As x approaches 2 from 2+ or 2- we see that the value is 1, thus allowing us to find f(t) as t approaches both negative and positive infinity. Mind Blown.
Because it doesn’t change anything. You’re just changing the variable, which is okay if you do it on both sides. The “x” at the end doesn’t have anything to do with the “x” at the beginning. The problem just asks for f(x).
Careful, when you multiply the function by some common factor with the intention of simplifying the whole expression you must divide as well by the same thing, it means that you are multiplying by 1, otherwise you can not say it is still the same equation.
that's not entirely correct the final x is not the same as the first one. t cannot be 1 or -1. and if f(x) = t^2/(t^2-1) then x cannot be 1 or -1. But t = x+2/(x-2), then whatever x t cannot be 1 so there is no problem here. t = -1 when x = 0 so you have one exception in common x = 0 is the same as t = -1. when x = 2 t is not defined so there is no problem. The first equation is not defined on 0 and -2 but the answer is not defined on 1 and -1
in the original statement f((x+2)/(x-2)) = (x+2)^2/8x you would _not_ input 2 into that function by replacing the x with 2, because x is not the input to the function. You would replace the x with a number such that (x+2)/(x-2) is equal to 2, because (x+2)/(x-2) is the input to the function. (6+2)/(6-2)=2 (6+2)^2)/8(6)=4/3 thus 2 is in the domain of the original function, you can watch him work out how 0 is in the domain of the original function in the beginning of the video. Changing the value you put into a function does not change the function or its domain. If we had a separate function g, defined so that g(x) = f((x+2)/(x-2)) then _that_ function, g, would not be defined at 2, but f still is, because when you feed 2 into f, it returns 4/3.
Yes u are almost right (I see what u were trying to say) - clearly plugging in x = 0 ==> there is a simple pole at t = -1 for f(t) and taking some limit e.g. let x -> 2+ ==> f(t) -> 1 as t-> +inf let x -> 2- ==> f(t) -> 1 as t -> -inf This can be seen all from the initial question (and clearly holds with the final answer!), but all he wanted to do was find the function, which he did - not specify the domain and range of the functional equation (which is an obvious 2 second job anybody can do). Slight mistake in your comment: the domain of f(x+2/x-2) has those problems, not the domain of f itself; domain of f only has a singularity at -1
It's a month late, but basically since f(t) means that t is a variable of that function, t could be replaced by other variable including x. However t ≠ x, like you said, t = x+2/x-2, t is not equal to x. And f(t) ≠ f(x). Basically means that t is not implied to be equal to x, the variable that is used in the function is changed, not made equal. For example if f(x) = x+2 Let's say x = 2 and t = 3 Then f(t) = t+2 = 3+2 = 5 And f(x) = x+2 = 2+2 = 4 the formula of (x+2) is still the same, but x ≠ t because 2 ≠ 3 and f(t) ≠ f(x) I hope anyone reading this comment understood my explanation.
@@fannyliem3536i got your point but aren’t we supposed to calculate the actual f(x)’s value? It is like saying i found the f(P) but not f(x) but since i can rename the variable let’s replace P by x…..the value which we got at last is not of the actual defined function.
I love you're enthusiasm. It makes me feel like I'm not crazy or left alone because sometimes I find math or science fascinating and when I try to talk to people about it they look at me weird. We need more teachers like you.
After substitution of t := (x+2)/(x-2), I found f(t) = 1 + 1/[(t - 1)(t + 1)] which can be reasonably defined for all real (or complex) values of t except for t = ±1. It's an even function with a double zero at t=0, two poles of order 1 at t=-1 and t=+1, and a horizontal asymptote y=1. 😃
Your content is so good that i think you deserve atleast a million subs. I am from India and i love watching your content. If for any reason you get depressed or think that you should stop making your videos, there's always me and my group of friends watching your vdos. Your teaching skills are fabulous. The way you make maths interesting. Thanks a lot my man. Love from india
Neat algebra - you might wish to explain how the original function won't given an answer at x = 2, whereas the revised function won't give an answer at x = 1 or -1 and how that works okay - as you have shifted the points where the function doesn't converge because of a divide by zero and why that would be allowed!
