@ 2:01 what???! who says always g(x) g(x)=3. Unless you have restricted yourself only to positie numbers which was not the case. There is a "correct" way to show that x=y gives one set of solutions
This is how I approached it. x^3+y=2 implies one of x^3 and y is at least one, and the other is at most 1. Since x^3>=1 implies x>=1 and x^3=y^3, so for x^3+y=x+y^3=2 to hold, equality must hold for both inequalities and (1, 1) is the only solution. If -18, but then y=2-x^3
@11vives it just means that one of the numbers is between 0 and 2, so is the other. Like, one may be 1.3 and other may be 0.7. It really means that the numbers are both positive.
@11vives One way to think of it is that the average of two numbers (x^3 and y) is 1, so the average is in between both numbers. Rigorously, you could just set up a proof by contradiction.
@@alainrogez8485 That doesn't follow at all. In fact, for the functions listed, when x< 1, (such x definitely exist because g takes all reals as its domain) g(x)> x. For example g(0) = 2 > 0. What is only true is that for a given decreasing function f, there exists some y such that if x>y, f(x) < x.
@@bencheesecake yes, my apologies. I am French and sometimes, it is difficult to understand math and English in the same time. You are right, when g is decreasing : x g(x)>g(y) and not g(x)
g(x) being decreasing does not imply g(x) ≤ x. Consider x=-1 for a counter example. I would tackle this through graph sketching and then algebra for a precise solution.
True, g(x) is not always less than x. if x=-2 g(x)=2-(-8)=10 also elsewhere someone says the inequality should be the other way but that's not true either because g(x) crosses the line y=x
@@Khalibi If 0 were a root, the constant term (-b^2) would have to be 0, but since b is a positive integer, the constant term can't be 0. This can be seen with the rational root theorem used in the video. 0 is a rational number, so if it were a root, it would have to be a factor of the constant term. 0 multiplied by anything is 0, so if 0 is a root, the constant term must be 0.
@@coc235 (A late reply). Yes, but doesn't this invalidate the proof? We have x=< y (from assumption), and g(x) >= x, ie y >-= x, which is the same as x=
@@coc235 It is a solution of the same kind of system: y=5/3-x^3/3 x=5/3-y^3/3 [The third real solution is x=-1;y=2.] This system shows the problem with presented solution, notice that g(z)=5/3-z^3/3 is also a decreasing function, and the system is symmetric in x,y.
@@srijanbhowmick9570 if you mean answer. Answer is correct. But reasoning towards answer is not clear to me. And to many other people, too, if check comments.
For the first problem, I just took the first equation minus the second. The obvious solution is to go ahead and factor (y-x) from both sides. This leads to your x=y solution immediately. Then for the case when y>x, go to the step before we said y=x and divide both sides by y-x leaving you with x^2+xy+y^2+1=0. It is very easy to manipulate this equation in two ways. First, complete the square to get (x+y)^2=xy-1. The second way is to isolate the squares on one side leaving you with x^2+y^2=-xy-1. Then it should be obvious to add these two equations to arrive at (x+y)^2+x^2+y^2=-2 which has no solutions in the real numbers. Thus we are finished.
I don't think this is quite right. When you factor out (y-x), you get x^2+xy+y^2=1. This is an ellipse with many solutions, including (0,1), (1, -1), which are not solutions to the original problem. That's expected, since subtracting the two equations gives us a third equation, but the overall solution still needs to be a simultaneous solution to 2 of those 3 equations. This does reduce the original problem from a 9th order polynomial (intersection of two cubics) to a 6th order one (the intersection of a quadratic and a cubic).
Great work . Could you help me with this? Let k be the positive root of x^2 +x-4=0 and the polynomial P(x)= ax^n +a'x^n-1 +a''x^n-2 ....+a0 which has non negative coefficients and an arithmetic value of P(k)=2017 1) prove that a0 +a1+......an =1 mod 2 (I can't find the right symbol with 3 slashes) 2) Find the least possible value of a0+a1+.....+an
@@coc235 Okay, but how does this get us to y being less than or equal to x which in turn is less than or equal to y? g(x) = y, so g(x) being greater than or equal to x is exactly what we have assumed before.
