between sqrt(2023) and sqrt (2025) and -sqrt(2023) and -sqrt(2025) hence if we add both we get zero. also coeffecient of x is zero here so the sum will eventually be zero!

[@evyatar100]

You need to prove to things: 1. There is a solution. 2. The number of solutions if finite. You didn't do any of them.

[@tessfra7695]

Thank you for another simple yet effective explanation.

[@mertunsal7335]

I think one also need to show that there are only finitely many solutions, otherwise we need to specify which kind of sum we are using. For this example, there are only finitely many solutions as x^2 + cos x >= x^2-1>2024 for x large enough. Thanks for the video!

[@evyatar100]

How this argument show that there are finite many solutions?

[@mertunsal7335]

You are right, I must have thought that the set of solutions is a discrete set so bounding it will be enough. However, the function is increasing on the positive real line and symmetric from the y-axis, thus it is 2 to 1 and there exist exactly two solutions which can be deduced from IVT

[@2L40K]

The solutions are symmetrical. What kind of sum you can use and not get zero even in the infinite case?

[@mertunsal7335]

@@2L40K If you use counting measure and define the sum as the integral associated to that measure the sum should be absolutely convergent in order that integral(sum) to be defined.

[@evyatar100]

@@2L40K S = 1-1+1-1+1-1+1-1... What is S in your opinion? Is it 0, 1 or 0.5?

[@ParasocialCatgirl]

Thats a pretty wild question

[@Unitedstatesofficialmedia]

Me: you're genius bro love from INDIA 🇮🇳❤️❤️

[@heartpiecegaming8932]

I'd like to add that (echoing a few other people) you *do* need that there are only finitely many solutions to the equation. This can be done in many ways, but the one I like is using compactness. The function R->R, x->x^2+cos(x) is a proper function since it goes to infinity as x->+- infinity (proper function means preimage of a compact set is compact). So then, the preimage (say K) of a singleton (say {2024}) is a compact subset. But note the function is increasing at each x in K (the preimage) (why?) so by the 1-dim inverse function theorem, this is a local diffeomorphism. In particular, the points in x are isolated. So we have a compact discrete set K which has to be finite. Sorry if this uses a bit of advanced math than is accessible. Maybe one can just do it more manually by carefully analyzing the derivatives, etc.? But I found this to be the slickest way to do it.

[@PrimeNewtons]

I agree. I just never knew about diffeomorphism until now. Thanks for sharing.

[@panagiotisvlachos6114]

Congratulations for loving your Maths science so much!! Greetings from Greece!!

[@davidbrisbane7206]

It seems like there might be an infinite number of positive solutions and a corresponding number of negative solutions. If there were only a finite number of positive and negative solutions, then the sum of the solutions would add to zero. But will an *infinite* number of positive solutions added to an infinite number of negative solutions add to zero? Is there an issue here with adding an infinite number of terms together in any order and still getting a finite answer which is the same in every case?

[@Grecks75]

If you analyze the equation further, you will see that there can be only one positive solution. Why? Because x must lie close to 45 (between sqrt(2023) and sqrt(2025) to be exact) and in that range the LHS is a strictly increasing function. So we have exactly one positive and one negative real solution to the given equation. Hence, their sum is a finite sum which yields zero because the LHS is an even function as was explained in the video.

[@PerthScienceClinic]

Since cosine is constrained between -1 and +1, this function behaves as x^2 as x increases... When x>sqrt(2025), x^2+cos(x) > 2024. This provides an upper bound for x (and also a lower bound for -x). So there is a finite number of solutions.

[@VicecrackVoldermort]

There is no way there is an infinite amount of solutions. When x^2 > 2025, x^2 + Cos x cannot intersect 2024 anymore. So there is a finite amount of solutions. This is if you understand that for any bounded interval on the x-axis, there can only be a finite amount of solutions in this specific question. I am glossing over this fact because I am assuming that you did not have an image of the graphs touching each other on a continuum in your head.

