This problem have restricted the domain & codomain of f being the reals, but since the thumbnail didn't include this, I also researched the case where f is a map over a ring(You might also find a lot of redundant conditions in the question if it is over reals, so I proposed the weakest possible form(that I am able to find) here): Here is my generalized claim: For any ring 𝔽(not necessarily a commutative ring, and not necessarily has the multiplication identity element) which contains a multiplication-non-degenerate element(i.e. it is bijection in terms of multiplication on 𝔽, its inverse map does not necessarily to be an element of the ring 𝔽) f(x)f(y)-a f(xy)=x+y holds, where x,y∈𝔽, f:𝔽→𝔽, a is a 𝔽-linear map 𝔽→𝔽 (a∈End(𝔽)), iff "a" commute with any element in 𝔽 as multiplication map in End(𝔽), a²=I (identity map on 𝔽 also the multiplication identity element in End(𝔽) ; this also implies that a⁻¹ exist) and f=xa⁻¹+a Proving the claim from right side to left is rather easy, just substitute the f(x) expression and use the commutation relation of a. To prove the claim from the left side to the right, let "g" be the non-degenerate element, then according to its bijection property, its inverse map exists, denoted as g⁻¹, is also an element of End(𝔽) since g and I are both elements of End(𝔽). Follows from f(x)f(y)-a f(xy)=x+y, let y be 0, we have (f(x)-a)f(0)=x Let x be g, then (f(g)-a)f(0)=g, left multiply g⁻¹, we have: g⁻¹(f(g)-a)f(0)=I, thus f(0)⁻¹ exist in End(𝔽) and takes the form g⁻¹(f(g)-a), f(0) is also non-degenerate. Now, right multiply f(0)⁻¹ to (f(x)-a)f(0)=x, we have f(x)-a=xf(0)⁻¹, thus f(x) takes the form xf(0)⁻¹+a Plugin x=0, then we have f(0)=a, thus f(x)=xa⁻¹+a Back to f(x)f(y)-a f(xy)=x+y, then the commutator [f(x),f(0)]=f(x)f(0)-f(0)f(x)=0, substitute f(x)=xa⁻¹+a and f(0)=a, we have [a,x]=ax-xa=0, thus a commute with any element in 𝔽. Substitute f(x)=xa⁻¹+a into f(x)f(y)-a f(xy)=x+y, using the commute relation of a, we would have the final constraint on a: 0=(a⁻²-I)xy. Since this holds for any x,y∈𝔽, a⁻²=I must hold. Inverse both sides, we have a²=I, and Q.E.D.
@@user-en5vj6vr2u a map which has its domain and codomain be a ring(𝔽 for instance here), could be regarded as a map from and to the underlying set of the ring.
7:29, 8:06 Errors are teacher's gold. They give you the opportunity to compliment the student who spots one, while admitting to your mistakes and show the importance of adjusting science to facts.
one trick commonly in doing functional equations is to make one side of the eq = 0, thus we sub y=-x: f(x)f(-x) - af(-x^2) = 0 and then to find certain values, we can make some terms equal [i.e. make f(?)=? or f(?)=? situation*] in this case, you can let -x^2 = x/-x, which leads to values x = 0/1/-1, in the video, Michael did x=0, thus here we will use x=1 as another example (x=-1 works identically). sub x=1, y=-1: f(1)f(-1) - af(-1) = 0 f(-1) = 0 or f(1) = a [*] if f(1) = a, sub x=y=1: 0 = a^2 - a^2 = f(1)f(1) - af(1) = 2, a contradiction. thus f(-1) = 0, sub y=-1: f(x)f(-1) - af(-x) = x-1 -af(-x) = x-1 -af(x) = -x-1 f(x) = (x+1)/a sub this to the original eq: (x+1)(y+1)/a^2 - a*(xy+1)/a = x+y (x+1)(y+1)/a^2 = xy+x+y+1 = (x+1)(y+1) a^2 = 1 a = 1 or a = -1 thus f(x) = x+1 or f(x) = -x-1. after checking, these are the only solutions to the problem.
This can be sorted out quickly if one puts y=1, which yields f(x) * (f(1)-a) = x + 1 . From here f(x) could only be A(x+1), nonexistent otherwise. Therefore we claim f(x) = A(x+1), and substitute back in the original equation, which yields A^2 = aA = 1. From here follows a = \pm 1.
I like this solution, very elegant! If you make initial assumption that f(x) is a polynomial it is quickly visible it must be of first degree (otherwise odd powers will never councel each other out) and then solving with y=Ax+B you will get a=A=B and AB = 1 giving the same result. But it doesn't prove that polynomial is the only possible solution...