@@coc235 Well, fuck. And according to WolframAlpha there are complex solutions where x and y aren't equal, so the supposed fact that x = y is just untrue, I guess.
In the second problem, if 'three integer roots' means 3 different values, we get (a,b)=(80,10) with roots (2,5,10) and (a,b)=(90,12) with roots (3,6,8) And for those considering 0 a natural number: we find eight more solutions: b=0 and a in {16,30,42,52,60,66,70,72} and roots (0, (17 +/- sqrt(289-4a^2))/2)
(x^2 + а)*x=17x^2 + b^2 Since the right side is positive, so should be the left side. It can only happen if x is nonnegative because a is natural number.
at 1:45, why is z a decreasing function? I've read some responses, but still don't understand why he/we can say that. I also noticed that Prof Penn hardly ever responds to comments....
For the first problem, you can show that the two curves only intersect at x=y=1 by considering a line segment that separates them. I picked the line segment that is tangent to one of the curves where they meet: y = (4-x)/3, where x1. So, the second curve also does not intersect the line segment. Since the curves are on opposite sides of the line segment, they cannot intersect each other in the region x1. A similar argument shows that they also do not intersect in the region x>1, y
"By natural numbers I mean positive integers." Might be the third most-said thing on this channel, right behind "And that's a good place to stop," and "okay."
It is good he says it means positive integers because there are some vids which mentions natural numbers but they include the zero. This way he avoids confusion.
@@williamperez-hernandez3968 +1. Mathematics is about communicating ideas, and both definitions of the natural numbers are hugely popular, so it's best to just specify.
I think the Lemma in the second question does not supply _sufficent_ evidence that the polynominal must have three positive roots. It only says if the polynomial has a real root, then it must be positive. Applying Decartes Rule of Sign to the polynomial, then it can have 1 to 3 real positive roots and no negative real roots. More work is required to show that the polynomial does indeed have 3 positive real roots. But of course, because the question says there are three integer roots, then we know we are dealing with 3 positive real roots. However, the problem can be modified to find solution when the cubic has integer roots, which would be one integer root (and two complex roots), or three integer roots as specified in the problem. For example, a cubic of the required form with one integer root is x^3 -17x^2 + 17x - 16 = 0 where (a,b) = (17,4). This polynomial has one integer root = 16. In fact, if b^2 = 17a, and a = c^2, where c is a natural number, then we can generate an infinite number of polynomials with different values for a and b, all having x = 17 as its root and no other positive real root. E.g. x^3 -17x^2 +612x - 10404 = 0, has x = 17 as its only real root. Here c = 6, so a = 612 and b = 102.
The curve y=2-x³ is a reflection of the curve x=2-y³ about the line y=x. Hence if there exists any point which satisfies both, it must lie on the line y=x. I think this is a less time-consuming method to get to the conclusion that y=x.
I thought that for any system were there is this type of symmetry then it is always the case that x=y. I have never seen anything different. I always just considered it an inherent characteristic of the symmetry. Is this what you think? Is there more going on here than I am unaware of?
@@anonymous_4276 Your language is very clear and I am sure you understand the situation very well. When I see things like in the video were something I thought was obvious and well understood being laboured, it makes me doubt my own understanding and I feel insecure.
@@tomasstride9590 when I first made that comment, I basically wrote down the way I reasoned y being equal to x. I saw your comment and thought maybe it was obvious to you (that y equals x in such systems). I was just reasoning out why that happens in the original comment.
We can say the two equations are the inverses of each other so that the solution lie along y=x...furthermore from graph we see that this is the only possible case(for Q1)
Around 2:00 that comment about g(x) being a decreasing function and so g(x)=1. However, if you graph g and g^-1 the coordinates of any intersections provide solutions to the problem and you will see that they indeed only intersect at one point. The solutions here have coordinates which are roots of g(g(z))=z which is a degree 9 polynomial equation; is the solution presented here a single root or is it multiple equal roots? For a more general g(z)=a-z^3 there could be 3 distinct intersection points. There certainly are when a=0. As a increases the three intersection points approach each other. Here's a problem: At what value of a do the three solutions all become equal? Beyond that value of a do they remain a triple of equal solutions, or do two of them become complex?
Ohh thank god someone else had the same problem , so does this mean his solution is incorrect ? and can we not solve this question with pure algebra and no graphs ???
