even funnier: Me: trying to solve this for a whole evening. Ok, there is one root from 0 to 1, but I, for the life of me, can't find it. I guess I'll give up and watch the rest of the video... Michel Penn: Don't look at me, I ain't have no solution either.
There isn't even a general equation for roots of a fifth degree polynomial. As my geometry teacher used to say, "Many great mathematicians 'wasted their lives' looking for it." A solution to a 70th order equation is definitely not guaranteed. I would be impressed if you could solve this example though.
@@asatellschow1908 yes, but there are plenty of special cases, when you can use some tricks to reduce it to a cubic equation of some such, I was hoping it is one of those, and I just can't see the trick.
@@Huxya exactly, just because there isn't a single-formula-catch-all like the quadratic formula, doesn't mean that all 5th-degree polynomials are "unsolvable" The point is to find the specal cases, like x⁵ − 2 = 0 is a 5th-degree polynomial, which is very easy x = ± |⁵√2| (plus-or-minus absolute-value fifth-root of 2)
To know exactly how many real root a polynomial has, you need to use Strum's theory. In this case there could have been 0 or 2 real roots, but because Michael found two roots, then there must be two real roots. I haven't checked, but if we apply Descartes rule of signs, then we will probably find that there are no negative real root. So, this is another reason why there are only three positive root to this polynomial.
@@davidbrisbane7206 Descartes rule of sign actually tells us there is exactly one negative root there. And it's quite obvious there must be one, this polynomial tends towards minus infinity at minus infinity, and evaluates to 1 at 0, so it must evaluate at 0 somewhere in between. But this negative root isn't useful anyway since it won't give a real value for x.
@@Falanwe If Michael had not shows there were 2 positive roots, then by Descartes rule of sign, there could have been *zero* positive roots and one negative root, with all the other 74 roots then being complex. Sturm's theorem would tell you there were two positive roots and one negative root.
@@GKinWor To learn? Do you mean to memorize? I don't think it's difficult to learn how to use it, as long as you think of it as a tool that you reach for and apply, instead of an item to memorize and apply. Our students need to learn how to read a mathematics book anyway. Descartes' Rule of signs is just complicated enough to make them appreciate being able to read and understand mathematical writing, rather than expect to memorize every trivial formula presented in class and four hundred examples of the nuances of their applications.
Can't we bypass most of the discussion? t^105-t^70-t^42+t^30 has 2 sign changes, so it has 0 or 2 positive roots, since 1 is a simple root, there must be exactly 2 positive roots. Together with 0, this gives 3 real roots to the original equation.
In case someone didn't follow the above explanation, read the first section of this link- en.m.wikipedia.org/wiki/Descartes'_rule_of_signs (The special case section is interesting too although it doesn't apply here)
@@angelmendez-rivera351 maybe explaining why the Rule of Signs works? (This is very difficult to do.) The point that Mike had is that the Rule of Signs is usually introduced with the parity condition, not just the upper bound, so the argument based on the IVT and the partial summation of a geometric series is unnecessary.
@@carsongbaker The leading term is t^75 so there probably isn't a "closed form", if there is it's not something we have tools for since there's no formula above 4th degree. It's also been proven that it is impossible to have a generalized formula above 4th degree so we'll never have a closed form formula for the general case. It might be possible for certain special cases to have closed forms under certain circumstances but even if that is the case it is unlikely anyone will ever discover such a case for a polynomial of degree 75.
It looks having 3 real solutions then? And infinite numbers get truncated inside black holes. “A black hole may seem to have a continuously infinite number of internal states, but [these get] cut off,” he said, due to quantum gravitational effects. “Real numbers can’t exist, because you can’t hide them inside black holes. Otherwise they’d be able to hide an infinite amount of information.” That doesn't look as difficult as black hole to be found... But wolfram alpha would not find the accurate/exact solution this case.
You can try to prove that the function f(t) is monotonically increasing between zero and one. If so, it’s easy to illustrate that there is only one root between 0 and 1.
8:37 Hum, if you don't factor out t^40, you don't even need to do a smart pairing. Each term on the left part will be bigger than each term on the right part (when t>1) and there are more terms on the left so the left part is bigger.
If there's a 100k video soon (hopefully) can you show us some of your solutions that you do on paper with all your scratch work and failed attempts etc before you present them on a board , love the content keep it up, easily the best math youtuber on the site 👍
At 3:50, it should be underlined that if t is real, then sqrt (t^210) = |t|^105, because both sides of this equation must be positive. Thus we have two cases: 1) sqrt (t^210) = t^105 if t >= 0; 2) sqrt (t^210) = - t^105 if t = 0; 2) t^30 - t^42 = t^70 + t^105 if t
13:33 Not really homework but still an interesting question I’ve found on Maths Stack Exchange. Link to the topic after the problem. A person is thinking of a number between 1 and 1000. What is the least number of yes/no questions that we can ask and know what that person's number is given that the person is allowed to lie on at most one of her answers. math.stackexchange.com/questions/329238/what-is-the-least-amount-of-questions-to-find-out-the-number-that-a-person-is-th
@@goodplacetostop2973 Okay, my first thought was to do it as a binary search where I start in the middle, ask each question twice (bigger than / less than midpoint), if I get two different answers, ask it a third time. That seems quite slow, maybe I'll build on that...
My first thought (or well second to be a bit more accurate) was that it could definitely be done in 2ceil(log2(1000))+1 or 21 questions by using all but one of the odd questions to halve the number of possible answers and using every even question to check if the answer given to the previous question was a lie, and if two answers contradict each other use an additional question to determine which one was the lie. I have no idea however as to whether a more clever method of checking which answer was the lie would be possible.
Wanted to say this as well. The only problem with the negative is the square root and by setting x=t^210 x is positive by design, so t can be any real number
Why is t only bigger than or equal to zero? Since you're raising it to an even power (210 in this case), it should not matter whether t is positive or negative, right?
x can't be negative because of the square root, and for t any negative solutions will also be positive because of what you said, so its easier to assume t is positive
@@danielflynn3067 No, that is obviously wrong. It's just that Descartes's Rule of Sign saved him and got the max of 2 positive roots, therefore no need to check for t
@@angelmendez-rivera351 I'm very sure that the function derived from that transformation, the t^30 (t-1) etc. is not an even function. Again, since we're talking about radical functions, we should not ignore a domain just because we can.