Your way of solving it is universal. Great! I found the numerator of RHS equals ( x + 2 )^2, and then I tried to express the denominator with ( x + 2) and ( x - 2 ). 8 x = ( x + 2 )^2 - ( x - 2 )^2 ∴ RHS = ( x + 2 )^2 ÷ { ( x + 2 )^2 - ( x - 2 )^2 } = { ( x + 2 ) / ( x - 2 ) }^2 ÷ [ { ( x + 2 ) / ( x - 2 ) }^2 - 1 ] Replace ( x + 2 ) / ( x - 2 ) with x, you can get x^2 / ( x^2 - 1 )
I also did this but the identity for the denominator might not be known by many so I will try make a story that might help finding solutions in the future. We will try to guess the function. First notice the x square in the numerator which means that there is some squaring involved. So try f(t) = t^2. You get the numerator of the Right Hand Side (RHS) but not the denominator. You can multiply and divide the denominator by the thing you want which is (x-2)^2. Then you have 8x/(x-2)^2 in the denominator. Issue is that there is no evident simplification unless you saw the relevant identity in the past and remembered it. So we will have to write 8x in some way that involves (x-2)^2. If in the end we want to write a function of (x+2)/(x-2) we will probably need to write 8x in terms of (x-2)^2 and (x+2). If we want to get rid of the x^2 in (x-2)^2 when that term is expanded then it might be interesting to look at (x-2)^2-(x+2)^2.
I am a pensioner and I alternate between doing math and the garden.Your presentation is just so captivating. I just can't imagine what I would be doing if I couldn't do math .Kudos from Johannesburg. Been thinking that functional equations were reserved for IMOs. 😅
I prefer a clear and simple formulation to avoid any confusion In the first and second lines, the letter 'x' is used in different ways. We're used to writing y=f(x), so it's easier to change the 'x' to 't' in the first line. This gives us the equations: f(x)=y x=(t+2)/(t-2) y=(t^2+4t+4)/(8t) Our task is to eliminate the variable 't' from these equations. (t-2)*x=(t+2) x*t-2x=t+2 x*t-t=2x+2 t=2(x+1)/(x-1) y=(t+2)^2/(8t)= ... etc
Well i actully solved this in my mind with a different solution. (X+2)²=X²+4X+4 (X+2+X-2)(X+2-X+2)=(2X)(4) = 8X so we can say: f(a/b) = (a²)/(a+b)(a-b) -> f(X/1) = X²/(X+1)(X-1) -> f(X) = X²/X²-1 😊 Pls like until he see this😢
The solution is actually easy. On the first sight, we can already see that 8x = (x+2)² - (x-2)², let u = (x+2)/(x-2), the eq becomes f(u)=1/(1-u⁻²) And that's the function we need to find.
You have very good content and scenic mastery. The form presented shows the equivalence with the change of variable It could also have been done like this x²+4x+4=(x+2)² (x²+4x+4)/8x=(x+2)²/8x, dividing numerator and denominator by (x-2)² =((x+2)²/(x-2)²)/(8x/(x-2)²), adding and subtracting 1 from the denominator =((x+2)/(x-2))²/(8x/(x-2)²+1-1) =((x+2)/(x-2))²/(((x+2)/(x-2))²-1) then the change f(x)=x²/(x²-1)
I have a very easy solution. in the RHS, the numerator can be written as (x+2) ^2 and denominator can be written as ((x+2) ^2 - (x-2) ^2) and then divide the numerator and denominator with (x-2) ^2. Then replace x + 2/x - 2 with x. The solution is x^2/x^2 - 1
Hey, really nice. I noticed something though, before 12:36 but at that time it's the step above the one you're pointing at. The top is the form a^2 + 2ab + b^2, so it equals (a+b)^2, which is ((t+1)+(t-1))^2 which evaluates to (2t)^2 then 4t^2, which is what you end up with as well. Just thought it was interesting, I immediately noticed it when I saw it
Sir I am an Indian student studying in class 12th (high school).. i substituted t = x+2/x-2, and then directly used COMPONENDO-DIVIDENDO to get t+1/t-1 = x/2.. so x = 2(t+1/(t-1)).. then I directly considered (x^2 + 4x + 4)/8x as (x+2)^2/8x and substituted x as 2(t+1/(t-1)) on both the sides to get the desired answer. Thanks a lot for this question sir..
Excellent video sir, i thoroughly enjoyed it. just by looking at the thumbnail. I guessed we would have to plug in another variable, But I made the mistake of substituting a directly into the equation. like, f(a) = (((x+1)/(x-1))^2 + 4(x+1)/(x-1) + 4 )/ 8((x+1)/(x-1))
Been WAAAAY too long since I looked at this stuff. I was always pretty good and keen on math, but once this stuff started to turn up, it made the subject loads more interesting. It's hard to describe, but the way these functions relate to one another, it almost feels like you're peeling away at the layers of how the universe as a whole operates. Some of the discoveries end up being more exciting than others, of course. Very similar vibes with how taking the derivative of a function, and then taking that derivative, and then taking that derivative, and all these functions you end up with all relate to one another. It's like the numbers behind the numbers behind the numbers. You're introduced to things like parabolas and other common graph shapes well before learning derivatives, so it just felt like a huge plot twist when you first learn that these derivatives were there 'driving' the shapes of the graph all along. I don't know, just always seemed very cool to me.