[@davidbrisbane7206]

@@VicecrackVoldermort No graph image. No further analysis either. I was just asking a hypothetical question. Still, it would be interesting to construct similar examples (if this is possible) with an infinite number of roots to see if they indeed always sum to zero. Clearly a finite number of roots would.

[@VicecrackVoldermort]

​@@davidbrisbane7206 I agree. Also, on closer inspection, the function can not have more than 2 roots. The second derivative is 2 - Cos x is positive for all of R which indicates a U shape.

[@jeroenvandorp]

I wish I could write as fast and legible on _paper_ as this man does on a chalkboard. Kudos. 👍

[@PhysicsDownUnder]

Fantastic problem. Particularly as a challenge problem for a 1st week Calc class learning about even and odd functions.

[@lapichfamily7595]

There is a problem. You must precise that the number of solutions is finite, because if not, yhe sum is not definite !

[@SGuerra]

Uau! Que bela questão. Eu cheguei a uma solução parecida utilizando a aproximação senx = x, para x muito pequeno que resulta em uma equação do segundo grau para análise. Parabéns! Brasil Agosto 2024. Wow! What a beautiful question. I arrived at a similar solution using the approximation sinx = x, for very small x which results in a quadratic equation for analysis. Congratulations! Brazil August 2024.

[@snowman2395]

For anyone curious x is about +/- 44.983

[@sharveshs5807]

+/- 63.60 I guess

[@meriemmoi8708]

It actually ±√2023

[@sharveshs5807]

@@meriemmoi8708 I got root 4046

[@Woah9394]

​@@meriemmoi8708roughly

[@LHH916]

anyone that plotted the graph on desmos would know the comment is slightly more correct than the replies... just saying

[@nainibrok9139]

a question / addition: If we have an infinite number of solutions (x1,x2,...) (possible because of the cosine), then we have an infinite sum. With infinite sums, the summands must not be interchanged arbitrarily, as the results can be different. (Riemann's series theorem) Isn't that a problem here?

[@Tom_TP]

x^2 isn't a periodic function, that means x^2 + cosx isn't a periodic function, so this equation doesn't have infinite solutions. We don't need to worry about 0*infinity form here. 0 times a finite number is 0.

[@nainibrok9139]

@@Tom_TP ah yeah true, my bad thanks mate

[@Tom_TP]

@@nainibrok9139 No problem. It was indeed an interesting thought.

[@DedenK]

You could have an infinite number of solutions without a periodic function... But it's right you have to prove first that you have here only 2 solutions.

[@2L40K]

@@Tom_TP cos(1/x) is not periodic, but has infinite number of zeros over a finite range of x.

[@nadonadia2521]

you have to study the variations of the function x²+cos(x)-2024 between sqrt(2023) and sqrt(2025), prove theat f(sqrt(2023))*f(sqrt(2025)) ≤0 and use (Bolzano's theorem) :If a continuous function has values of opposite sign inside an interval, then it has a root in that interval.

[@evyatar100]

Correct. This shows that there is a solution. How can you show that the number of solutions (is this interval) is finite?

[@no_name_fifa_mobile]

I have a question. Is it "mathematically accurate" or "get a perfect score on a test" to demonstrate by drawing a graph of y=-cosx and y=x^2-2024??

[@ccdsah]

Stopped at 0:39. Since cos(-x)=cos(x) and sqr(-x)=sqr(x). The sum of all solutions (if there are any) must be zero)

[@renyxadarox]

The only thing you forgot to show, that the equation has at least one real solution. What if there is no real solutions at all? The sum of empty set of real solutions is undefined, I guess... )))

[@profesorleonardo1645]

The sum of no solutions is zero. But it's easy to show there's at least one solution: if x=45, then f(x)>2024

[@nanamacapagal8342]

Easy to show that there are solutions by the intermediate value theorem. If f(x) = x^2 + cos(x), f(44) = 2115. f(x) is continuous, so by the IVT there exists some value x between 44 and 46 such that f(x) = 2024

[@2L40K]

@@profesorleonardo1645 The sum of no solutions is not zero. You have an empty sum over an empty set. This thing is undefined.