3:40 It isn't entirely obvious to me that we need x>=0. Although we want real x, that doesn't eliminate the possibility that some of the roots involved could be the complex non-principal roots (in addition to the square root which would be pure imaginary) and all the imaginary parts cancel out. Perhaps there is an understanding that the other three roots are the principal root only.
@@HASHIM2542 For an easier explanation, IMO: For x < 0, x^(1/2) is imaginary but x^(1/3), x^(1/5), and x^(1/7) are real. So, the left-hand side of the equation is definitely real, but the right-hand side is definitely not real.
when you make x = t^210, the right hand side of the equation looks like t^70 - t^(105). If t < 0, t^70 - t^105 > 0 because t^70 > 0 and t^105 < 0 => -t^105 > 0 => t^70 - t^105 > 0, while on the left hand side t^30 - t^42 < 0 because t^30 < t^42 => t^30 - t^42 < 0. In conclusion, we must have t > 0 (or = 0).
@@HASHIM2542 Remember that, even if sqrt(x) could be a complex number, we must consider that the value of the cube root, fifth root and seventh must be the complex number with the smallest argument, and we know what this means: theese three roots are real numbers (odd index) while the square root is complex, wich breaks the equality at the right-hand-side. This is why your comment is correct. But, if u doubt that a odd root can be a complex number, let me show something: suppose y³=1(y is the cube root of one). This means that, by the De'Moivre formula, y³=exp(2(pi)i*k) and k is any integer. Divide both exponents by 3, and you'll find out that y=exp(2(pi)i*k/3). There are 3 solutions for y: 1 real and 2 complex conjugates.
Subtract 40 from exponents is the same as dividing by t raised to 40. However, you don't actually divide. Instead you factor out to keep the overall equality true.
2 number theory from thailand Question1 TMO2020: proof that phi(2n)|n! for all natural number n Question10 TMO2020: Find all integer polynomial P such that P(n)|n!+2 for all integer n
1. Let n = 2^a b, b odd. Then phi(2n) = phi(2^(a+1) b) = 2^a phi(b) ≤ n, so it is a factor of n!. 10. Let a be the constant coefficient. Since P works iff -P does, we can assume a≥0. If a≠0, then a|P(a)|a! + 2, so a|2 -> a = 1,2. If a = 0, then n|P(n)|n!+2 for all n>0, hence taking n=3 gives 3|8, contradiction. Not sure what to do next.
Making the substitution u = -t in t^75 - t^40 - t^12 +1 = 0 gives -u^75 - u^40 - u^12 + 1 = 0 and then u^75 + u^40 + u^12 - 1 = 0 which, again by Descartes Rule of Signs, has one positive root for u = -t (exactly, as root reductions must occur in pairs) and thus exactly one additional negative root for t^75 - t^40 - t^12 +1 = 0 Thus I get 4 roots for x, not 3.
When I multiply by x^210 the exponents that result on my tablet are (210+(1/7)), (210+(1/5)), (210+(1/3)), and (210+(1/2)). I agree that the roots number three and they are zero, one, and an intermediate.
How does non negative value of x imply to onpy look for non negative values of t? Even if t is negative, it's even powered so x will still be jon negative
How did we guard against the possibility that t=1 is a multiple solution? Is it because f(1) is not equal to zero? I would think this needed to be stated from the onset when the factorization is attempted.
WE ONLY LOOK FOR X>=0 (POSITIVE ) because x shall be positive: sqrt of x condition of existence BUT variable t CAN BE NEGATIVE WHEN YOU HAVE EVEN POWERS =CORRECT BUT WE ALSO HAVE t WITH ODD POWERS UNDER RADICAL, SO FINALLY the variable t MUST BE ONLY POSITIVE or zero , in order to cover both situations!!!Here prof. Penn did not explain the requirement (many didnt get to point why t must be positive, t>=0), but I am doing myself now….wow I am still so good…And actually I solved the equation without looking how prof did it…
you could have also justified the number of roots using the strict concavity of the initial functions in x. if you move the negative terms to the others sides then there are 2 concave functions on both sides with can have at most 3 common points, the initial one (0,0) if it exists, and two more along the way.
@@gabriel7233 you can assume the existence of 3 or more comm pts and then prove by its 2nd derivative that it changes convexity on a given interval on the domain, which is a contradiction as we have assumed strict concavity
@@gabriel7233 yes or convex but strictly ones; you can also check geometrically that you cannot match those curves to have more than 2 sol; it also works when you have functions of opposite concavities (one convex one concave)
Can you explain why you assumed that t cannot take negative values? x= t^even number allows for negative values of t to transform to positive values of x.
@@Artur_Stoll If t was negative, the result of t^210 = (t^2)^105 would be positive, so no contradiction with the definability of x. The question of Amit Gupta still stands.
The reply is that t can be negative, but it will produce the same values as the positive t, so we can simply look at only the nonegative side in this case.
@@SamiVasileiosAmiris root(x) isnt equal some degree of x. Calcularing degrees is operating. But root is a form of transcription of number. P.S. sry for bad english. I hope u can understand what i mean.
How do you know the accessory function in t is continuous over (0,1), which is needed in order to apply the IVT? Maybe I’m not catching something here...
(t-1) just divides (t^75-t^40-t^12+1) because the sum of their coefficients is 0. From there you can just divide the whole polynomial by (t-1) and search roots from there.
x=t^210 is not an equation, t is defined to be x^(1/210). Since x must be >=0, there is a one-to-one match between non-negative x solutions and non-negative t solutions.
I was confused cause the one in the thumbnail is extremely easy. The left hand side is always on the real number line while right hand side only in the non-negatives. Also the roots on the left are always smaller for x>1 and larger for 0
I don't know that rule but by drawing the massive odd term (looks like a cubic) to the negative even curves (look like parabolas) we can get to the same conclusion. A max at X=0 then when the ^105 term takes over, the curve crosses at X=1.
@@riseciv7991 i check and it is a mistake. In my books, it is the "Théorème des Valeurs Intermédiaires" aka TVI. It seems Michael uses the English name instead.