I havent really thought about it but can someone tell me if this approach is wrong if we replace 2 with y+x^3 in the second equation we get x=y+x^3-y^3 which means x-y-x^3+y^3=0 which means that x-y-(x-y)(x^2+xy+y^2)=0 which means (x-y)(1-x^2-xy-y^2)=0 so x=y or x^2+xy+y^2-1=o which means x=(-y±√(-3y^2+4))÷2 for -√(4÷3)≤y≤√(4÷3)
I followed this line of approach too. I got the same possible values for x in terms of y as you did. However, when you find the values for x from the x^2 + xy + y^2 = 1 equation and place then back into x = 2 - y^3, then when you get rid of the square root, you end up with a polynomial of degree 6 with integer coefficients. You can in fact prove that this 6th degree polynomial has no real value of y which is a root, which is harder to do that the original problem, or you can just graph it in Desmos and see it has no real roots. So, even although x^2 + xy + y^2 = 1 gives two values of x that satisfy it in terms of y, it turns out that these solutions for y don't satisfy the original equations in real numbers. So, basically, x^2 + xy + y^2 = 1 yields *no* more real solutions.
@@davidbrisbane7206 ok got it but I also have another question if we have two polinomials p(x)=p(y) does it mean that for those two to be equal always as fanctions x=y or is there another imput lets call it a(a is variable) that can satisfy this
No, actually, x = 2-y^3 and y = 2-x^3. So, y - x = 2-x^3 - 2 + y^3 = y^3 - x^3 = (y-x)(x^2 + y^2 + xy)=> Either x = y or x^2 + y^2 + xy -1 = 0. Note that it is -1 and not + 1 here. Now your argument does not hold true.
This is a fix of the argument. if x!=y, WLOG x < y and it holds x^2 + xy + y^2 = 1 (*) case 1: xy >= 0: It holds x^2 < y^2 1 which is inconsistent to y^2 0 y = 2 - x^3 > 2, however form (*) (x+y/2)^2 = 1 - 3y^2/4 < -2 which is inconsistent to x+y/2 is real.
By the way, I just brute forced it and solved the cubic equation resulting from solving for (x+y). Indeed, the solution for x^2 + xy + y^2 = 1 leads to a complex root for x and y. Once I saw the result, I knew there must be way to prove it and you have certainly shown the way.
At 2:00 the fact that g(x) is a decreasing function only tells us that when x= g(y). That's the step he missed out. Now, if we *also* know that x and y satisfy y = g(x) and x = g(y), *then* we can substitute x for g(y) and arrive at his assertion g(x)
@@christophniessl9279 i know (8, 3, 6) and (8, 6, 3) are the same. I did not argue with that. My concern is that he only gave (10, 5, 2), (9, 4, 4), and (8, 8, 1). I hope you get my point.😊
Decreasing does NOT mean that the function is less than its variable! decreasing means that as the variable increases the function decreases, nothing else!
The more I look at the first question the more I realize that graphing gives the fact that there’s only one solution right away and that it’s in the first quadrant. This kinda supports 2-x^3 leq x since x must be positive to be a solution of the system.
If x≤y then g(x)=y ≥ x = g(y). Unfortunately, this gives nothing new. It would work if g to be increasing. Trying to get Michael's reasoning: from x=g(y)≤y=g(x) follows x≤g(x), g(y)≤y. However, g(x) ≥ g(y) for decreasing function. Hm. I can't close these inequalities to get something useful... Michael, help! 😅
@@anshumanagrawal346 Apparently noboby agrees on 0 being a natural number or not. N* is the set (not sure about the word in english) of all positive integers excluding 0. and N is the set of all positive integers including 0. Looks like the definition of N differs between colleges and countries ^^
btw if p(x)= x³+x-350 , then the required expression is k = p(0)p(1)p(2)...p(350) but p(x) = (x-7)(x²+7x+50), hence p(7)=0 {yeah I was blind to the fact that 7³-343 is a factor in the product} and k = 0
I like it when Michael doesn’t edit out mistakes or misspoken bits. Shows us all that he too does what we all do: mistakes when lecturing.
It makes it more real 👍
>be michael
>pick an ordering for l, m, n at 9:49
>use the exact opposite ordering instead
(love you michael, just poking fun!)