[@2L40K]

@@nanamacapagal8342 Much easier than that. F goes to infinity at both sides (e.g. can be made bigger than any finite number) and is one at zero.

[@pojuantsalo3475]

The problem can be written as cos x = 2024 - x², but -1 ≤ cos x ≤ 1 which gives -1 ≤ 2024 - x² ≤ 1 that can be manipulated the give √2023 ≤ ∣x∣ ≤ √2025 => 44.9777722 < ∣x∣ ≤ 45. It doesn't really matter if x is in radians or degrees, because x² grows so fast around x ≈ 45. Assuming x is in radians, this is a really narrow angle (about 0.02 rads) where the positive and negative solutions are. 45/2𝜋 ≈ 7.16. This means we must "unwind" the angles by ±14𝜋: 45-7*2𝜋 ≈ 1.018 rads and √2023 - 7*2𝜋 ≈ 0.995 rads. This means the value of cos x is same as cos y where y = ± (1.0065±0.0115) rads Assuming x is in degrees, we instantly have the solution: x ≈ ± 45 degrees. Either way the sum of all real solution is zero.

[@myfyrmadocjones]

In calculating an actual solution, how does one define Cos x - does the x represent degrees, radians or gradians?

[@PerthScienceClinic]

By default, angles in mathematics are measured in radians. This allows us to use the relatively simple Taylor series expansions for the trig functions.

[@myfyrmadocjones]

@@PerthScienceClinic thank you 👍

[@himanshukafaltiya7823]

I did it this way: The eq can be written as x^2+0x+cos(x)-2024, Here a=1, b=0, c=cos(x)-2024 The sum of solutions of a quad. eq = -b/a =-0/1 =0 Im only a student, can someone confirm whether my way is correct. THANK YOU!!

[@lumina_]

no since this is not a quadratic equation. cos(x) is not a constant

[@kimberlyvaldez0423]

​@@lumina_ Are you sure???

[@lumina_]

@@kimberlyvaldez0423 👊👊💥💥🦶👊💥🦶💥💥👊💥💥🦶💥👊👊🦶💥👊💥🦶

[@kimberlyvaldez0423]

​@@lumina_jzjdk

[@kimberlyvaldez0423]

​@@lumina_huoco

[@MD_Areeb]

To solve the equation x2+cos⁡x=2024x2+cosx=2024 and find the sum of all real solutions, we need to analyze the equation step by step. Step 1: Analyze the components x2x2 is a parabola that opens upwards and takes all real values from 00 to ∞∞. cos⁡xcosx is a trigonometric function with a range of [−1,1][−1,1]. Thus, the equation can be rewritten as: x2=2024−cos⁡x x2=2024−cosx Given the range of cos⁡xcosx, the range of 2024−cos⁡x2024−cosx is: 2023≤2024−cos⁡x≤2025 2023≤2024−cosx≤2025 Thus, x2x2 must satisfy: 2023≤x2≤2025 2023≤x2≤2025 Step 2: Solve for xx Taking the square root of both sides: 2023≤∣x∣≤2025 2023 ​≤∣x∣≤2025 ​ This implies: −2025≤x≤−2023or2023≤x≤2025 −2025 ​≤x≤−2023 ​or2023 ​≤x≤2025 ​ Given that x2=2024−cos⁡xx2=2024−cosx, xx must be close to ±2024±2024 ​ because cos⁡xcosx is small in magnitude compared to 2024. So the approximate solutions for xx are: x≈2024,−2024 x≈2024 ​,−2024 ​ where cos⁡x≈0cosx≈0. Step 3: Calculate the exact values 2024=44.98 2024 ​=44.98 Thus the real solutions are x≈44.98x≈44.98 and x≈−44.98x≈−44.98. Step 4: Sum of all real solutions Adding these two solutions gives: 44.98+(−44.98)=0 44.98+(−44.98)=0 Final Answer The sum of all real solutions is 00​.

