Proof of existence by I.V.T.

Рет қаралды 23,639

Prime Newtons

Күн бұрын

In this video, I showed how to use Intermediate Value Theorem to prove the existence of a number.

Пікірлер: 88

@oraculum_ 11 ай бұрын

You're one of those few people I would actually love to click every video of theirs and immediately give a like without a further thought

@clionekimura9604 4 ай бұрын

I feel you, that’s exactly what I just did!

@sinanmertulucay9427 2 ай бұрын

I think it could also be proved like this: x = x^3 + 3 x^3 - x + 3 = 0 let P(x) = x^3 - x + 3 Inspecting the coefficients: 1 ∈ R, -1 ∈ R and 3 ∈ R. Hence, P(x) is a real cubic polynomial. Therefore it either has: a. 3 real solutions (x - R1)(x - R2)(x - R3) b. 1 real and 2 complex solutions (x - c)(x - a - bi)(x - a + bi), a ∈ R and b ∈ R So, there exists at least one real x such that P(x) = 0.

@demongeminix Жыл бұрын

Awesome demonstration of the use of the IVT.

@Necrozene 4 ай бұрын

I immediately answered "yes". In my mind I saw the graphs of y = x, y = x^3, and saw there were differences above and below +3, and I did not know exactly where. So, inituitively, I used the IVT. Cheers!

@magefreak9356 10 ай бұрын

I think you could also say that: f(x) is continuous on all real numbers. As x approaches negative infinity, f(x) approaches negative infinity. As x approaches positive infinity, f(x) approaches positive infinity. I don't know if the intermediate value theorem would apply here as infinity and negative infinity isn't a number. This would also apply to all polynomials which have an odd number for the highest power of x.

@9adam4 10 ай бұрын

I don't think you're technically allowed to do that. But what you can do is prove that f(-c) is negative and f(c) is positive, where c is any integer greater than the sum of the absolute values of the coefficients of the polynomial.

@magefreak9356 10 ай бұрын

@@9adam4 why do you need the extra restriction of the coefficients?

@9adam4 10 ай бұрын

@magefreak9356 The coefficients will determine how far you need to go to assure you have applicable points. For instance, f(x) = x^3 - 1000x^2, you need to get past 1000 before f(-c) is negative.

@manpreetkhokhar5318 8 ай бұрын

As per asked question, x^3 + 3 = x; I.e x^3 - x + 3 = 0 Let's define f(x) = x^3 - x + 3 So, question now transforms to checking if f(x) = 0 has any real root. We can clearly examine that, f(-2) = -3 and f(-1) = 3. Since, 0 lies between -3 and 3, so, by IVT, there exists atleast one c between -2 and -1, such that f(c) = 0. This proves, that f has atleast one root.Hence, for some c between -2 and -1, c^3 - c + 3 = 0. Also, by further analyzing the limits at -inf and its increasing,decreasing nature, we can deduce that,it is the only root.

@Ignoranceisbliss-i2e 4 ай бұрын

@@9adam4 yes you can because the graph is continuous and always increasing...

@markmajkowski9545 Жыл бұрын

??? By observation - Every cubic (odd polynomial as well) has at least one real solution. Plus there is a cubic “formula” despite its modern perceived complexity - especially when the x^2 term is 0 which was first solved before the generalized cubic.

@ramennoodle9918 3 ай бұрын

well yeah i think this is for more general proofs of this type

@kragiharp Жыл бұрын

Your videos are great, sir. I really appreciate them. Best wishes to you.

@PrimeNewtons Жыл бұрын

Glad you like them!

@tcoren1 3 ай бұрын

You can also use the fundamental theorem of arithmatic, which is arguably more advanced but is a bit more generic and is more fundamental to the question. The idea is to prove every odd degree polynomial with real coefficients has a real root. Every n degree polynomial has exactly n roots, and can be decomposed as p(x)=a(x-x1)(x-x2)...(x-xn), with x1,x2,...xn the roots. Taking the complex conjugate should result in the same polynomial since it's real, so p(x)=a(x-x1*)(x-x2*)...(x-xn*) Equating the roots of the two versions of the polynomial shows that each coefficient is either real or comes in a pair with another coefficient such that each is the other's coefficient. Since it is impossible to arrange an odd number of numbers into pairs at least one of them must be real

@boguslawszostak1784 Ай бұрын

And how will you prove this property

@tcoren1 Ай бұрын

@@boguslawszostak1784which property?