His answer is wrong imo. Subbing in to check fails miserably too.1/8 (x+2)(x-2) or something I got. Did it in my head and forgot.while checking his answer in my head. Might be wrong, dont know, dont care.
10:49 You could have factorised the numerator (t+1)^2 + 2(t+1)(t-1) + (t-1)^2 = ((t+1) + (t-1))^2 = (2t)^2 = 4t^2 since it is of the form (a + b)^2 = a^2 +2ab +b^2
(x^2+4x+4)/8x=(x+2)^2/((x+2)^2-(x-2)^2). After dividing both numerator and denominator of the fraction by(x-2)^2, the result is: f(z)=z^2/(z^2-1), where z=(x+2)/(x-2). It is always a pleasure to watch your enthusiastic presentations.
Great videos you make, they are super useful. For me personally i have, in the last couple of days, learned a bunch of new techniques from your videos.
You have great "board-side" manner. Cool...But sometimes shorter methods are easier to follow. Put x+2=a, x-2=b and a/b=c, then, f((x+2)/(x-2)) is f(a/b) or f(c) and RHS = a^2/(a^2-b^2) = 1/(1-(b/a)^2) = 1/(1-(1/c)^2) = c^2/(c^2-1) Now, as f(c)=c^2/(c^2-1) Substituting x for c, gives f(x)= x^2/(x^2-1)
there is a simple way altogether, even we can do it on fly and I have told on fly , that f(x) = x^2/(x^2-1) as follows. Let y/1=x+2 by x-2 then x/2=y+1 by y-1 -> x = 2 times y+1 by y-1. Right side, numerator = (x+2) ^2. so. first find out x+2. x+2 = 2 times ( y+1 by y-1 plus 1) = 2 ties 2y by y-1 = 4y by y-1. so numerator = 16 y^2 by (y-1) ^2. Dr = 8x = 8 * 2 * y+1 by y-1. Nr/Dr -> 16 cancels. y-1 , one cancels .. so, remains y^2 / (y+1) *y-1) which is y^2/ (y^2-1). so f(y) = y^2/ (y^2-1) so, f(x) = x^2 / (x^2-1)
That big niominator can be simplify without expanding by noticing that it is a prefect square (t+1)^2 +2(t+1)(t-1) + (t-1)^2 =(t+1+t-1)^2 =(2t)^2 =4t^2
I am from Bangladesh. And mymother langyage is not english. But your lecture is incredible. Despite being a bangali i can understand your solution so easily.your way of teaching is not boring at all. You are a really great teacher
My thought was to consider the expression as f(g(x))= E(X) and then see if g(x)=(x+2)/(x-2) is bijective to see if and inverse function exists. I see you just jumped to obtaining the inverse function and plugging it in. It means that we need to check if any value of the codomain will have a value in the domain of g(x) which is not specified directly but by sketching a rough graph we get the horizontal asymptote y=1 and vertical asymptote for x=2 with g(x) taking values from -inf to +inf for x real. Secondarily we need to prove that if we have g(a)=g(b) then it means a=b. Or alternatively just show g'(x) is not zero for any x or the function admits no minimum/maximum. Just then can we go trough and get the inverse to plug in. Not sure why we just glossed over the problem of an inverse existing
I would argue that this problem is ill defined. (x+2)/(x-2) can never be one so the functional equation does not restrict f at one. You have solved for f(x) at all x different from one. f(1) can be arbitrary. A functional equation like this will only be solvable on the range of the argument of the LHS.
شكرا لكم على المجهودات يمكن استعمال x réel différent de 2 et de 0 On pose y=x+2/x-2 ,y différent de 1 et de -1 x=2(y+1)/y-1 ..... f(y)=y^2/y^2-1 Pour x différent de 1 et de -1 f(x)=x^2/x^2 -1 😂
When you get (t+1)^2 + 2(t+1)(t-1) + (t-1)^2 in numerator you can use the formula for (a+b)^2, where a= t+1 and b= t-1. If you do that you will get (t+1 + t-1)^2. This is equal (2t)^2 and this is 4t^2.
Could someone please explain the concept of functional equations? What I don't understand is: say one has a function F(x) = y. X is a independent variable, So one chooses an X and the function gives a Y. straightforward. Say, If one wants to know what f(x+4)is? ,then one adds 4 to x and plugs it into the same function, right? I really don't know why the function needs to be reworked? ie F(x) = x^2 + 3, if x= 2, then F(2) = 7. then F(x+2) = F(4) = 19. What am i missing?