Then he says 9+4+4=25
🤣
Okay, so it wasn’t just me...came here after the normal Groups video and was all too prepared to be totally confused
@ 2:01 what???! who says always g(x) g(x)=3. Unless you have restricted yourself only to positie numbers which was not the case. There is a "correct" way to show that x=y gives one set of solutions
This is how I approached it. x^3+y=2 implies one of x^3 and y is at least one, and the other is at most 1. Since x^3>=1 implies x>=1 and x^3=y^3, so for x^3+y=x+y^3=2 to hold, equality must hold for both inequalities and (1, 1) is the only solution. If -18, but then y=2-x^3
@@Examinx Nice!
@@Examinx Brilliant use of inequalities (I don't use the word "Brilliant" often )
I wish I would have thought of this method to approach the problem
@11vives it just means that one of the numbers is between 0 and 2, so is the other. Like, one may be 1.3 and other may be 0.7. It really means that the numbers are both positive.
@11vives One way to think of it is that the average of two numbers (x^3 and y) is 1, so the average is in between both numbers. Rigorously, you could just set up a proof by contradiction.
02:04 why g(x) is less than or equal to x??
I guess he forgot to say that that's true for x geq 1.
@@alainrogez8485 That doesn't follow at all. In fact, for the functions listed, when x< 1, (such x definitely exist because g takes all reals as its domain) g(x)> x. For example g(0) = 2 > 0.
What is only true is that for a given decreasing function f, there exists some y such that if x>y, f(x) < x.
@@bencheesecake yes, my apologies. I am French and sometimes, it is difficult to understand math and English in the same time.
You are right, when g is decreasing : x g(x)>g(y) and not g(x)
g(x) being decreasing does not imply g(x) ≤ x. Consider x=-1 for a counter example. I would tackle this through graph sketching and then algebra for a precise solution.
True, g(x) is not always less than x. if x=-2 g(x)=2-(-8)=10 also elsewhere someone says the inequality should be the other way but that's not true either because g(x) crosses the line y=x
I think what he meant was that since g(z) is a decreasing function, g(x) ≤ g(y) since x≤y
But since g(y) = x, we get g(x)≤x
@@anantkerur557 g(z) is a decreasing function means that for any x,y if x=g(y), not the opposite.
@@michelesetnikar8863 oh yeah! Thanks for catching that
There is also 8, 6, and 3 (a=90, b=12). So, there's 4 solutions.
yes, basically he messed up in this video in many places
@@amirb715 Yep!
Yeah, I saw that too.
i have a question , why did he eliminate all 0 roots ? Cant 0 be counted as a root to an equation ?
@@Khalibi If 0 were a root, the constant term (-b^2) would have to be 0, but since b is a positive integer, the constant term can't be 0.
This can be seen with the rational root theorem used in the video. 0 is a rational number, so if it were a root, it would have to be a factor of the constant term. 0 multiplied by anything is 0, so if 0 is a root, the constant term must be 0.
If g was increasing, then x
"Particularly for g(x)=2-x^3 the result happens to be true" : ... but no ! g(-2)=10 which is > -2
@@egillandersson1780 You are right. I mean "the end result", i.e. that the only intersection point is (1,1). Of course g(x)
Can somebody explain where the g(x)=1
Same, i don't understand it...
@@eta.tauri32 same here also!
We suppose x=g(y), in other words, g(x)>=x
He made a mistake in the sign, it should be greater or equal
@@coc235 (A late reply). Yes, but doesn't this invalidate the proof? We have x=< y (from assumption), and g(x) >= x, ie y >-= x, which is the same as x=
At about 4:00 no need to check for multiple roots as g(x) is monotonic.
At 8:00 the last step of the lemma, i didn't get. If c
He didn't say c
@3:30 if the value of the coefficients add to zero, then x = 1 is a solution.
The reasoning at 1:57 is completely wrong, from x
Ohh thank god someone else had the same problem , so does this mean his solution is incorrect ?
@@srijanbhowmick9570 Yes, he finds only those solutions where x=y, or in general x1=x2=...=xk [if you have more than two variable].