As a French student I learnt it as the TVI indeed but IVT is the correct English traduction (I don’t know who has discovered it by the way but if it is written in French in your English book... must be a French guy 🤣)
Hi, my name is Adam and I just discoverd your channel a week ago. It's fantastic, on the level of Dr Peyam and others. as mid 30s adult who wants to get into math olympiads for recreation, what's a good course plan or syllabus? What books should I go through? I have studied up to single variable calculus and the foundations analysis(set theory, constructing reals from peano, proving stuff using induction, diagonalisation and some rigorous mathematical argumentation).
Several noticed that he should not have assumed t was nonnegative, but it's actually a safe assumption despite him not outwardly thinking it through properly. If a negative value of t were a solution, its absolute value would give the same value for x=t^210. Therefore, without loss of generality, we can assume t is nonnegative. Also, a few people have thrown complex analysis in the comments. While it's not entirely clearly stated, this appears that he is speaking strictly of the real root functions with a real input and a real output. In addition, a complex value of t would, at best, create a negative value of x or a duplicated value of x already accounted for by its sole positive real 210th root. It's a similar argument to why we don't need to consider whether t is negative albeit more complicated.
couldnt you technically have complex numbers on either side of the given equation which then sum together to something real? i dont think it can happen in this case, but is that something you would need to show before you can assume x (and t) has to be positive?
Yes, that can't happen here because x^(1/7) and x^(1/5) are real for any real x, therefore their sum is real. On the other hand, for x < 0, x^(1/2) must be imaginary and x^(1/3) must be real, and the sum of an imaginary number and a real number must be imaginary.
@@nathanisbored Yes, that was pointed out to me in another thread. In the first few seconds of the video, Prof. Penn does say "how many REAL x..." when he poses the question, so he did eliminate this possibility.
@@zanti4132 just because x has to be real doesnt mean the square root, cube root, 5th root or 7th root have to be. if x is negative, sqrt(x) would be strictly complex, (the others could be either real or complex), and i guess my question was: is there another solution where the complex stuff on one side of the equation might cancel with the complex stuff on the other side, in cases where the roots are complex? i am pretty sure in this setup it is not possible, but im just saying in the video he glossed over that, and continued on assuming that x was nonnegative
We don't consider t < 0 because we are interested in x and the function that maps t to x, namely x(t) = t^210, is even, which is to say x(t_1) = x(-t_1) for any given t_1, so the sign of values of t doesn't matter--might as well ignore all the negative ones.
We don't consider t nonreal because although x = t^210 may have many complex solutions for t given some x, admitting complex t does not increase the number of values x can have, which is what we're interested in.
@@Merssedes x cannot be negative, so any values of t that produce negative x are irrelevant. We know x cannot be negative because we are told x is real and -√x is nonreal for x < 0, while the other 3 terms in the original equation are real for x < 0.
I'm not sure why this had to go on at such length. Once you get to the polynomial and hoist out t^30 you have t = 0 or (t^75-t^40-t^12+1) = 0. The rule of signs already tells you that there can only be 2 positive roots of what's left. You don't have to factor it. You happen to know that one of those roots is 1 but that's just a side-track. There's some other root. And we don't care about negative roots anyway so we're done right there. You only need to go on if you don't want to use da rule of signs hamma. Well I guess you still need to use IVT to show that there is a root...
Negative values for t give the same values for x as the positive values. Therefore we can restrict t to be positive without any loss of generality. Put another way, if some negative value of t works, so will the positive value, since the value of x will be the same.
@@TedHopp *ding* many people were saying you can't assume that t is positive, and I questioned that too. But as you said, we can assume it is positive without loss of generality. I don't think he thought about that in the making of this video and just made a hasty assumption though.
would like for you to have at least attempted to find that third value with a little bit more specificity than (0, 1) might have been a nice time to introduce some of your viewers to any of the variety of numerical methods at your disposal.
Newton Raphson method with an initial guess between 0 and 1. Best to try x = 0.5 so that the method doesn't end up gravitating toward the root at 0 or the root at 1. It might even be possible to apply the fixed point theorem. mat.iitm.ac.in/home/sryedida/public_html/caimna/transcendental/iteration%20methods/fixed-point/iteration.html
Alternate solution but seems too simple to be true. Am I missing something glaring? If you write the radicals as rational exponents, then square bith sides, you get a polynomial equation. Then you reorder the monomials so that you have the ones with even exponents on one side and the ones with odd exponents on the other. You get x^7 + x^5 - x^3 - x = 2x^6 - 2x^2 Then look for the difference 4 in exponents: on the right hand side you can extract 2x^2 and on the left you can extract from the 5er and the 1er x^1 and from the 7er and 3er x^3. On both sides youre left with multiples of (x^4 -1) that you can cancel out on both sides, and youre left with a simple equation. You can cancel another x off then solve the quadratic equation. The three positive answers are 0 1 and 1+sqrt2.
Help how do you solve a quadnomial with degree 75 when grouping doesn’t work? (I’m trying the problem before the video and figured others might experience this as well)
I continued banging my head against the wall trying to continue factoring. I have now watched the video and was devastated to see no more roots. I found something interesting regarding the complex roots: t^24+t^12+1=(t^11+t^10+...+t^2+t+1)^(-1) + t^-39 I’m thinking there should be something we can do to determine the argument of the complex roots using mod here, but I’m too heartbroken on this problem already
I don't know what are you referring to as a "quadnomial" but if a polynomial isn't specifically constructed then you probably won't be able to solve it. Moreover, it will more likely be impossible to find roots of a 5-th root polynomial if you choose coefficients by random.
@@angelmendez-rivera351 i know, so the question on the thumbnail should have asked for the number of solutions. OR the presenter could have introduced some numerical solutions.
My solution: Plug both equations into desmos. Graph has three real solutions. The third at x = 0.011715. It really is the case that visualising something makes it much easier to prove oml. No idea how you would get the third answer though.
@@angelmendez-rivera351 considering that this video ends with merely "a solution exists", it would indeed seem that visualisation has defeated proof in this case.
I divided out the x = 0 and x = 1 roots and got the 74 degree polynomial b^74 + b^73 + ... + b^40 - b^12 - b^11 - ... - b - 1 = 0, from which the rule of signs gives the answer (1 remaining root), but before this I had an error that made all the coefficients positive, so I thought the answer was that x=0 and x=1 were the only solutions with b >= 0...if only...