[@nadonadia2521]

There is no x≈44.98 and x≈−44.98 in mathematics, what is x≈ what does it mean. In mathematics there is no small or big number, these ideas came from physics and engineering. If 0.001 is small, what about 0.000000000001.

[@PaulMiller-mn3me]

How is x^2 always even? 3 or 5 squared, for example, are not

[@iskandar_em]

Even function means that for x1,x2,x3,….xn the value of f(xn)=f(-xn). For example : f(x)=x^2, then f(5)=25 and f(-5)=25 also because 5^2 and (-5)^2 are equal

[@NewYoutubeisstupid]

Not that type of even. He's talking about even functions @PaulMiller-mn3me. An function f is said to be even if f(x)=f(-x). Since x^2 = (-x)^2 and cos(x) = cos(-x), the function can be said to be even. If I recall correctly, an even function has reflectional symmetry about the y-axis.

[@lumina_]

a function is called even if it satisfies the criterion f(-x) = f(x)

[@kimberlyvaldez0423]

​@@lumina_MISTER

[@lumina_]

@@kimberlyvaldez0423 omg what

[@epsilonxyzt]

Good! Never stop learning!

[@Shin_Siyun]

cosx는 -1에서 1까지의 값을 가집니다. 그래서 cosx를 우변으로 넘기면 x²은 2025와 2023 사이의 값을 가질 것 입니다. x²은 y축 대칭인 우함수이므로 2025일때의 2개의 x의 합이 0이고 2023일때의 2개의 x의 합 또한 0입니다. 그 사의 값들 또한 대칭성으로 인해 합이 0이 됩니다. 따라서 모든 x의 합은 0이 됩니다.

[@Grecks75]

If I get the English translation of your answer right, I need to correct one thing: There are only *two* real solutions of the equation, one positive and one negative lying symmetrically around the y-axis. Not four.

[@surendrakverma555]

Very good. Thanks Sir 🙏🙏🙏🙏

[@promixinc.8434]

Hell no. It went on a totally unexpected path. New trick achieved however. Amazing explanation though. ❤

[@RabinSaidÖsteränggymnasietNA1C]

So what is x?

[@nmanoharreddy2525]

I have a doubt that can't we use expansion of cosx

[@abdelmajid3409]

And what is the solution s

[@abhisheksaini3489]

Great question i wonder how graphically can we see it

[@ramunasstulga8264]

Is it possible to solve this equation though?

[@canyoupoop]

With calculus you can see how the graph looks like "a cosine curve curling with y=x²" so this helps in telling us there are exactly. Another argument could be this is even AND for x∈ℝ+ this is strictly increasing. If you want to compute x (which is ≈√2024 btw pointed out by with an error 0.006) you could use maclaurin expansion of cos (x) and solve the polynomial maybe. Netwons approximation also could be used. But the answer is most probably transcendental so idk.

[@muneebahmad7729]

you can by Taylor expansion maybe

[@DedenK]

2 problems, here: To be sure that the sum is defined, you have to check first that you have not an infinite number of solutions. Moreover, the x_n you let, have to be positive (or negative). If not, you calculate 2*S at the end. So, I would have first checked the variations to count the solutions, and then, applied your method.

[@objectifmaths]

I agree. No solution on [0,1] and the derivative is >0 on [1,infty[ : only 1 solution on [0,+infty[

[@simmmr.9040]

The sum of all roots (even if they are infinite) will still be zero if for every positive root there is always a corresponding negative root.

[@objectifmaths]

@@simmmr.9040 1-1+2-2+3-3 ... is divergent. I do not master the sums of such series.