I very rarely make suggestions, so please accept my words in the spirit in which they are offered. Your errors are instructional for beginning and intermediate students, and I recommend not editing them out; you should even draw attention to them when they are recognized. Editing mistakes produces a sense of discontinuity in the reasoning, especially for those students who need more guidance in constructing their own reasoning. There is never a shame in being wrong; it is the shortest path to correctly modifying one's hypothesis. Of course, your on-screen annotations of errors not caught during filming have been satisfactory. I wanted to let you know what I thought works well also.
@@gregorymorse8423 Naturally, the above comment is my opinion. But habitually checking oneself and rectifying one's own errors are key components of the mathematical discipline, and are as worthwhile to demonstrate as any core concept in math.
You said you “squared in yellow” the two cases for f(0), but as you drew yellow rectangles and “encircled” means “has a circle drawn around” I would propose that you have “entangled” both cases in yellow.
Maybe it's just me but my first instinct is not to restrict both variables at the same time. For example I'd just do x=0 and x=1 yielding f(y)=y/f(0)+a and f(y)=(y+1)/(f(1)-a). This shows that 1/(f(1)-a)=a so f(y)=a(y+1) and from here you get a^2=1. Restricting both variables also loses both variables and leaving one in actually allows you some nice maneuverability.
I agree, this is exactly what I did. Additionally, this yields the solution f(x)=ax+a, whereas Michael’s solution is equivalent to f(x)=(1/a)x+a, yielding a=1/a. Immediately, we see the only solutions are +1 and -1
Is checking necessary? We got these functional equations using the previous form of them in terms of a and finding what a must be. How could these possibly fail?
I say it's a good practice to check, in case you made any mistake in your calculation. Anyway, I just checked that both answers are ok. It took me less than five minutes.
@@luisaleman9512 I’m not saying it’s a bad thing to do or would take long I’m asking is it mathematically necessary? As in, assuming one hasn’t done any mistakes, is it possible one would be left with incorrect solutions after this process without checking?
@@Happy_Abe I agree that in this specific problem doesn't seem to be any possibility of incorrect solutions. Still doesn't hurt to check if you don't have much experience with this kind of problems
@@luisaleman9512 yeah again I agree checking on s good Just Michael made it sound like the step was never and I’m trying to make sure I’m not wrong in thinking it’s not
As long as every step is reversible *and* has been done without mistakes, then you don't need to go back and check. I, for one, am never sure that I've done everything correctly so checking is always good, and it's also easy to not realise that a step only works in one direction.
It was a first time I'm watching your videos I've seen a thumbnail picture, I tried to do this by my own without even clicking on a video and I've done it correctly :) I'm proud of myself!
another way is to let x=1, and so we see that f(x) is a linear function with slope and y intercept both equal to 1/(f(1)-a). then set x=0 so we get that the slope is just 1/f(0) and the y intercept is a. then using these we see that f(0)=f(1)-a and f(0)=a=1/(f(1)-a). so a= 1 or -1
Haven’t watched the video- from the thumbnail alone: By putting y=0, we get c*f(x) = x+d for some constants c and d. c can’t be zero by varying x, so f is linear - and we can just sub in and eg compare coefficients. There’s probably some intermediary work you can do to simplify the final substitution. EDIT: yeah ok, this is basically what Michael does
About your very last point - How could they not work? You basically plugged in that form to determine for the formula for a? I don't see anyway in which they are not going to work.
@@praharmitra there could be a situation where we've made an implication in one direction which is not necessarily true in the other direction. Not the case here, but it's certainly possible.
I didn't have to do power of 4 or fractional multiplications to get a² = 1 f(0)f(1) - a f(0) = 1 f(0)² - a f(0) = 0 f(1)² - a f(1) = 2 Three equations for f(0), f(1) and a f(0) [ f(1) - a ] = 1 f(0) [ f(0) - a ] = 0 f(1) [ f(1) - a ] = 2 Second equation implies either f(0) or [ f(0) - a ] is 0, but then f(0) cannot be 0 in order to satisfy the first equation. So we are left with [ f(0) - a ] = 0, i.e. f(0)=a Then we get a [ f(1) - a ] = 1 f(1) [ f(1) - a ] = 2 So, f(1) = 2a, which we then substitute to the first equation to give a² = 1 Therefore, a can only be ±1 Last step (find actual functions): f(x) f(0) - a f(0) = x + 0 f(x) a - a² = x f(x) = x/a + a For a=1: f(x) = x+1 f(x) f(y) - a f(xy) = (x+1)(y+1) - (xy+1) = x+y For a=-1: f(x) = -(x+1) f(x) f(y) - a f(xy) = (x+1)(y+1) + [-(xy+1)] = x+y Both functions work. So, even though we at first obtained a only from the restrictions on f(0) and f(1), we know that the result works for all other values as well.