X=2 and y=-1 is not the solution, 2 does not equal to 2-(-1)^3
He made a mistake in the sign, it should be greater or equal
@@coc235 It is a solution of the same kind of system:
y=5/3-x^3/3
x=5/3-y^3/3
[The third real solution is x=-1;y=2.]
This system shows the problem with presented solution, notice that g(z)=5/3-z^3/3 is also a decreasing function, and the system is symmetric in x,y.
9:55 opposite order; should be: l >= m >= n.
2:00 I don't get it , why did you take g(x)
This is reasoning mistake. Happens.
@@danielmilyutin9914 So does this means that his solution is incorrect ?
@@srijanbhowmick9570 no, it is correct but Mike doesn't do well in this one. Read other comments, some members proved y as to be x.
@@srijanbhowmick9570 if you mean answer. Answer is correct. But reasoning towards answer is not clear to me. And to many other people, too, if check comments.
We suppose x=g(y), in other words, g(x)>=x
For the first problem, I just took the first equation minus the second. The obvious solution is to go ahead and factor (y-x) from both sides. This leads to your x=y solution immediately. Then for the case when y>x, go to the step before we said y=x and divide both sides by y-x leaving you with x^2+xy+y^2+1=0. It is very easy to manipulate this equation in two ways. First, complete the square to get (x+y)^2=xy-1. The second way is to isolate the squares on one side leaving you with x^2+y^2=-xy-1. Then it should be obvious to add these two equations to arrive at (x+y)^2+x^2+y^2=-2 which has no solutions in the real numbers. Thus we are finished.
I don't think this is quite right. When you factor out (y-x), you get x^2+xy+y^2=1. This is an ellipse with many solutions, including (0,1), (1, -1), which are not solutions to the original problem. That's expected, since subtracting the two equations gives us a third equation, but the overall solution still needs to be a simultaneous solution to 2 of those 3 equations. This does reduce the original problem from a 9th order polynomial (intersection of two cubics) to a 6th order one (the intersection of a quadratic and a cubic).
@@matthewself2811 Oh jeez, you're right, I made a sign error 🤦
For the first problem showing that x = y you could do it like this:
Assume x
Nice!
Or we can realize that both the curves are reflections of each other along the line y=x. So if a point of intersection exists, it lies on y=x.
You started from x = x, which is the same thing...
@@federico4475 exactly. If g(z) to be increasing we could close these inequalities to imply x=y. But it is decreasing.
Wait a minute, here. Assuming (1) x
Great work . Could you help me with this?
Let k be the positive root of x^2 +x-4=0 and the polynomial P(x)= ax^n +a'x^n-1 +a''x^n-2 ....+a0 which has non negative coefficients and an arithmetic value of P(k)=2017
1) prove that a0 +a1+......an =1 mod 2 (I can't find the right symbol with 3 slashes)
2) Find the least possible value of a0+a1+.....+an
Where is this task from?
Why does g(z) being decreasing imply that g(x) is less than or equal to x?
For example, g(0) equals 2 , which is strictly greater than 0 .
He made a mistake in the sign, it should be greater or equal
We suppose x=g(y), in other words, g(x)>=x
@@coc235 Okay, but how does this get us to y being less than or equal to x which in turn is less than or equal to y?
g(x) = y, so g(x) being greater than or equal to x is exactly what we have assumed before.
@@xCorvus7x I guess that mistake in the sign ruined the whole proof, im not sure though
@@coc235 Well, fuck.
And according to WolframAlpha there are complex solutions where x and y aren't equal, so the supposed fact that x = y is just untrue, I guess.
In the second problem, if 'three integer roots' means 3 different values, we get
(a,b)=(80,10) with roots (2,5,10) and
(a,b)=(90,12) with roots (3,6,8)
And for those considering 0 a natural number: we find eight more solutions: b=0 and a in {16,30,42,52,60,66,70,72} and roots (0, (17 +/- sqrt(289-4a^2))/2)
(x^2 + а)*x=17x^2 + b^2
Since the right side is positive, so should be the left side. It can only happen if x is nonnegative because a is natural number.
It in fact has to be positive, it can't be other because then the Left Hand Side is 0
you missed the solution where roots are (3,6,8)
at 1:45, why is z a decreasing function? I've read some responses, but still don't understand why he/we can say that. I also noticed that Prof Penn hardly ever responds to comments....