[@anupn3rmie565]

it's actually easy if you just create a function f(x)=x^2+cosx-2024 and find roots of f(x)=0...the function increase for x>0 and is even

[@ODOk-vs8ut]

I wrote X²+cosx=45²-1 =>X²+2cos²(x/2)=45² As 2cos²(x/2)€[0,2] Its save to say 44

[@uggupuggu]

I applied vieta's formula and the taylor series of cos x doing so results in -2023 + x^2/2 - x^4/24 + x^6/720 ... = 0 If you consider partial sums, with degree 2n The sum of zeroes is the limit of (2n)!/(2n+2)! as n tends to infinity which is 0

[@user-sb9ho5jz3e]

good video. x^2+cosx is a function shapes like a parabola, there are only 2 real roots; there is no complex solutions. x^2 is dominating over cosx when compounding two basic funcions

[@evyatar100]

Why are there only two roots?

[@jarikosonen4079]

Only numerical solutions? x1 and -x1. Good observation. Can you get more solutions than x1 and -x1? x^2+cos(k*x)=2024, with some value of k1.

[@muhammadhussainsarhandi9928]

Hi Prime Newtons, one thing that you need for correction is the use of an ellipsis, which has 3 dots (...). Whenever you write an infinite sequence or a series, for example 1,2,3, . . . or 1 + 2 + 3 + . . . , you should only put 3 dots and not more than that. I have seen in many of your videos that you put 4 dots or more.

[@shiningwhiffle]

Wouldn’t you also need to show that there are at most finitely many solutions? If there were infinitely many then you would have a conditionally-convergent series that could be reordered to produce any sum.

[@PrimeNewtons]

You make a good point. How do I do that. That would actually be a good video.

[@shiningwhiffle]

​@@PrimeNewtons Here's what I've come up with. It involves a little bit of calculus, which takes away a little from the simplicity of your approach, but there's a way around that. Calculus-based approach: Step 1. Since -1 ≤ cos(x) ≤ 1, then 2023 ≤ x^2 ≤ 2025. This gives two intervals where we might find solutions to the equation: -sqrt(2025) ≤ x ≤ -sqrt(2023) and sqrt(2023) ≤ x ≤ sqrt(2025). Step 2. Focusing on the case where x > 0, we see that d/dx(x^2 + cos x) = 2x - sin x > 0 for all positive x except possibly near 0 (which is nowhere near where the solutions might be so it doesn't concern us). This means that x^2 + cos x is strictly increasing on the interval sqrt(2023) < x < sqrt(2024). Step 3. Since it's strictly increasing, there can only be one value for x in that interval satisfying the equation. Step 4. The proof for the case where x < 0 is similar. (The derivative on that side is always negative except possibly near 0 so the function is strictly decreasing.) Conclusion: There are at most 2 solutions to the equation, one for each of the intervals. (In fact there are exactly 2.) Non-calculus based approach: Step 1 is the same, but then just look at the graph of x^2 + cos x: it's pretty clearly strictly increasing/decreasing on the relevant intervals. Hope this is what you're looking for. Thanks for replying!

[@shiningwhiffle]

​@@PrimeNewtons I don't think my reply went through, so if not let me do it again, briefer this time Here's what I came up with: since -1 ≤ cos x ≤ 1, obviously x^2 is doing most of the work, so any solutions must be somewhere around either sqrt(2024) or -sqrt(2024). Next we show that x^2 + cos x is strictly decreasing when x > 0 (it might wiggle up and down when x is near zero but that's irrelevant since there are no solutions near zero). You can do that two ways: either take the derivative d/dx(x^2 + cos x) = 2x - sin x and note that it's always negative when x > 0 -- or just look at the graph of x^2 + cos x since it's pretty obvious just from looking at it. Since it's strictly decreasing/increasing anywhere we might have a solution, there can be at most two solutions: one when x < 0, one when x > 0. (In fact there are exactly two solutions.) So the number of solutions is finite. Then from your argument, since x^2 + cos x is even, the solutions must be additive inverses of each other. I hope this helps and that I didn't make any major mistakes. Thanks for replying!

[@2L40K]

So, you define a canonical order - by absolute value. It, for example, preserves the symmetry of the function. Mathematics is not that rigorous and super accurate thing many people consider. In fact, it's quite messy and incompatible... And you can do many things as you wish, if it serves a purpose...