Not Michael (obviously), but I think his notation comes from the contradicting arrow notation (→←). He draws it quick by drawing a horizontal line (the stems of the two arrows) then an x (which makes the heads of the two arrows).
Not that simple, alas, because we don't know what f(x) is initially, so we don't know that the expression on the right-hand side always has the same value regardless of the values of x and y.
If F(x) is x+1, Therefore F(y) is y+1 and F(xy) is xy+1. If you plug these three in the eqn F(x)F(y)-1(F(xy))=x+y (Since a is posi-nega 1 (But I labeled variable a as 1)). That Eqn will be simplify as. (?) labels equal or not equal (x+1)(y+1)-1(xy+1) (?) x+y (xy+x+y+1)+(-xy-1) (?) x+y x+y (xy and -xy adds up to 0, same as 1 and -1 which adds to 0) = x+y Thus F(x) = x+1 is a true function to satisfy this eqn
The proof showed that the only possible solutions were 1 or -1, but sometimes we might introduce false solutions by squaring a term without noticing. It's usually sensible to check both solutions, particularly when they arise from taking a square root.
Hello, im a high school student and the problems i see on your page seem hard and i wonder what level of maths your problems could be given. Are these high school maths?
It depends, but aside specific theorems and some calculus type questions, its high school level stuff, (by high school level I mean contest level obviously)
Solution before watching the video (eq.1) : f(x)f(y) - a f(xy) = x+y Substituting x=0, y=0 in eq.1 (eq.2) : f²(0) - a f(0) = 0 Which means that either f(0)=0 or f(0)=a Substituting y=1 in (eq.1), and letting f(1) = z (eq.3) : f(x) = (x+1)/(z-a) Substituting x=0 in (eq.3) we see that f(0) = 0 was not a valid option, and therefore f(0) = a. We can express z in terms of a by substituting x=0 in (eq.3) (eq.4) : f(0) = a = 1/(z-a) (eq.5) : z = (a²+1)/a We can also substitute x=1 in (eq.3), yielding (eq.6) : f(1) = z = 2/(z-a) (eq.7) : a = (z²-2)/z Now we have two equations relating a and z, and we can solve them by substituting (eq.5) into (eq.7) (eq.8) : az = z²-2 (eq.9) : a²+1 = z²-2 (eq.10) : a²+3 = z² (eq.11) : a⁴+3a² = a²z² (eq.12) : a⁴+3a² = (a²+1)² (eq.13) : a⁴+3a² = a⁴+2a²+1 (eq.14) : a² = 1 (eq.15) : a = ±1 And we can verify by substituting both a=1 and a=-1, and (eq.3) into (eq.1) to see that those are the two values that are allowed
I.e., b³ - b - 5 = 0 A cubic with one real & 2 complex roots. Graphing that, shows the real root just below 2. Writing b = ∛(b+5) you can start with b=2 and iterate, to find b = 1.904160859134920603676... which can be "divided out" of the cubic, giving a quadratic from which the (approximate) complex roots can be found. Could also apply the cubic formula to get exact answers. Or try submitting the cubic to Wolfram Alpha, and it might give all 3 roots in both symbolic & numeric form, to many places. Of course, once you have b, you simply get a = b-1, c = b+1. Fred
Let f(1)=b, then we have f(x)(b-a)=x+1. Hence b is not a (take x=0) and f(x)=c(x+1), where c=1/(b-a) is not zero. Thus f(x)f(y)-af(xy)=c²(x+1)(y+1)-ac(xy+1)=x+y. Hence c²=ac (from x=y=0) and then c²(x+y)=x+y for all x,y (take x=1, y=0). So a=-1,1. And (x+1)(y+1)-(xy+1)=x+y, and (-(x+1))(-(y+1))+(-(xy+1))=x+y.
I found that only a that makes it for all x's and y's is a=1, and then f(t)=t+1. Now it's time to view the video. // and I was wrong. I lost the a=-1 case and f(t)=-t-1.