@boguslawszostak1784 Ай бұрын

@@tcoren1 fundamental theorem of arithmatic

@tcoren1 Ай бұрын

@@boguslawszostak1784I mean it depends on what you are allowed to start with. I do not recall a proof of it over the top of my head. For what it's worth it's usually taught & proven before the intermediate value theorem in university, and is IIRC easier to prove, even if it makes less intuitive sense

@jesusthroughmary 6 ай бұрын

Every cubic function (in fact every polynomial of any odd degree) must have at least one real root because complex roots come in pairs

@okaro6595 4 ай бұрын

Or if one goes towards plus or minus infinity he x³ term dominates and the value also goes towards the same infinity.

@jesusthroughmary 4 ай бұрын

@okaro6595 true, and therefore since it tends to both infinities it must pass through 0 on the way

@김민국-c9f 4 ай бұрын

If the coefficients of a polynomial are complex numbers, it doesn't need to have a real number root

@anilbera9499 4 ай бұрын

@@김민국-c9f but they are not here..

@pk2712 7 ай бұрын

Great demonstration of the use of the intermediate value theorem .

@rceretta 9 ай бұрын

Beautiful dialect solution!!!! I simply love it!

@davidcawthorne7115 5 ай бұрын

It is real important to remember that the function has to be continuous for IVT to hold or at least continuous over the interval a b. And to always wear a cool hat when doing mathematics. 😊❤

@JayTemple 11 ай бұрын

I used the Sign Rule (I think that's the name) to show that there is exactly one solution that's negative. As originally stated, the number the problem asks for is a solution of x = x^3 + 3, which works out to x^3 - x + 3 = 0. If x is a solution, let y = -x. Then -y^3 + y + 3 = 0. The polynomial has exactly one sign change, so it has exactly one positive zero, but if y is positive, then x, which is -y, is negative. ETA: Also, I find 10 and -10 to be easier to use than smaller numbers (other than 1). My test here would be (-10)^3 - (-10) + 3 = -1000 + 100 + 3

@jan-willemreens9010 Жыл бұрын

... A good day to you Newton despite the bad weather, Didn't I tell you ... BLACKBOARD --> SIR. NEWTON

@PrimeNewtons Жыл бұрын

😊😊😊😊 Thank you!

@michaelbujaki2462 3 ай бұрын

The number turns out to be approximately -1.67169988165716.

@davidbrisbane7206 3 ай бұрын

x³ - x + 3 = 0 has one real root and two complex conjugate roots. In fact, Descartes' Rule of Signs tells us that one negative real root.

@abdikadirsalad1572 Жыл бұрын

Thanks sir . Please make a video on mid term and final exam reviews calculus 1

@Faroshkas Жыл бұрын

I love this channel

@davidbrisbane7206 6 ай бұрын

Cardano's cubic formula always find a real solution to a cubic equation, so at least one such c exists and we know f(c) = 0.

@artandata 8 ай бұрын

I kept waiting for the numerical real value of the answer!

@markgraham2312 2 ай бұрын

Excellent!

@TheTrx3richie 8 ай бұрын

Good demonstration!

@kianushmaleki Ай бұрын

Love the "natural " joke. 😅🤣

@sjn7220 8 ай бұрын

Is there a solution that doesn’t involve guessing?

@tanjilsarker7678 Жыл бұрын

Thanks for the help!!

@Osaode-g5j 3 күн бұрын

Please can you use the cubic formula to solve this question?