For the first problem, you can show that the two curves only intersect at x=y=1 by considering a line segment that separates them. I picked the line segment that is tangent to one of the curves where they meet: y = (4-x)/3, where x1. So, the second curve also does not intersect the line segment.
Since the curves are on opposite sides of the line segment, they cannot intersect each other in the region x1. A similar argument shows that they also do not intersect in the region x>1, y
What about (8,6,3) with 8+6+3 = 17 and 8*6*3 = (8*2)*3*3 = 12 squared?
Why can't we have 8,6,3?
For the triples of (1,8,8) and (9,4,4), isn't that really 2 integer roots and not part of the desired solution?
"By natural numbers I mean positive integers." Might be the third most-said thing on this channel, right behind "And that's a good place to stop," and "okay."
I know, right?
It is good he says it means positive integers because there are some vids which mentions natural numbers but they include the zero. This way he avoids confusion.
@@williamperez-hernandez3968 +1. Mathematics is about communicating ideas, and both definitions of the natural numbers are hugely popular, so it's best to just specify.
As compared to the “most said thing” when you walk into a room: Oh NO!
I think the Lemma in the second question does not supply _sufficent_ evidence that the polynominal must have three positive roots.
It only says if the polynomial has a real root, then it must be positive.
Applying Decartes Rule of Sign to the polynomial, then it can have 1 to 3 real positive roots and no negative real roots.
More work is required to show that the polynomial does indeed have 3 positive real roots. But of course, because the question says there are three integer roots, then we know we are dealing with 3 positive real roots.
However, the problem can be modified to find solution when the cubic has integer roots, which would be one integer root (and two complex roots), or three integer roots as specified in the problem.
For example, a cubic of the required form with one integer root is
x^3 -17x^2 + 17x - 16 = 0
where (a,b) = (17,4).
This polynomial has one integer root = 16.
In fact, if b^2 = 17a, and a = c^2, where c is a natural number, then we can generate an infinite number of polynomials with different values for a and b, all having x = 17 as its root and no other positive real root.
E.g. x^3 -17x^2 +612x - 10404 = 0, has x = 17 as its only real root. Here c = 6, so a = 612 and b = 102.
The curve y=2-x³ is a reflection of the curve x=2-y³ about the line y=x. Hence if there exists any point which satisfies both, it must lie on the line y=x. I think this is a less time-consuming method to get to the conclusion that y=x.
I thought that for any system were there is this type of symmetry then it is always the case that x=y. I have never seen anything different. I always just considered it an inherent characteristic of the symmetry. Is this what you think? Is there more going on here than I am unaware of?
@@tomasstride9590 I don't see anything more going on. I've just used extra steps or my language might be a bit confusing.
@@anonymous_4276 Your language is very clear and I am sure you understand the situation very well. When I see things like in the video were something I thought was obvious and well understood being laboured, it makes me doubt my own understanding and I feel insecure.
@@tomasstride9590 when I first made that comment, I basically wrote down the way I reasoned y being equal to x. I saw your comment and thought maybe it was obvious to you (that y equals x in such systems). I was just reasoning out why that happens in the original comment.
Oui, et les2courbes ont 1 seul point commun , c’est (1,1)
1st one
I would have started by subtracting the two equations
But you actually proved that x=y
5:50 - By natural numbers, I mean positive integers
Me too.
Don't quite get the point at 2:04 that g(x) decreases --> g(x) 0.01
How c³ -17c² + ac -b² is lessthan or equalto -b² ?!?
c^3 - 17c^2 = c^2(c-17) < 0 because c^2 is positive and c
@@matthewryan4844 oh ,,, ty...
12:58 Good Place To Stop
why is g(x)
We suppose x=g(y), in other words, g(x)>=x
He made a mistake in the sign, it should be greater or equal
Good video, thanks
I was shocked too about the g(x) if x>=y then g(x)
If g is decreasing x = g (y). You have the second inequality the other way round.
@@tommike2548 thanks. I edited my answer
He made a mistake in the sign, it should be greater or equal
Clearly x=y is a solution but the "all solution" part would be the hard part, i.e proving no other solutions exist
(8,6,3) is a solution
We can say the two equations are the inverses of each other so that the solution lie along y=x...furthermore from graph we see that this is the only possible case(for Q1)
Around 2:00 that comment about g(x) being a decreasing function and so g(x)=1.