I m not surr about the correctness of your solution cause t is not necessarily non negative because X=t^(2*105), so any real value of t would not cause an issue. That would change the last part of your solution as well
Here is a problem from the 2020 JBMO(junior balkan mathematical Olympiad) Solve the following system of equations over the real numbers: a+b+c=1/a+1/b+1/c a^2+b^2+c^2=1/a^2+1/b^2+1/c^2
I've plotted y = x**(1/7)-x**(1/5)-x**(1/3)+x**(1/2) using matplotlib and numpy, and it gives a root in between 0.0117 and 0.0118. Take x = 1/10**(7*5*3*2) = 1/10**210, then because 5*3*2 = 30 < 7*3*2 = 42, 7*5*2 = 70, 7*5*3 = 105, y > 0 But for x = 1+e, y = e/7-e/5-e/3+e/2 + o(e) = 23e/210 + o(e), so for e < 0, y < 0 thus there is another zero in between 0 and 1.
You say there is only 2 roots! Tracing a graph I have a third which is around 0.0117155 ???? x=0.0117155; x^(1/7)-x^(1/5)=0.118881 x^(1/3)-x^(1/2)=0.118881
I simply made the substitution at the start: Let t be defined from x by: x^(1/n) = t^n This transforms the entire problem into a degree 7 polynomial. After some factoring, I got it down to a 4th degree polynomial (happy that I can see Wolfram work to give me the closed form).
Was it a vaiable way to take derivatives and see that the derivative of the left term of the equation was always positive/negative in (-inf,0) and (1,+inf)?
The question in the thumbnail is find x suchat that \sqrt[7]{x) + \sqrt[5]{x) = \sqrt[3]{x) + \sqrt{x) (which is quite easier than the question in the video).
why x should to be positive : x can't be negative, because if we suppose x = - t²¹⁰ with t ∊ ℝ⁺ , we have : In ℝ : t³⁰ - t⁴² - t⁷⁰ = - j t¹⁰⁵ ; (with j=√-1 ) otin ℝ ⇒ Contradiction
Putting "qed" at the end of proofs gets boring. I'm going to start putting "atagpts".
😂
What does that stand for?
@@ritam8767 and that's a good place to stop
@@kaishang6406 oh alright. Thanks
Et locus bonus cessare ille est
Nobody:
Michael Penn: spend weekend trying to solve 70th order polynomial
even funnier:
Me: trying to solve this for a whole evening. Ok, there is one root from 0 to 1, but I, for the life of me, can't find it. I guess I'll give up and watch the rest of the video...
Michel Penn: Don't look at me, I ain't have no solution either.
There isn't even a general equation for roots of a fifth degree polynomial. As my geometry teacher used to say, "Many great mathematicians 'wasted their lives' looking for it."
A solution to a 70th order equation is definitely not guaranteed. I would be impressed if you could solve this example though.
@@asatellschow1908 yes, but there are plenty of special cases, when you can use some tricks to reduce it to a cubic equation of some such, I was hoping it is one of those, and I just can't see the trick.
@@Huxya exactly, just because there isn't a single-formula-catch-all like the quadratic formula,
doesn't mean that all 5th-degree polynomials are "unsolvable"
The point is to find the specal cases, like x⁵ − 2 = 0
is a 5th-degree polynomial, which is very easy
x = ± |⁵√2|
(plus-or-minus absolute-value fifth-root of 2)
@@chasemarangu But negative fifth root of two isn't a solution to that equation, only the positive one is since the index of the root is odd
I haven't heard about descartes rule of signs, something new to me) Interesting staff)
it sucks! super annoying to learn and super annoying to teach. what can u do
To know exactly how many real root a polynomial has, you need to use Strum's theory. In this case there could have been 0 or 2 real roots, but because Michael found two roots, then there must be two real roots. I haven't checked, but if we apply Descartes rule of signs, then we will probably find that there are no negative real root. So, this is another reason why there are only three positive root to this polynomial.
@@davidbrisbane7206 Descartes rule of sign actually tells us there is exactly one negative root there. And it's quite obvious there must be one, this polynomial tends towards minus infinity at minus infinity, and evaluates to 1 at 0, so it must evaluate at 0 somewhere in between.
But this negative root isn't useful anyway since it won't give a real value for x.
@@Falanwe
If Michael had not shows there were 2 positive roots, then by Descartes rule of sign, there could have been *zero* positive roots and one negative root, with all the other 74 roots then being complex. Sturm's theorem would tell you there were two positive roots and one negative root.
@@GKinWor
To learn? Do you mean to memorize? I don't think it's difficult to learn how to use it, as long as you think of it as a tool that you reach for and apply, instead of an item to memorize and apply. Our students need to learn how to read a mathematics book anyway. Descartes' Rule of signs is just complicated enough to make them appreciate being able to read and understand mathematical writing, rather than expect to memorize every trivial formula presented in class and four hundred examples of the nuances of their applications.
Can't we bypass most of the discussion? t^105-t^70-t^42+t^30 has 2 sign changes, so it has 0 or 2 positive roots, since 1 is a simple root, there must be exactly 2 positive roots. Together with 0, this gives 3 real roots to the original equation.
In case someone didn't follow the above explanation, read the first section of this link- en.m.wikipedia.org/wiki/Descartes'_rule_of_signs
(The special case section is interesting too although it doesn't apply here)
I think he explains that in the video too. Originally he was trying to do without the Descartes rule but got stuck.
@@angelmendez-rivera351 he also used descartes rule of signs, so be didn't "explain the logic"
@@angelmendez-rivera351 maybe explaining why the Rule of Signs works? (This is very difficult to do.) The point that Mike had is that the Rule of Signs is usually introduced with the parity condition, not just the upper bound, so the argument based on the IVT and the partial summation of a geometric series is unnecessary.
It only gives two solutions to the original equation cause x>=0
According to Wolfram Alpha: t≈0.979047; x≈0.0117155 . The graph of the function of t is very spiky due to the high powers.
With the t polynomial all the interesting stuff happens close to 1.0. With the original x equation, it all happens close to 0.0 .
Any way to rationalize it or just a wild decimal?