@@gregorymorse8423 no, no. x and y both are reals, but don't depend on each other. Likewise, another real a may or may not depend on any of them. The fact all three of them are reals doesn't say absolutely anything about their dependencies. If we allow a to depend on x and y (that makes is not a parameter, but just a placeholder in the equation), then yes, there may be such "solutions". But it is called a parameter, which means it must be completely free and requires an independence. The question was: what are values of a, for which the functional equation holds for any x and y? I understand it this way: first we fix some a, and then independently vary x and y, but the functional equation should be still true. Which value of a we had fixed? With dependend a this is, of course, impossible, because we aren't able to fix a in the first place. All the problems with parameter I used to have so far (I had many) assume this parameter could not depend on x or y, so why this particular problem suddenly become an exception?
@@gregorymorse8423 Ok, so it is held as a constant. We choose a and hold it constant, then vary x and y. The only way for this to be possible is if a doesn't depend on x and y (remember, those both also vary independently of each other). How can you depend on something and stay fixed when it changes? This means "to not depend". The "slowly changing" passage is for very interesting special case, a variation methods or iterative approximation methods. For example, consider a linear nonhomogeneous differential equation solving process. First you solve a corresponding homogeneous equation, and get a general solution with some parameters. Then you vary those parameters, considering them as functions and resolve which functions they could be. Consider an oscillator with decay very far from being damped, so there'll be a periodic sine (fast changing) with the slowly exponentially decaying amplitude. We sometimes call this quasiperiodic solution, because it is almost periodic, so we consider the amplitude as slowly changing parameter in this case. All in all, this is not about the problem in the video and your suggested "solution", because in this case a will be changing exactly at the same rate as variables, not being any slower. Yes, the definition you've found forbids simple dependence. It forbids strictly in problems like this.
Michael Penn showed from 1:38 to 2:31 that f(0) = 0 or f(a) = 0. He then (by around 3:30 ) showed that f(0) = 0 leads to a contradiction. So f(a) must be equal to 0. It should be clear that if f(a) = 0, and f(0) is not equal to 0, then a cannot be 0. The nature of these sort of problems is that x and y must be free to take on any real value, but the parameter a is constrained to a particular set of values - in this case only 1 or -1.
If a=0, the equation is reduced to f(x)f(y) = x+y, which doesn't have any solutions, to see this, just plug in x=y=0 first, and then x=0 and y=1. In fact, this case has been covered in the solution because he shows that f(a) must be 0 and f(0) must not be 0
as "a" is in nominator a cannot be 0 in final case so a=0 has to be cosidered seperately since all previous operation would be ilegal. But is needed to Remind we earlier called f(0) = a, but for a=0 we have first case
Possibly he is the only one who has such an attractive thumbnail but not that attractive black board 🤣🤣.I think you should rather make some animation kinda thing.
@@kemalkayraergin5655 but I think animation creates a much bigger visual impact on the viewer and it also helps visualise the problem itself.i think it wouldn't be very good for this channel as it mostly posts number theory problems and not geometry problemo
This problem have restricted the domain & codomain of f being the reals, but since the thumbnail didn't include this, I also researched the case where f is a map over a ring(You might also find a lot of redundant conditions in the question if it is over reals, so I proposed the weakest possible form(that I am able to find) here):
Here is my generalized claim:
For any ring 𝔽(not necessarily a commutative ring, and not necessarily has the multiplication identity element) which contains a multiplication-non-degenerate element(i.e. it is bijection in terms of multiplication on 𝔽, its inverse map does not necessarily to be an element of the ring 𝔽)
f(x)f(y)-a f(xy)=x+y holds, where x,y∈𝔽, f:𝔽→𝔽, a is a 𝔽-linear map 𝔽→𝔽 (a∈End(𝔽)), iff "a" commute with any element in 𝔽 as multiplication map in End(𝔽), a²=I (identity map on 𝔽 also the multiplication identity element in End(𝔽) ; this also implies that a⁻¹ exist) and f=xa⁻¹+a
Proving the claim from right side to left is rather easy, just substitute the f(x) expression and use the commutation relation of a.
To prove the claim from the left side to the right, let "g" be the non-degenerate element, then according to its bijection property, its inverse map exists, denoted as g⁻¹, is also an element of End(𝔽) since g and I are both elements of End(𝔽).
Follows from f(x)f(y)-a f(xy)=x+y, let y be 0, we have (f(x)-a)f(0)=x
Let x be g, then (f(g)-a)f(0)=g, left multiply g⁻¹, we have:
g⁻¹(f(g)-a)f(0)=I, thus f(0)⁻¹ exist in End(𝔽) and takes the form g⁻¹(f(g)-a), f(0) is also non-degenerate.