[@shiningwhiffle]

@@PrimeNewtons Oh no, it looks like KZbin ate my reply to you. Maybe it was too long? The key idea was that if you treat f(x) = x^2 + cos x as a function, it's easy to show that it's strictly decreasing when x < 0 and strictly increasing when x > 0, except possibly near 0 itself, which doesn't matter because there's clearly no solution to f(x) = 2024 there. Then because it's strictly decreasing/increasing there can be at most one solution on each side of zero.

[@simmmr.9040]

The proof of existence of at least one solution is missing for the completeness of the solution ;)

[@LouisEmery]

I would have missed that one. Funny and tricky.

[@memer5629]

couldn’t there be one solution giving that x(cos + x) = 2024 meaning that x = 2024 and cos + x dosent exist

[@konraddapper7764]

in case any one cares f(x)=x^2 + cos(x) -2024 is strictly ly than 0 for all |x|

[@Mvgof242up1]

This equation are four solutions by approximation +_44,98 and +_ 45

[@Ron_DeForest]

Don’t know how to feel about this one. It’s both inspired and cheesy. While I was fully expecting some of your mathematical magic, you do this. 🤣 Congrats on figuring out the true key. Still was hoping for an actual solution. Perhaps you could do one where you don’t need to add up the answers. Just a thought.

[@potentialofprotonis3]

1:55 there will only be 2 real solutions, and one will be the modulus of the other *-1

[@カインなのよ-t5g]

偶関数だから、解の総和が 0になるのは納得。面白い！

[@kelvinella]

But you have to show that there are finitely many solution.

[@vitowidjojo7038]

x^2 is always increasing, so it would hit 2024 finite amount of times before it blasts to infinity.

[@benjamintroncoso8105]

very nice solution

[@nadonadia2521]

Half answer, you did not prove that the equation has a solution, you are not a mathematics teacher.

[@maburwanemokoena7117]

How about “How many solutions are there?” well for x^2> 2024 there are no more solutions, so the solutions lie between -45

[@Mathguy1729]

Let f(x)=x²+cos(x) f’(x)=2x-sin(x) ≥2x-1 >0 if x>1/2 f’(x) ≤2x+1

[@xinpingdonohoe3978]

The cosine fluctuations are negligible. There are two solutions, and they're only about 0.006 off of the actual values of √2024.

[@laoltomsk]

We may easly find a derivative of the function, which is 2x-sinx. Obviously, it's always positive for x>0.5 (since sinx1), so the initial function is ascending for x>0.5. If it is ascending, it may intersect a constant only once for x>0.5. Obviously, there are no solutions for 0

[@aidencrooks]

Good quote, Ecclesiastes 5:10. Praise the Lord Jesus. Let all creation praise the glory and love of the Lord God. Amen.

[@zachariastsampasidis8880]

Should probably also show that it has a finite number of solutions which isn't difficult mind you

[@akultechz2342]

2024-x² = cosx Cosx, x² both even hence x = ±n

[@РоманФилиппов-о4ф]

The equation has only two real solutions. ±14.3185*pi. Sum of it is 0.

[@nikogoat14563]

Solve it using graphs

[@micahell4179]

x = 0 because f(x) is an even function.

[@burak4537]

nope. x is 45 and -45

[@Maths_3.1415]

0:52 I was unable to solve this problem 😂.

[@sepehrhaghverdi8977]

This is one of cool one not hard one

[@師太滅絕]

Brilliant.

[@kadappadevarmani9616]

Plz give mathematical. Reasoning ..n write big clear handwriting

[@9ybruhwt491]

beautiful!

[@BrawlN64]

x-2024/x-cos(pin)/x 1-1/2025=2024/2025

[@slippinchillin]

I would have checked, whether x = 0 is a solution, cause if x = 0 were a solution it would only be counted once, BUT x = 0 is NOT, thus not an issue here

[@2L40K]

And if you add 0 to the sum will it change?