@@carsongbaker The leading term is t^75 so there probably isn't a "closed form", if there is it's not something we have tools for since there's no formula above 4th degree. It's also been proven that it is impossible to have a generalized formula above 4th degree so we'll never have a closed form formula for the general case. It might be possible for certain special cases to have closed forms under certain circumstances but even if that is the case it is unlikely anyone will ever discover such a case for a polynomial of degree 75.
I wonder if it's hitting some limitations.
It looks having 3 real solutions then?
And infinite numbers get truncated inside black holes. “A black hole may seem to have a continuously infinite number of internal states, but [these get] cut off,” he said, due to quantum gravitational effects. “Real numbers can’t exist, because you can’t hide them inside black holes. Otherwise they’d be able to hide an infinite amount of information.”
That doesn't look as difficult as black hole to be found...
But wolfram alpha would not find the accurate/exact solution this case.
You can try to prove that the function f(t) is monotonically increasing between zero and one. If so, it’s easy to illustrate that there is only one root between 0 and 1.
Good Job!
I once said to a mathematician "if you use Descartes' rule of sign . . ". He said "What's That?".
8:37 Hum, if you don't factor out t^40, you don't even need to do a smart pairing. Each term on the left part will be bigger than each term on the right part (when t>1) and there are more terms on the left so the left part is bigger.
At 3:35, how does it follow that t is nonnegative? A negative value of t corresponds to a positive value of x since x = t ^ an even power.
Due the square root of X. If X is negative, you'll get a complex number
If there's a 100k video soon (hopefully) can you show us some of your solutions that you do on paper with all your scratch work and failed attempts etc before you present them on a board , love the content keep it up, easily the best math youtuber on the site 👍
such a good idea
At 3:50, it should be underlined that if t is real, then sqrt (t^210) = |t|^105, because both sides of this equation must be positive. Thus we have two cases:
1) sqrt (t^210) = t^105 if t >= 0;
2) sqrt (t^210) = - t^105 if t = 0;
2) t^30 - t^42 = t^70 + t^105 if t
Underrated comment (exactly what I was looking for!)
@@pblpbl3122 thanks :-)
@ゴゴ Joji Joestar ゴゴ I'm using the principal root too.
As mentioned in the video and already here, necessarily x>=0 so there is no point in considering t
@@pedroteran5885 What are you saying exactly? If t < 0 still x > 0, so your "observation" is pointless.
13:33
Not really homework but still an interesting question I’ve found on Maths Stack Exchange. Link to the topic after the problem.
A person is thinking of a number between 1 and 1000. What is the least number of yes/no questions that we can ask and know what that person's number is given that the person is allowed to lie on at most one of her answers.
math.stackexchange.com/questions/329238/what-is-the-least-amount-of-questions-to-find-out-the-number-that-a-person-is-th
ua-cam.com/video/o0QikBfTGFA/v-deo.html
I guess it depends on whether you can ask the same question twice? That would help out with my method anyway
@@malignusvonbottershnike563 You can ask twice, three times, n times... the only restriction is being a yes/no question.
@@goodplacetostop2973 Okay, my first thought was to do it as a binary search where I start in the middle, ask each question twice (bigger than / less than midpoint), if I get two different answers, ask it a third time. That seems quite slow, maybe I'll build on that...
My first thought (or well second to be a bit more accurate) was that it could definitely be done in 2ceil(log2(1000))+1 or 21 questions by using all but one of the odd questions to halve the number of possible answers and using every even question to check if the answer given to the previous question was a lie, and if two answers contradict each other use an additional question to determine which one was the lie.
I have no idea however as to whether a more clever method of checking which answer was the lie would be possible.
i'm not sure about t being non negative as the power 210 is even so t can also be negative.
Wanted to say this as well. The only problem with the negative is the square root and by setting x=t^210 x is positive by design, so t can be any real number
0:56 imagine if he says "that's a good place to stop"
While using glasses, wearing a white shirt and holding a huge wall clock.
We love Michelle whatever😅
Decartes Law is something new for me aswell, very nice practical usage
In what sense is this "practical"?
9:45 why are there no roots when t>1?
So what is the value of t_0?
12:34 t>1?
What is your best Bench press?
Why is t only bigger than or equal to zero? Since you're raising it to an even power (210 in this case), it should not matter whether t is positive or negative, right?
x can't be negative because of the square root, and for t any negative solutions will also be positive because of what you said, so its easier to assume t is positive
@@danielflynn3067 No, that is obviously wrong. It's just that Descartes's Rule of Sign saved him and got the max of 2 positive roots, therefore no need to check for t
@@angelmendez-rivera351 I'm very sure that the function derived from that transformation, the t^30 (t-1) etc. is not an even function. Again, since we're talking about radical functions, we should not ignore a domain just because we can.
3:40 It isn't entirely obvious to me that we need x>=0. Although we want real x, that doesn't eliminate the possibility that some of the roots involved could be the complex non-principal roots (in addition to the square root which would be pure imaginary) and all the imaginary parts cancel out.
Perhaps there is an understanding that the other three roots are the principal root only.
Why can’t t be negative? Missed that point. Since x shouldn’t be negative but t has an even power (210)
Because t and -t give the same value for x, it is easier to define t such that t>0.
t can be negative but x can't be.. Because in the equation there is sqrt(x) as well and even power of x is not defined for negative values.. Hence x>0
@@HASHIM2542 For an easier explanation, IMO:
For x < 0, x^(1/2) is imaginary but x^(1/3), x^(1/5), and x^(1/7) are real. So, the left-hand side of the equation is definitely real, but the right-hand side is definitely not real.
when you make x = t^210, the right hand side of the equation looks like t^70 - t^(105). If t < 0, t^70 - t^105 > 0 because t^70 > 0 and t^105 < 0 => -t^105 > 0 => t^70 - t^105 > 0, while on the left hand side t^30 - t^42 < 0 because t^30 < t^42 => t^30 - t^42 < 0. In conclusion, we must have t > 0 (or = 0).