Now, right multiply f(0)⁻¹ to (f(x)-a)f(0)=x, we have f(x)-a=xf(0)⁻¹, thus f(x) takes the form xf(0)⁻¹+a
Plugin x=0, then we have f(0)=a, thus f(x)=xa⁻¹+a
Back to f(x)f(y)-a f(xy)=x+y, then the commutator [f(x),f(0)]=f(x)f(0)-f(0)f(x)=0, substitute f(x)=xa⁻¹+a and f(0)=a, we have [a,x]=ax-xa=0, thus a commute with any element in 𝔽.
Substitute f(x)=xa⁻¹+a into f(x)f(y)-a f(xy)=x+y, using the commute relation of a, we would have the final constraint on a:
0=(a⁻²-I)xy. Since this holds for any x,y∈𝔽, a⁻²=I must hold. Inverse both sides, we have a²=I, and
Q.E.D.
Thansk👍 but what is a map over a ring
@@user-en5vj6vr2u a map which has its domain and codomain be a ring(𝔽 for instance here), could be regarded as a map from and to the underlying set of the ring.
7:29, 8:06
Errors are teacher's gold. They give you the opportunity to compliment the student who spots one, while admitting to your mistakes and show the importance of adjusting science to facts.
one trick commonly in doing functional equations is to make one side of the eq = 0,
thus we sub y=-x: f(x)f(-x) - af(-x^2) = 0
and then to find certain values, we can make some terms equal [i.e. make f(?)=? or f(?)=? situation*]
in this case, you can let -x^2 = x/-x, which leads to values x = 0/1/-1, in the video, Michael did x=0, thus here we will use x=1 as another example (x=-1 works identically).
sub x=1, y=-1: f(1)f(-1) - af(-1) = 0
f(-1) = 0 or f(1) = a [*]
if f(1) = a, sub x=y=1: 0 = a^2 - a^2 = f(1)f(1) - af(1) = 2, a contradiction.
thus f(-1) = 0, sub y=-1: f(x)f(-1) - af(-x) = x-1
-af(-x) = x-1
-af(x) = -x-1
f(x) = (x+1)/a
sub this to the original eq:
(x+1)(y+1)/a^2 - a*(xy+1)/a = x+y
(x+1)(y+1)/a^2 = xy+x+y+1 = (x+1)(y+1)
a^2 = 1
a = 1 or a = -1
thus f(x) = x+1 or f(x) = -x-1.
after checking, these are the only solutions to the problem.
I did exactly this.
Thanks, very cool
@@ahzong3544 no you're cooler bro
@Gregory_Morse You can't make a=1-x-y because it's not a function and its value cannot change.
I started with "Assume f() is the identity function" and found that "a" can't be any numeric constant.
8:07 smooth correction!
This can be sorted out quickly if one puts y=1, which yields f(x) * (f(1)-a) = x + 1 . From here f(x) could only be A(x+1), nonexistent otherwise. Therefore we claim f(x) = A(x+1), and substitute back in the original equation, which yields A^2 = aA = 1. From here follows a = \pm 1.
8:05 Sneaky correction of - into +
Saw that too
Now that's real math wizardry
I like this solution, very elegant! If you make initial assumption that f(x) is a polynomial it is quickly visible it must be of first degree (otherwise odd powers will never councel each other out) and then solving with y=Ax+B you will get a=A=B and AB = 1 giving the same result. But it doesn't prove that polynomial is the only possible solution...
I very rarely make suggestions, so please accept my words in the spirit in which they are offered. Your errors are instructional for beginning and intermediate students, and I recommend not editing them out; you should even draw attention to them when they are recognized. Editing mistakes produces a sense of discontinuity in the reasoning, especially for those students who need more guidance in constructing their own reasoning. There is never a shame in being wrong; it is the shortest path to correctly modifying one's hypothesis.
Of course, your on-screen annotations of errors not caught during filming have been satisfactory. I wanted to let you know what I thought works well also.
@@gregorymorse8423 Naturally, the above comment is my opinion. But habitually checking oneself and rectifying one's own errors are key components of the mathematical discipline, and are as worthwhile to demonstrate as any core concept in math.
Outstanding video, Michael. Well done.
10:01
Very interesting solution!
Thank you for making those great videos!
You said you “squared in yellow” the two cases for f(0), but as you drew yellow rectangles and “encircled” means “has a circle drawn around” I would propose that you have “entangled” both cases in yellow.