[@slippinchillin]

@@2L40K what do you mean? When x = 0, LHS is not 2024, thus it’s not part of the solutions, you wouldn’t want to add this case into the sum

[@panjak323]

Lol that's easy... Both functions are symetric along y axis thus: x^2 + cosx = (-x)^2 + cos(-x) = 2024 with solution somewhere around sqrt 2024... The contribution of cos x should be insignificant at tha point. So the two potential roots must be same in magnitude, just with opposite sign Meaning the sum of roots is 0. This could be applied to any linear combination of symmetrical functions, even with more than 2 roots, because they would always cancel out.

[@jonthebob2159]

From 2024 HMMT February guts round

[@brawlstars-xh7xf]

Great❤

[@face-good]

解の個数が有限個であることは書かないとダメでしょ

[@AZ-ow9wy]

zero because of the Y-axial symmetry

[@shigototravaillez9972]

Beautiful

[@Nothingx303]

It was really juicy 😋

[@johnkabila6617]

Wonderbar!

[@Viki13]

That was insane haha

[@saravana4217]

❤❤❤

[@Ryan95Channel]

0 is NOT a real result

[@evgenylavrentiev2512]

X=44.9829 and x=-44.9829. Author of this video should learn math…

[@evyatar100]

We need to prove two things: 1. There is a solution. 2. The number of solutions is finite. Define f(x)=x^2 + cos(x) Proof of (1): As cos(s) is bounded, notice that, f(44)

[@Krmpfpks]

Elegant solution for the two missing proofs in the video

[@emanuellandeholm5657]

The empty sum is 0 by definition, so you actually don't need a solution.

[@Krmpfpks]

@@emanuellandeholm5657 if the function had no solutions, a sum of the solutions would not exist. That is very different from 0

[@2L40K]

Clearly, you have a problem with overcomplicating trivial matters. 2) There is no need for the number of the solutions to be finite. A sum with an infinite number of addends can easily be a finite number. 1) There are at least 2 solutions, since the left side: - goes to plus infinity both ways and is one at zero - and is a continuous function

[@2L40K]

Furthermore, get: f(x)=x^2 + 3cos(x) And your proof fails, but no matter that nothing changes in the solution.

[@КД.Димка]

First you have to show that there is at least one real solution to the equation and then your solution will work...

[@Cold_Guy69]

x² + cosx = 2024 x² + (cosx - 2024) = 0 Comparing the general equation, ax² + bx + c = 0 Here in our original eq, b = 0 So, sum of solutions would be, α + β = -b/a = 0

[@shravan8292]

General equation is assuming the constant c is independent of x but here it's dependent in the form of cosx so I don't think this works

[@SomeOne2027]

I need a shower after watching this 🥵

[@barryzeeberg3672]

But there is another alternative, namely that there is no (exact) solution. If there is no solution, then there cannot be a sum of 2 solutions. If the "solution" happens to be an irrational number that is not representable in the same way as pi or e (ie, an infinite sum) then it does not really exist, since we cannot write an infinite decimal expansion that has no formula. Numerically, the (positive) solution is an infinite expansion between 44.9941 and 44.99415. That is, if you have 2 things that both do not converge, then you cannot get rid of the lack of convergence by taking a linear combination of them.

[@Tritibellum]

interesting, why would an irrational number that cannot be represented as an infinite sum not exist in the real numbers? i mean, there are methods of approximating the number (like newton's approximation method), isn't there a way to do this infinitely many times?