However, if you graph g and g^-1 the coordinates of any intersections provide solutions to the problem and you will see that they indeed only intersect at one point.
The solutions here have coordinates which are roots of g(g(z))=z which is a degree 9 polynomial equation; is the solution presented here a single root or is it multiple equal roots?
For a more general g(z)=a-z^3 there could be 3 distinct intersection points. There certainly are when a=0. As a increases the three intersection points approach each other. Here's a problem: At what value of a do the three solutions all become equal? Beyond that value of a do they remain a triple of equal solutions, or do two of them become complex?
Ohh thank god someone else had the same problem , so does this mean his solution is incorrect ?
and can we not solve this question with pure algebra and no graphs ???
@@srijanbhowmick9570 He made a mistake in the sign, it should be greater or equal
@@srijanbhowmick9570 We suppose x=g(y), in other words, g(x)>=x
I havent really thought about it but can someone tell me if this approach is wrong if we replace 2 with y+x^3 in the second equation we get x=y+x^3-y^3 which means x-y-x^3+y^3=0 which means that x-y-(x-y)(x^2+xy+y^2)=0 which means (x-y)(1-x^2-xy-y^2)=0 so x=y or x^2+xy+y^2-1=o which means x=(-y±√(-3y^2+4))÷2 for -√(4÷3)≤y≤√(4÷3)
I followed this line of approach too. I got the same possible values for x in terms of y as you did.
However, when you find the values for x from the x^2 + xy + y^2 = 1 equation and place then back into x = 2 - y^3, then when you get rid of the square root, you end up with a polynomial of degree 6 with integer coefficients.
You can in fact prove that this 6th degree polynomial has no real value of y which is a root, which is harder to do that the original problem, or you can just graph it in Desmos and see it has no real roots.
So, even although x^2 + xy + y^2 = 1 gives two values of x that satisfy it in terms of y, it turns out that these solutions for y don't satisfy the original equations in real numbers.
So, basically, x^2 + xy + y^2 = 1 yields *no* more real solutions.
@@davidbrisbane7206 ok got it but I also have another question if we have two polinomials p(x)=p(y) does it mean that for those two to be equal always as fanctions x=y or is there another imput lets call it a(a is variable) that can satisfy this
@@mrhatman675
Let p(z) = (z - 1)(z - 2).
Let x = 1 and y = 2.
So, p(x) = p(y), but x y.
Taking differences of the equations, we get (y-x)(x^2+xy+y^2+1)=0. Because x^2+xy+y^2+1 = (x+y/2)^2+3y^2/4+1 > 0, we get x=y.
No, actually, x = 2-y^3 and y = 2-x^3. So, y - x = 2-x^3 - 2 + y^3 = y^3 - x^3 = (y-x)(x^2 + y^2 + xy)=> Either x = y or x^2 + y^2 + xy -1 = 0. Note that it is -1 and not + 1 here. Now your argument does not hold true.
Oh, good catch.
This is a fix of the argument.
if x!=y, WLOG x < y and it holds
x^2 + xy + y^2 = 1 (*)
case 1: xy >= 0:
It holds x^2 < y^2 1 which is inconsistent to y^2 0
y = 2 - x^3 > 2, however form (*)
(x+y/2)^2 = 1 - 3y^2/4 < -2 which is inconsistent to x+y/2 is real.
@@HideyukiWatanabe Yes. That does make sense.
By the way, I just brute forced it and solved the cubic equation resulting from solving for (x+y). Indeed, the solution for x^2 + xy + y^2 = 1 leads to a complex root for x and y. Once I saw the result, I knew there must be way to prove it and you have certainly shown the way.
try to solve if we replace 2 with parameter a. and find for which a values the equations has 1, 3, 5 solutions. this is more interesting problem/
The fact that all three roots of the cubic must be positive can be also seen using the Descartes' rule of signs
Use 1 = ( 1 + x + x^2)(1 + y + y^2) >= 3 * 3/4 -> 1 or y > 1
At 2:00 the fact that g(x) is a decreasing function only tells us that when x= g(y). That's the step he missed out.