Maybe it's just me but my first instinct is not to restrict both variables at the same time. For example I'd just do x=0 and x=1 yielding f(y)=y/f(0)+a and f(y)=(y+1)/(f(1)-a). This shows that 1/(f(1)-a)=a so f(y)=a(y+1) and from here you get a^2=1. Restricting both variables also loses both variables and leaving one in actually allows you some nice maneuverability.
I agree, this is exactly what I did. Additionally, this yields the solution f(x)=ax+a, whereas Michael’s solution is equivalent to f(x)=(1/a)x+a, yielding a=1/a. Immediately, we see the only solutions are +1 and -1
Is checking necessary?
We got these functional equations using the previous form of them in terms of a and finding what a must be.
How could these possibly fail?
I say it's a good practice to check, in case you made any mistake in your calculation. Anyway, I just checked that both answers are ok. It took me less than five minutes.
@@luisaleman9512 I’m not saying it’s a bad thing to do or would take long
I’m asking is it mathematically necessary?
As in, assuming one hasn’t done any mistakes, is it possible one would be left with incorrect solutions after this process without checking?
@@Happy_Abe I agree that in this specific problem doesn't seem to be any possibility of incorrect solutions. Still doesn't hurt to check if you don't have much experience with this kind of problems
@@luisaleman9512 yeah again I agree checking on s good
Just Michael made it sound like the step was never and I’m trying to make sure I’m not wrong in thinking it’s not
As long as every step is reversible *and* has been done without mistakes, then you don't need to go back and check. I, for one, am never sure that I've done everything correctly so checking is always good, and it's also easy to not realise that a step only works in one direction.
It was a first time I'm watching your videos I've seen a thumbnail picture, I tried to do this by my own without even clicking on a video and I've done it correctly :) I'm proud of myself!
..michael penn ..never can stop watching your amazing videos....
another way is to let x=1, and so we see that f(x) is a linear function with slope and y intercept both equal to 1/(f(1)-a). then set x=0 so we get that the slope is just 1/f(0) and the y intercept is a. then using these we see that f(0)=f(1)-a and f(0)=a=1/(f(1)-a). so a= 1 or -1
Haven’t watched the video- from the thumbnail alone:
By putting y=0, we get c*f(x) = x+d for some constants c and d. c can’t be zero by varying x, so f is linear - and we can just sub in and eg compare coefficients.
There’s probably some intermediary work you can do to simplify the final substitution.
EDIT: yeah ok, this is basically what Michael does
Can confirm both values hold, nice bit of cancelling of xy and 1 terms in there leaving x+y.
About your very last point - How could they not work? You basically plugged in that form to determine for the formula for a? I don't see anyway in which they are not going to work.
it's always good to double check. there might be an unexpected consequence of plugging them.
@@nirajmehta6424 double checking makes sense so that is fine. But there can never be any unexpected consequence.
@@praharmitra there could be a situation where we've made an implication in one direction which is not necessarily true in the other direction. Not the case here, but it's certainly possible.
I didn't have to do power of 4 or fractional multiplications to get a² = 1
f(0)f(1) - a f(0) = 1
f(0)² - a f(0) = 0
f(1)² - a f(1) = 2
Three equations for f(0), f(1) and a
f(0) [ f(1) - a ] = 1
f(0) [ f(0) - a ] = 0
f(1) [ f(1) - a ] = 2
Second equation implies either f(0) or [ f(0) - a ] is 0, but then f(0) cannot be 0 in order to satisfy the first equation. So we are left with [ f(0) - a ] = 0, i.e. f(0)=a
Then we get
a [ f(1) - a ] = 1
f(1) [ f(1) - a ] = 2
So, f(1) = 2a, which we then substitute to the first equation to give a² = 1
Therefore, a can only be ±1
Last step (find actual functions):
f(x) f(0) - a f(0) = x + 0
f(x) a - a² = x
f(x) = x/a + a
For a=1:
f(x) = x+1
f(x) f(y) - a f(xy) = (x+1)(y+1) - (xy+1) = x+y
For a=-1:
f(x) = -(x+1)
f(x) f(y) - a f(xy) = (x+1)(y+1) + [-(xy+1)] = x+y
Both functions work.
So, even though we at first obtained a only from the restrictions on f(0) and f(1), we know that the result works for all other values as well.
Hey Michael, I just wanted to ask about that contradiction symbol. What exactly is it supposed to be?
Not Michael (obviously), but I think his notation comes from the contradicting arrow notation (→←). He draws it quick by drawing a horizontal line (the stems of the two arrows) then an x (which makes the heads of the two arrows).