@@HASHIM2542 Remember that, even if sqrt(x) could be a complex number, we must consider that the value of the cube root, fifth root and seventh must be the complex number with the smallest argument, and we know what this means: theese three roots are real numbers (odd index) while the square root is complex, wich breaks the equality at the right-hand-side. This is why your comment is correct. But, if u doubt that a odd root can be a complex number, let me show something: suppose y³=1(y is the cube root of one). This means that, by the De'Moivre formula, y³=exp(2(pi)i*k) and k is any integer. Divide both exponents by 3, and you'll find out that y=exp(2(pi)i*k/3). There are 3 solutions for y: 1 real and 2 complex conjugates.
6:42 i didn't understand please explain
Subtract 40 from exponents is the same as dividing by t raised to 40. However, you don't actually divide. Instead you factor out to keep the overall equality true.
@@lee7956 Thanks 💜
How would one solve this for floating point numbers? Those are also a field, right? Just occurred to me that maybe it would be interesting to compare
For anyone interested, t0 > 0.9790472 (just barely), very close to 1, x = 0.01171554 (approx).
2 number theory from thailand
Question1 TMO2020:
proof that phi(2n)|n! for all natural number n
Question10 TMO2020:
Find all integer polynomial P such that P(n)|n!+2 for all integer n
1. Let n = 2^a b, b odd. Then phi(2n) = phi(2^(a+1) b) = 2^a phi(b) ≤ n, so it is a factor of n!.
10. Let a be the constant coefficient. Since P works iff -P does, we can assume a≥0. If a≠0, then a|P(a)|a! + 2, so a|2 -> a = 1,2. If a = 0, then n|P(n)|n!+2 for all n>0, hence taking n=3 gives 3|8, contradiction. Not sure what to do next.
Making the substitution u = -t in
t^75 - t^40 - t^12 +1 = 0
gives
-u^75 - u^40 - u^12 + 1 = 0
and then
u^75 + u^40 + u^12 - 1 = 0
which, again by Descartes Rule of Signs, has one positive root for u = -t
(exactly, as root reductions must occur in pairs)
and thus exactly one additional negative root for
t^75 - t^40 - t^12 +1 = 0
Thus I get 4 roots for x, not 3.
The negative root you found is the negative of his t_0 in this case, which leads to a duplicate x.
@@mike1024. Careless of me. Thank you.
When I multiply by x^210 the exponents that result on my tablet are (210+(1/7)), (210+(1/5)), (210+(1/3)), and (210+(1/2)). I agree that the roots number three and they are zero, one, and an intermediate.
What about the complex roots?
So far the best problem you have solved! I really like your explanations.
How does non negative value of x imply to onpy look for non negative values of t? Even if t is negative, it's even powered so x will still be jon negative
But how we can find the root beetween (0;1) ???????
@@angelmendez-rivera351 thank uu
Is there a way to determine how many roots the function has between 0 and -1?
It doesn't have any, because x is non-negative.
where is the third solution?
How did we guard against the possibility that t=1 is a multiple solution? Is it because f(1) is not equal to zero? I would think this needed to be stated from the onset when the factorization is attempted.
WE ONLY LOOK FOR X>=0 (POSITIVE ) because x shall be positive: sqrt of x condition of existence BUT variable t CAN BE NEGATIVE WHEN YOU HAVE EVEN POWERS =CORRECT BUT WE ALSO HAVE t WITH ODD POWERS UNDER RADICAL, SO FINALLY the variable t MUST BE ONLY POSITIVE or zero , in order to cover both situations!!!Here prof. Penn did not explain the requirement (many didnt get to point why t must be positive, t>=0), but I am doing myself now….wow I am still so good…And actually I solved the equation without looking how prof did it…
you could have also justified the number of roots using the strict concavity of the initial functions in x. if you move the negative terms to the others sides then there are 2 concave functions on both sides with can have at most 3 common points, the initial one (0,0) if it exists, and two more along the way.
how do you know that they have at most 3 common points?
@@gabriel7233 you can assume the existence of 3 or more comm pts and then prove by its 2nd derivative that it changes convexity on a given interval on the domain, which is a contradiction as we have assumed strict concavity
this is also a well known lemma taught for high school olympiad maths in Romania, so my proof is just a sketch
@@michuosas so it works with 2 arbitrary concave functions?
@@gabriel7233 yes or convex but strictly ones; you can also check geometrically that you cannot match those curves to have more than 2 sol; it also works when you have functions of opposite concavities (one convex one concave)
But what is t0??
Me, at the start of the video, feeling very proud: One! One! One!
Yes, I even forgot zero.
Can you explain why you assumed that t cannot take negative values? x= t^even number allows for negative values of t to transform to positive values of x.
210th root(x) cannot to be equal t less than zero
@@Artur_Stoll If t was negative, the result of t^210 = (t^2)^105 would be positive, so no contradiction with the definability of x. The question of Amit Gupta still stands.
The reply is that t can be negative, but it will produce the same values as the positive t, so we can simply look at only the nonegative side in this case.
@@SamiVasileiosAmiris root(x) isnt equal some degree of x. Calcularing degrees is operating. But root is a form of transcription of number.
P.S. sry for bad english. I hope u can understand what i mean.
Can we use Newton’s method to approximate the third solution?
What does 'radical' mean in this context?
How do you know the accessory function in t is continuous over (0,1), which is needed in order to apply the IVT? Maybe I’m not catching something here...
You know the accessory function is continuous because it's a polynomial on the Reals, and those are continuous everywhere.
(t-1) just divides (t^75-t^40-t^12+1) because the sum of their coefficients is 0. From there you can just divide the whole polynomial by (t-1) and search roots from there.
What about complex solutions for t that yield a real x? Shouldn´t we check those?
x=t^210 is not an equation, t is defined to be x^(1/210). Since x must be >=0, there is a one-to-one match between non-negative x solutions and non-negative t solutions.
12:40 ... I guess you meant "no roots for t>1"
What you've written in the "recall" rectangle, I wouldn't trust it for... t = 1 !!!
The curious case of the youtuber who kept miswriting his signs on thumbnails.
It is most definitely not on purpose... I am really that sloppy!!
@@MichaelPennMath There's only one way to solve (no pun intended) this: do the one on the thumbnail too 🙏
I was confused cause the one in the thumbnail is extremely easy.
The left hand side is always on the real number line while right hand side only in the non-negatives. Also the roots on the left are always smaller for x>1 and larger for 0
Also, the thumbnail asks to find all x, not how many.