[@barryzeeberg3672]

@@Tritibellum newton's approximation is obviously an approximation. that might be fine for engineering or in numerical analysis or in the everyday world, but in mathematics we are seeking the exact value - that is, an infinite number of digits to the right of the decimal point. If there is no rule or formula to specify that sequence of digits, then representing that number can only be done by enumeration, which is a process that will never terminate, and is therefore not an algorithm. if this is incorrect, then I would be happy to be corrected :)

[@Tritibellum]

@@barryzeeberg3672 i think enumeration and approximations formulas are actually the same, since they both tend to do 1 thing, which is approximating the desired value. now, i wouldn't say that there is a moment where enumeration of an irrational number can terminate, but isn't approximations formulas (i.e infinite sum) essentially the same thing? like a process that never ends? which is basically like limits are? infinite sums are basically derived from limits, so technically speaking pi and e wouldn't exist in the real numbers since there isn't a way to express them numerically, which doesn't feel right to me.

[@barryzeeberg3672]

@@Tritibellum good point about pi and e. but both of those can be expressed in a convergent closed form formula, which is of course an infinite sum. but if you have a convergent closed form formula, then that is sufficient to uniquely and specifically define the particular irrational number - you do not need to actually evaluate it numerically, you are ok as long as there is a convergent closed form that uniquely defines it. on the other hand, if you have something that does not have a closed form AND it would take infinitely long to compute by brute force, then that seems to me to be a non-computable quantity? and you don't actually know if it really exists, you are just guessing/assuming that it exists?

[@barryzeeberg3672]

@@Tritibellum and even if pi did not have a closed form representation, it exists as the ratio of two quantities that do exist in the real world or in the mathematical world. so we know that pi really exists - there is an explicit unambiguous definition of what it is. the definition of e seems to be more implicit, not explicit like pi, so e really does need to have a closed form formula?

[@알쓰-k2z]

you did not prove that there is a solution first

[@nubidubi23]

My mind is blown

[@gokalpgorduk7685]

amazing

[@smabedi]

I mean, really?! 😂

[@zeroone7500]

cosx with x is radian or degree unit You can't sum two different unit variable to make a number. What unit of 2024 ?

[@dukespider]

putting a number in a trig function means that you are using radians. cos(x) would mean cosine at x radians, cos(1) would be 1 radian, and cos(pi) would be pi radians or 180 degrees

[@RabinSaidÖsteränggymnasietNA1C]

Only me who thinks about quadratic formula?

[@kevinvanmieghem1432]

False... For instance : 1-1+1-1+1-1+... is not convergent... So isn't 0.

[@evyatar100]

You are right. It is possible to prove here that the number of positive solutions is finite - But he didn't.

[@Rakesh_kumar711]

Because it's a quadratic equations. Sum of the roots=-b/a=0

[@objectifmaths]

it's not a quadratic equation.

[@Issac-ff2ec]

It is a quadratic equation so it is going to have two solutions. Now we know the sum of the roots of the quadratic equation = -b/a . here there is no term with only x as a variable so -b=0 therefore 0/1 = 0. I am not sure if i am correct though

[@ElVerdaderoAbejorro]

You can't use Girard's relations, this isn't a polynomial.

[@xinpingdonohoe3978]

​@@ElVerdaderoAbejorro it can be a polynomial.

[@ElVerdaderoAbejorro]

@@xinpingdonohoe3978 How? By using Taylor's expansion?

[@davidbrisbane7206]

@@xinpingdonohoe3978 Not really, as cos(x) can be expressed as an infinite power series of powers of x.

[@xinpingdonohoe3978]

@@davidbrisbane7206 I know.

[@tubetigeerr]

😆😆😆😆😆

[@mohamedchabraoui6433]

Noooo

[@thaerthaer1120]

But also solution is easy by Newton Ruffson or just ignoring the value of cosx bz its so small compare with 2024 and say x approx sqrt(2024)

[@PerthScienceClinic]

That sounds like engineer logic to me.

[@thaerthaer1120]

@PerthScienceClinic Yes but I'm an electronic engineer

[@PerthScienceClinic]

@@thaerthaer1120 Oh sorry, I should have typed it slower :D

[@mooncake2665]

0 0 0 0 0 0 0 0

[@DulingElegant]

Tu parles trop

[@user-wl4zu2ok1e]

x ≈ ±44.9829

[@ramunasstulga8264]

Blud using graphing calculator 😭🙏