Could also be an asterisk
a=(f(x)f(y)-x-y)/f(xy) ? not that simple i guess?
Not that simple, alas, because we don't know what f(x) is initially, so we don't know that the expression on the right-hand side always has the same value regardless of the values of x and y.
Thank you, professor!
Nice. Greeting from Turkey.
If F(x) is x+1, Therefore F(y) is y+1 and F(xy) is xy+1. If you plug these three in the eqn F(x)F(y)-1(F(xy))=x+y (Since a is posi-nega 1 (But I labeled variable a as 1)).
That Eqn will be simplify as. (?) labels equal or not equal
(x+1)(y+1)-1(xy+1) (?) x+y
(xy+x+y+1)+(-xy-1) (?) x+y
x+y (xy and -xy adds up to 0, same as 1 and -1 which adds to 0) = x+y
Thus F(x) = x+1 is a true function to satisfy this eqn
Great problem. Thank you.
I am confused why we have to check the cases for a =+-1 for a second time. We already limited our choices of a to valid solutions
We checked that *if* they hold, a has to be 1 or -1, but not the reciprocal, which in this case is true if I'm not mistaken
@@orionmartoridouriet6834 but we only used eqiuvaletzes in that step.
So the inverse should be free
The proof showed that the only possible solutions were 1 or -1, but sometimes we might introduce false solutions by squaring a term without noticing. It's usually sensible to check both solutions, particularly when they arise from taking a square root.
@@RexxSchneider yeah, the extra solutions are called extraneous solutions
Hello, im a high school student and the problems i see on your page seem hard and i wonder what level of maths your problems could be given. Are these high school maths?
It depends, but aside specific theorems and some calculus type questions, its high school level stuff, (by high school level I mean contest level obviously)
The question can be answered with just high school mathematics, but I wouldn't expect most high schoolers to actually be able to solve it.
Solution before watching the video
(eq.1) : f(x)f(y) - a f(xy) = x+y
Substituting x=0, y=0 in eq.1
(eq.2) : f²(0) - a f(0) = 0
Which means that either f(0)=0 or f(0)=a
Substituting y=1 in (eq.1), and letting f(1) = z
(eq.3) : f(x) = (x+1)/(z-a)
Substituting x=0 in (eq.3) we see that f(0) = 0 was not a valid option, and therefore f(0) = a.
We can express z in terms of a by substituting x=0 in (eq.3)
(eq.4) : f(0) = a = 1/(z-a)
(eq.5) : z = (a²+1)/a
We can also substitute x=1 in (eq.3), yielding
(eq.6) : f(1) = z = 2/(z-a)
(eq.7) : a = (z²-2)/z
Now we have two equations relating a and z, and we can solve them by substituting (eq.5) into (eq.7)
(eq.8) : az = z²-2
(eq.9) : a²+1 = z²-2
(eq.10) : a²+3 = z²
(eq.11) : a⁴+3a² = a²z²
(eq.12) : a⁴+3a² = (a²+1)²
(eq.13) : a⁴+3a² = a⁴+2a²+1
(eq.14) : a² = 1
(eq.15) : a = ±1
And we can verify by substituting both a=1 and a=-1, and (eq.3) into (eq.1) to see that those are the two values that are allowed
Yeahh, his solution was way simpler
Can you find a solution for such abc=5 in whitch b=a+1 and c=b+1?
I.e., b³ - b - 5 = 0
A cubic with one real & 2 complex roots. Graphing that, shows the real root just below 2. Writing
b = ∛(b+5)
you can start with b=2 and iterate, to find
b = 1.904160859134920603676...
which can be "divided out" of the cubic, giving a quadratic from which the (approximate) complex roots can be found.
Could also apply the cubic formula to get exact answers.
Or try submitting the cubic to Wolfram Alpha, and it might give all 3 roots in both symbolic & numeric form, to many places.
Of course, once you have b, you simply get a = b-1, c = b+1.
Fred
You end up with (xy+1)(1-a)=0 and (xy+1)(1+a)=0 respectively. So what does this mean?
Because this has to hold for all values of x and y, and a is a constant, you can just assume the factors with a are to be set equal to 0
Let f(1)=b, then we have f(x)(b-a)=x+1. Hence b is not a (take x=0) and f(x)=c(x+1), where c=1/(b-a) is not zero. Thus f(x)f(y)-af(xy)=c²(x+1)(y+1)-ac(xy+1)=x+y. Hence c²=ac (from x=y=0) and then c²(x+y)=x+y for all x,y (take x=1, y=0). So a=-1,1. And (x+1)(y+1)-(xy+1)=x+y, and (-(x+1))(-(y+1))+(-(xy+1))=x+y.