I don't know that rule but by drawing the massive odd term (looks like a cubic) to the negative even curves (look like parabolas) we can get to the same conclusion. A max at X=0 then when the ^105 term takes over, the curve crosses at X=1.
As a HS Calc teacher, I completely nerded out when I saw the IVT coming.
12:00 I always learn this theorem as "TVI" instead of "IVT". Michael did some mistakes ?
no
@@riseciv7991 i check and it is a mistake. In my books, it is the "Théorème des Valeurs Intermédiaires" aka TVI. It seems Michael uses the English name instead.
As a French student I learnt it as the TVI indeed but IVT is the correct English traduction (I don’t know who has discovered it by the way but if it is written in French in your English book... must be a French guy 🤣)
@@alainrogez8485 so not a mistake.
Easy to understand. Great solution!
Hi, my name is Adam and I just discoverd your channel a week ago. It's fantastic, on the level of Dr Peyam and others.
as mid 30s adult who wants to get into math olympiads for recreation, what's a good course plan or syllabus? What books should I go through? I have studied up to single variable calculus and the foundations analysis(set theory, constructing reals from peano, proving stuff using induction, diagonalisation and some rigorous mathematical argumentation).
number theory is a good topic to start with and there's loads of materials and resources available on the internet
Several noticed that he should not have assumed t was nonnegative, but it's actually a safe assumption despite him not outwardly thinking it through properly. If a negative value of t were a solution, its absolute value would give the same value for x=t^210. Therefore, without loss of generality, we can assume t is nonnegative.
Also, a few people have thrown complex analysis in the comments. While it's not entirely clearly stated, this appears that he is speaking strictly of the real root functions with a real input and a real output. In addition, a complex value of t would, at best, create a negative value of x or a duplicated value of x already accounted for by its sole positive real 210th root. It's a similar argument to why we don't need to consider whether t is negative albeit more complicated.
couldnt you technically have complex numbers on either side of the given equation which then sum together to something real? i dont think it can happen in this case, but is that something you would need to show before you can assume x (and t) has to be positive?
Yes, that can't happen here because x^(1/7) and x^(1/5) are real for any real x, therefore their sum is real. On the other hand, for x < 0, x^(1/2) must be imaginary and x^(1/3) must be real, and the sum of an imaginary number and a real number must be imaginary.
@@zanti4132 x^(1/3) could be real or complex
@@nathanisbored Yes, that was pointed out to me in another thread. In the first few seconds of the video, Prof. Penn does say "how many REAL x..." when he poses the question, so he did eliminate this possibility.
@@zanti4132 just because x has to be real doesnt mean the square root, cube root, 5th root or 7th root have to be. if x is negative, sqrt(x) would be strictly complex, (the others could be either real or complex), and i guess my question was: is there another solution where the complex stuff on one side of the equation might cancel with the complex stuff on the other side, in cases where the roots are complex? i am pretty sure in this setup it is not possible, but im just saying in the video he glossed over that, and continued on assuming that x was nonnegative
@@angelmendez-rivera351 How so? Please provide one source which defines the cubic root sign as denoting a function on R with non-real values.
Try watching his videos at 2x speed, it's more fun.
i watch them all at 2x
Can somebody explain, please, why we don't consider t < 0?
Also, why we bound t by real numbers?
We don't consider t < 0 because we are interested in x and the function that maps t to x, namely x(t) = t^210, is even, which is to say x(t_1) = x(-t_1) for any given t_1, so the sign of values of t doesn't matter--might as well ignore all the negative ones.
We don't consider t nonreal because although x = t^210 may have many complex solutions for t given some x, admitting complex t does not increase the number of values x can have, which is what we're interested in.
@@keyboard_toucher but it does not mean that if t < 0 is solution then -t > 0 is solution and vice versa
@@keyboard_toucher it can give us negative values of x
@@Merssedes x cannot be negative, so any values of t that produce negative x are irrelevant. We know x cannot be negative because we are told x is real and -√x is nonreal for x < 0, while the other 3 terms in the original equation are real for x < 0.
Very informative. First time I have ever heard of Descartes rule of signs.
Can you try APMO 2012 Q3??
How about x ∈ ℂ ?
Can t be complex and the power 210 is a real number?
0:18 that was a pf snare and now I am extremely sure he knows how to beatbox asit was to clean pf snare to be a fluke
Check the roots of g'(t)
I'm not sure why this had to go on at such length. Once you get to the polynomial and hoist out t^30 you have t = 0 or (t^75-t^40-t^12+1) = 0. The rule of signs already tells you that there can only be 2 positive roots of what's left. You don't have to factor it. You happen to know that one of those roots is 1 but that's just a side-track. There's some other root. And we don't care about negative roots anyway so we're done right there. You only need to go on if you don't want to use da rule of signs hamma. Well I guess you still need to use IVT to show that there is a root...
If t is negative t^210=x is not negative. So why do not consider negative values of t?
It's pointless to use negative values because they yield the same x
I just didn't understand why t is positive.
x needs to be greater than 0, but since x = t^210, t can be either positive or negative
Negative values for t give the same values for x as the positive values. Therefore we can restrict t to be positive without any loss of generality. Put another way, if some negative value of t works, so will the positive value, since the value of x will be the same.
@@TedHopp *ding* many people were saying you can't assume that t is positive, and I questioned that too. But as you said, we can assume it is positive without loss of generality. I don't think he thought about that in the making of this video and just made a hasty assumption though.
would like for you to have at least attempted to find that third value with a little bit more specificity than (0, 1)
might have been a nice time to introduce some of your viewers to any of the variety of numerical methods at your disposal.
Newton Raphson method with an initial guess between 0 and 1. Best to try x = 0.5 so that the method doesn't end up gravitating toward the root at 0 or the root at 1.
It might even be possible to apply the fixed point theorem.
mat.iitm.ac.in/home/sryedida/public_html/caimna/transcendental/iteration%20methods/fixed-point/iteration.html
So many cheeky little tricks to make this such a simple methodical process... I learned a lot :D
Alternate solution but seems too simple to be true. Am I missing something glaring?