@Ahmed-kg2gf 6 ай бұрын

The question is , "is x real" in order to solve that u need to get the notation for the equation and it is x³+3=x Reenraging the terms X³-x+3=0 Any cubic polynomial with real coefficient must have at least one real solution so the answer is ye , no need to solve

@tcmxiyw 2 ай бұрын

The question is “Is x real?”. You have answered the question “Can x be real?”. In order to answer the given question, you must rule out option 2, in which two of the possibilities for x are imaginary. For a cubic equation with real coefficients, there are these possibilities: three real solutions, two real solutions (one with multiplicity two), or one real solution and two imaginary solutions. So the answer to the question could be “no”. More work to do. You did, however, answer the question in the thumbnail for the video.

@tudorsafir2766 Жыл бұрын

Isn't that also called Bolzano's Theorem?

@PrimeNewtons Жыл бұрын

Yes, it is!

@kennethgee2004 5 ай бұрын

well the question changed a little from the thumb nail to the question on the board. This one feels like a no, but it is the proof they want.

@denniskisule8131 2 ай бұрын

what about iteration?

@moeberry8226 9 ай бұрын

Good video brother but you haven’t shown that x^3-x+3 is continuous even though I know it’s a polynomial and it’s continuous on R. Taking the derivative would also be good to show. Because differentiability implies continuity.

@tcoren1 3 ай бұрын

That's circular logic. The proof for the derivative of a polynomial goes through proving it is continuous. It's like using L'hopital's rule to find the limit of sinx/x

@manitubergaming Жыл бұрын

U soooo intelligent

@9adam4 10 ай бұрын

It's about -1.6717

@senseof_outrage9390 9 ай бұрын

Can you post the exact format of the solution. Did you solve the problem yourself... 😅

@neevhingrajia3822 8 ай бұрын

Thats a transcendental number right? So how can the a transcendental number be 3 more than its own cube?

@9adam4 8 ай бұрын

@neevhingrajia3822 What leads you to believe it's a transcendental number? As you correctly point out, a transcendental number can't be the solution to an equation like this.

@neevhingrajia3822 8 ай бұрын

@@9adam4 so are you saying that -1.6717 cubed would be 3 less than -1.6717?

@9adam4 8 ай бұрын

@neevhingrajia3822 Yes I am. Put it into the calculator yourself.

@amiinmohammed8552 16 күн бұрын

You are Soo good but you like to talk more please make short

@williamspostoronnim9845 Жыл бұрын

I like Your English very much!

@yaronbracha4923 7 ай бұрын

Who's Dad?

@holyshit922 Жыл бұрын

It is not so difficult to calculate x Assume that x is sum of two unknowns,plug in into the equation use binomial expansion , rewrite as system of equations Transform this system of equations to Vieta formulas for quadratic Check if solution of quadratic satisfies system of equations before transformation

@m.h.6470 Жыл бұрын

Solution: is there a real solution to x = x³ + 3? x = x³ + 3 |-x³ x - x³ = 3 |therefore x³ < x, which means that, as the result is an integer, x *has* to be negative! (x³ < x would also be possible, if 0 < x < 1, but then the result of x - x³ would not be an integer) Since x³ is growing very fast, x has to be quite small. testing left term assuming x = -1 -1 - (-1)³ = -1 - (-1) = -1 + 1 = 0 testing left term assuming x = -2 -2 - (-2)³ = -2 - (-8) = -2 + 8 = 6 Therefore there is a real solution of x between -1 and -2.

@BB-sc8ed Жыл бұрын

Thank you for saving my ass

@boguslawszostak1784 Ай бұрын

x = 1/3 (81/2 - (3 sqrt(717))/2)^(1/3) + (1/2 (27 + sqrt(717)))^(1/3)/3^(2/3) so exists :p

@Dr_piFrog 2 ай бұрын

Approximately: -1.67...

@physife 2 ай бұрын

I just used -10 and 0, these are easy enough

@Harrykesh630 7 ай бұрын

Using Derivatives would have been a better approach 🤔

@NathanSibali Жыл бұрын

Wow

@sunil.shegaonkar1 11 ай бұрын

Nearest answer is - 1.672, it is still an approximate. Question remains: is there an exact solution ? Probably not in rational numbers, but may be an irrational one. Oh, That is why they are called irrational number.