They both work, right?
Yes
yup; everything on the last board is a chain of if and only if statements (since a=/=0 has already been ruled out)
Yep
I found that only a that makes it for all x's and y's is a=1, and then f(t)=t+1. Now it's time to view the video. // and I was wrong. I lost the a=-1 case and f(t)=-t-1.
@@gregorymorse8423 is this even legal? I thought a must be a constant.
@@gregorymorse8423 no, no. x and y both are reals, but don't depend on each other. Likewise, another real a may or may not depend on any of them. The fact all three of them are reals doesn't say absolutely anything about their dependencies. If we allow a to depend on x and y (that makes is not a parameter, but just a placeholder in the equation), then yes, there may be such "solutions". But it is called a parameter, which means it must be completely free and requires an independence. The question was: what are values of a, for which the functional equation holds for any x and y? I understand it this way: first we fix some a, and then independently vary x and y, but the functional equation should be still true. Which value of a we had fixed? With dependend a this is, of course, impossible, because we aren't able to fix a in the first place. All the problems with parameter I used to have so far (I had many) assume this parameter could not depend on x or y, so why this particular problem suddenly become an exception?
@@gregorymorse8423 Ok, so it is held as a constant. We choose a and hold it constant, then vary x and y. The only way for this to be possible is if a doesn't depend on x and y (remember, those both also vary independently of each other). How can you depend on something and stay fixed when it changes? This means "to not depend".
The "slowly changing" passage is for very interesting special case, a variation methods or iterative approximation methods. For example, consider a linear nonhomogeneous differential equation solving process. First you solve a corresponding homogeneous equation, and get a general solution with some parameters. Then you vary those parameters, considering them as functions and resolve which functions they could be. Consider an oscillator with decay very far from being damped, so there'll be a periodic sine (fast changing) with the slowly exponentially decaying amplitude. We sometimes call this quasiperiodic solution, because it is almost periodic, so we consider the amplitude as slowly changing parameter in this case. All in all, this is not about the problem in the video and your suggested "solution", because in this case a will be changing exactly at the same rate as variables, not being any slower.
Yes, the definition you've found forbids simple dependence. It forbids strictly in problems like this.
@@gregorymorse8423 But then the question "what values of a make this possible" wouldn't make sense
What if a = 0?
Michael Penn showed from 1:38 to 2:31 that f(0) = 0 or f(a) = 0. He then (by around 3:30 ) showed that f(0) = 0 leads to a contradiction. So f(a) must be equal to 0.
It should be clear that if f(a) = 0, and f(0) is not equal to 0, then a cannot be 0.
The nature of these sort of problems is that x and y must be free to take on any real value, but the parameter a is constrained to a particular set of values - in this case only 1 or -1.
If a=0, the equation is reduced to f(x)f(y) = x+y, which doesn't have any solutions, to see this, just plug in x=y=0 first, and then x=0 and y=1. In fact, this case has been covered in the solution because he shows that f(a) must be 0 and f(0) must not be 0
@@gregorymorse8423 What?
@@gregorymorse8423 I know sqrt is square root, what is z here?
@@gregorymorse8423 But f is a function of a single variable here
Yay nice video
as "a" is in nominator a cannot be 0 in final case so a=0 has to be cosidered seperately since all previous operation would be ilegal. But is needed to Remind we earlier called f(0) = a, but for a=0 we have first case
f(x) x+1....f(x)=-x-1
Very clear explanation, also for non-English speaking viewers :-)
Easy
TURKIYE!! (means Turkey in Turkish :-)
sen ne anlatıyon dayı
@@muratcan-k6x dur bi, ecnebiler de ogrensin
@@cihant5438 :D
Sanene olm biz öyle seviyok
Possibly he is the only one who has such an attractive thumbnail but not that attractive black board 🤣🤣.I think you should rather make some animation kinda thing.
I think blackboard is the best for that thing
@@kemalkayraergin5655 but I think animation creates a much bigger visual impact on the viewer and it also helps visualise the problem itself.i think it wouldn't be very good for this channel as it mostly posts number theory problems and not geometry problemo
2,🇹🇷
As bayraklari
@@alperenaydin6139 ne as bayrakları ulkeyi kurtardınız sanki adam soru cozdu alt tarafi
Yeet
I am firs like
so ?
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