If you write the radicals as rational exponents, then square bith sides, you get a polynomial equation. Then you reorder the monomials so that you have the ones with even exponents on one side and the ones with odd exponents on the other. You get
x^7 + x^5 - x^3 - x = 2x^6 - 2x^2
Then look for the difference 4 in exponents: on the right hand side you can extract 2x^2 and on the left you can extract from the 5er and the 1er x^1 and from the 7er and 3er x^3. On both sides youre left with multiples of (x^4 -1) that you can cancel out on both sides, and youre left with a simple equation. You can cancel another x off then solve the quadratic equation. The three positive answers are 0 1 and 1+sqrt2.
But I guess the assumption x itself needs to be positive is used without proof and that aint good.
Squaring both sides does not yield a polynomial equation.
Very good, thank you!
Help how do you solve a quadnomial with degree 75 when grouping doesn’t work?
(I’m trying the problem before the video and figured others might experience this as well)
I continued banging my head against the wall trying to continue factoring. I have now watched the video and was devastated to see no more roots.
I found something interesting regarding the complex roots:
t^24+t^12+1=(t^11+t^10+...+t^2+t+1)^(-1) + t^-39
I’m thinking there should be something we can do to determine the argument of the complex roots using mod here, but I’m too heartbroken on this problem already
I don't know what are you referring to as a "quadnomial" but if a polynomial isn't specifically constructed then you probably won't be able to solve it. Moreover, it will more likely be impossible to find roots of a 5-th root polynomial if you choose coefficients by random.
Let x = y ^210
1- x ^ (1/5-1/7)
= x ^ (1/3-1/7)- - x ^ (1/2-1/7)
1- x° (2/35)= x° (4/21)-- x° (5/14)
1- y°12 = y°40- y^75
the question on the video thumbnail asks us to find all values of x, you have only shown the number of solutions.
x~=0.117155 , 0, 1
@@angelmendez-rivera351 i know, so the question on the thumbnail should have asked for the number of solutions.
OR the presenter could have introduced some numerical solutions.
@@angelmendez-rivera351 my complaint is technical, the thumbnail question is not the question on the board.
Wait didn't the equation used to have a t^12 + 1? Why is it t^12 - 1 now?
I was disturbed by that too.
My solution:
Plug both equations into desmos.
Graph has three real solutions. The third at x = 0.011715.
It really is the case that visualising something makes it much easier to prove oml. No idea how you would get the third answer though.
@@angelmendez-rivera351 considering that this video ends with merely "a solution exists", it would indeed seem that visualisation has defeated proof in this case.
I divided out the x = 0 and x = 1 roots and got the 74 degree polynomial b^74 + b^73 + ... + b^40 - b^12 - b^11 - ... - b - 1 = 0, from which the rule of signs gives the answer (1 remaining root), but before this I had an error that made all the coefficients positive, so I thought the answer was that x=0 and x=1 were the only solutions with b >= 0...if only...
I m not surr about the correctness of your solution cause t is not necessarily non negative because X=t^(2*105), so any real value of t would not cause an issue. That would change the last part of your solution as well
Here is a problem from the 2020 JBMO(junior balkan mathematical Olympiad)
Solve the following system of equations over the real numbers:
a+b+c=1/a+1/b+1/c
a^2+b^2+c^2=1/a^2+1/b^2+1/c^2
ua-cam.com/video/o0QikBfTGFA/v-deo.html
just today ive seen that many people were asking Balkan MO questions on aops , was the exam today or something ?
@@prithujsarkar2010 l think that the JBMO was at September or something. I found it on the internet yesterday.
actually t doesn't have to be >=0 because x=t raised to power 210 and even if t is negative x will still be positive.
Thanks For Providing Us Such quality content :)
I've plotted y = x**(1/7)-x**(1/5)-x**(1/3)+x**(1/2) using matplotlib and numpy, and it gives a root in between 0.0117 and 0.0118.
Take x = 1/10**(7*5*3*2) = 1/10**210, then because 5*3*2 = 30 < 7*3*2 = 42, 7*5*2 = 70, 7*5*3 = 105, y > 0
But for x = 1+e, y = e/7-e/5-e/3+e/2 + o(e) = 23e/210 + o(e), so for e < 0, y < 0 thus there is another zero in between 0 and 1.
You say there is only 2 roots!
Tracing a graph I have a third which is around 0.0117155 ????
x=0.0117155;
x^(1/7)-x^(1/5)=0.118881
x^(1/3)-x^(1/2)=0.118881
we can also show that there is only one t0 between 0 and 1 (bijection theorum) since g(x) is strictly increasing
I simply made the substitution at the start: Let t be defined from x by: x^(1/n) = t^n
This transforms the entire problem into a degree 7 polynomial. After some factoring, I got it down to a 4th degree polynomial (happy that I can see Wolfram work to give me the closed form).
That doesn't seem to make much sense. You'd be making a different substitution for each n.
Was it a vaiable way to take derivatives and see that the derivative of the left term of the equation was always positive/negative in (-inf,0) and (1,+inf)?
Good
The question in the thumbnail is find x suchat that \sqrt[7]{x) + \sqrt[5]{x) = \sqrt[3]{x) + \sqrt{x) (which is quite easier than the question in the video).
Why do we specify t>=0? 210 is odd so t^210 is positive no matter the real value of t.
The thing you click on for the video has the equation with a + sign
Not to me. 7th root minus 5th equals cube root minus square,
Radical yupee 🤘🤘🙂
this must be true for any aVx - bVx = cVx - dVx , a>b>c>d>1 for the similar reason
Came here for the biceps
confusing that the thumbnail says find all solutions, but the blackboard says how many solutions. Very different questions.
why x should to be positive :
x can't be negative, because if we suppose x = - t²¹⁰ with t ∊ ℝ⁺ , we have :
In ℝ : t³⁰ - t⁴² - t⁷⁰ = - j t¹⁰⁵ ; (with j=√-1 )
otin ℝ ⇒ Contradiction
what happens between x=0.0117 and x=0.0118?
i think this deserves a second edition.
( x = approx 0,011715 ) = ( t = approx 0,979045 ) ^ 210
interesting to play with the indices and see how the curves behave.
wish you best
Why do we want t to be nonnegative?
Won't raising t to the power of 210 make it nonnegative anyway?
t can be negative because x=(t^105)^2
could we have just drawn it kind of
Solutions (for x in R) are x=0 and x = +/- 1
Sqrt(-1) isn't R.
Yes, but -1 is