Now, if we *also* know that x and y satisfy y = g(x) and x = g(y), *then* we can substitute x for g(y) and arrive at his assertion g(x)
Well, 17 = 13+4+1 and then if I get the lmn = 52 is a perfect squares that can be allowed.
f decreasing x = g(y)
g(y) = 2 - y^3= x
Then g(x) >= x
Not g(x)
@ 6:50 Descartes rule of signs tells you straight away that all three roots must be positive, no need to prove.
I think Descartes rule is not known by some wievers. So, Michael explains this point very well.
2:00 Not true; e.g. g(-100) = 1000002 > -100.
He made a mistake in the sign, it should be greater or equal
It may be not true for all x, but we have to use given equations which restrict the value of x
10:51 "9+4+4=25"???????
I was wondering the same thing
Why can you assume that g(X)≤x? For x = 0 you get 2≤0 which is not true.
I guess you are excluding 0 from N, because (l, m, n) = (16, 1, 0) seems the simplest solution.
that would give a = 16, b = 0.
Yes, he says as much at 5:49.
{8, 3, 6} is also possible. So, there are actually 4 solutions.
No, as (8,6,3) and (8,3,6) for (l,m,n) result in the same solution for a and b; (a,b) = (90,12)
@@christophniessl9279 i know (8, 3, 6) and (8, 6, 3) are the same. I did not argue with that. My concern is that he only gave (10, 5, 2), (9, 4, 4), and (8, 8, 1). I hope you get my point.😊
Decreasing does NOT mean that the function is less than its variable! decreasing means that as the variable increases the function decreases, nothing else!
Good Place To Start at 0:01
Don't get the g(x) < x. Say x is 0 then g(x) = 2 which is greater than x.
Just looked at other comments ; lots of other people noticed it too,
10:52 9+4+4 is not equal to 25!!! 😫
Got lost at min.8...gnight
please make a video about how we dont know whether pi+e is rational or not, and how pathetic that is
Good
The more I look at the first question the more I realize that graphing gives the fact that there’s only one solution right away and that it’s in the first quadrant. This kinda supports 2-x^3 leq x since x must be positive to be a solution of the system.
If x≤y then g(x)=y ≥ x = g(y). Unfortunately, this gives nothing new. It would work if g to be increasing.
Trying to get Michael's reasoning:
from x=g(y)≤y=g(x)
follows x≤g(x), g(y)≤y.
However, g(x) ≥ g(y) for decreasing function. Hm. I can't close these inequalities to get something useful...
Michael, help! 😅
You're right, just substitute g(y)=x into the last inequality. Michael actually messed up with the sign, it should be >=
thanks for your videos, although sometimes you make certain mistakes, the comments help me to know more.
thanks for the trick bro, very well done
If a=0 and b=0, you also end up with 3 integers roots :)
a and b are natural numbers, 0 is not a natural number
@@anshumanagrawal346 If 0 were excluded from the solutions, I would have expected a,b to be included in |N* and not in |N
@@laurentchi1062 what's N* ?
@@laurentchi1062Just google "is 0 a natural numbers?" You'll realise that the natural numbers DO NOT Include 0, you're thinking of Whole Numbers
@@anshumanagrawal346 Apparently noboby agrees on 0 being a natural number or not.
N* is the set (not sure about the word in english) of all positive integers excluding 0.
and N is the set of all positive integers including 0.
Looks like the definition of N differs between colleges and countries ^^
Are you building a house?
He's refinishing his basement.
I get headache when I listen to him speaking.
Get yourself together. So many mistakes in a single video.
"Always a trick " only give 2 examples no explanation how to do otherwise so bad title
HOMEWORK : Express, as concisely as possible, the value of the product (0^3 − 350)(1^3 − 349)(2^3 − 348)(3^3 − 347) ... (349^3 − 1)(350^3 − 0).
SOURCE : 2002 HMMT - Guts Round
0 because 7^3 = 343
0
@@mrpenguin815 good spot
@@mrpenguin815 Well you didn’t fall into the trap 😛
btw if p(x)= x³+x-350 , then the required expression is k = p(0)p(1)p(2)...p(350)
but p(x) = (x-7)(x²+7x+50), hence p(7)=0 {yeah I was blind to the fact that 7³-343 is a factor in the product} and k = 0