There is a mistake. You have arrived at the equality x = (2t + 2)/(t - 1) that is correct. But when you substitute x in the functional equation, you had a mistake when you put the factor 2 multiplying the whole fraction, i.e., multiplying both numerator and denominator. Really, factor 2 applies to the numerator only.
For this specific case I found It easier to write It as f(g(x))=p(x) find the inverser of (x+2)/(x-2) and them plug It into the equation. Ok the result I got was wrong but thats only because among all the simplifications I missed a negative sign.
Why can you replace t with x at the end? I understand replacing t with another variable like z or y. But t had already been defined in terms of x (t=(x+2)(x-2)); so replacing t with x directly (t=x) seems weird to me because you are overwriting the previous assumption?
For the step where you distribute, you can actually use the formula a² + 2ab + b² = (a + b)², so the numerator will simplify to ((t + 1) + (t - 1))² which further simplifies to (t + 1 + t - 1)² = (2t)² = 4t²
13:18 Why can you switch the f(t) = (t^2)/( (t^2) - 1) to f(x) = (x^2)/( (x^2) - 1) when we defined t as (x + 2) / (x - 1)? It doesn't make sense to me because I think that is like saying t = x, when at the start it did not. Was the goal just to get the input to be one letter?
Easier solution give x the value x+2 and you will have 1+4/x. Then give x the value 4/x-1 and you will end up with x. Dont forget the numerator on the solution side is (x+2) square whick makes things easier
Sorry if this answer chosed you with No prior decision Who are thé countries strong inMaths :Algèbra and Geometry ? Level of industrialisation?? Tell me once Studies done
This is, in my opinion, the most complexe part of this exercise, if you do that you would go back to the begining and would have undo every thing. The final x is not the same as the first one.
@@glorrinThat's the only possible explanation for this in my understanding too. "Final X is not the same as the first one". But this method works for other functions too which has same X. If anyone knows any other explanation , please enlighten me.
If you multiply a fraction with a fraction, you'll get numeratorxnumerator over denominatorxdenominator. In contrast to that, you multiplied numerator x fraction by denominator x fraction which ist oviously incorrect. Nevermind, we're all fallible Humans.
Aren’t you supposed to put the value of “t” i.e. (x+2)/(x-2) in the last step cause just by simply replacing the the letter t by x won’t give you the actual f(x) do let me know if I’m incorrect.
Great channel, I really appreciate what you're doing and how you explain math concepts. Regarding this algebra the only thing I miss is to determine the function domain which is also part of the solution.
I try like this Put x=m+2 , then take 4 common from lhs both in numerator and denominator, then put 4/m =t , and then put 1+t =p we get f(p) , then put p=x to get the same result
salam alaykom, your expression is ok fo x not equal to 1 or -1 but the right expression that you should write all possibility then you can discuss the continuity of function f(x)
Non je ne suis pas d'accord vous posez un changement de variable t=x+2/x-2 après vous vous dites tt simplement que t=x parce qu'il la solution vient de satisfaire une condition limite
let's say you have g(y) = y. You can also say g(z) = z, without much of an explanation right ? You can also say g(t) = t or g(x) = x. y z t and x are defined localy, they only exis fort this function and do not depend on anything else around. It is the same thing at the end, you just replace the variable inside without any trouble. f(t) = t^2 / (t^2 -1) is the same as f(x) = x^2/(x^2-1) The last x is not the same as the first one. Moreover the first x has not the same domain as the last one.
Hello! At the beginning of the video we are given a function f. We know that if the input is x+2/x-2 then the output is x²+4x+4/8x. We want to know what will be the output if the input is just x. And of course, x is just some number. Sio if we know the output od the function for some number, we can subatitute x as input value and find the result! If we create a new variable t such that t=x+2/x-2, we can say that f(x+2/x-2)=f(t)=x²+4x+4/8x. As you can see, we now have one single number as an input, instead of expression. However, function of t is equal to the x²+4x+4/8x, how do we calculate that? Well, t is a number that we defined as t=x+2/x-2, and that implies that x=2t+2/t-1. So now we can replace every x in x²+4x+4/8x with 2t+2/t-1, because that's what x is equal to, if x+2/x-2=t. After simplifying we get f(t)=t²/t²-1. I remind you that t is just some number. So if we plug some number into the function, we get number²/number²-1. Just plug x as the number and the result is f(x)=x²/x²-1
I never studied functional equations and I have a degree in math with 4 semesters of calculus under my belt. I did not focus on algebra, more on probability and statistics and this sort of mathematics does not come up much in that area of math.
He is a very patient teacher with a very sympathic voice and charisma.
I love the fun you have with maths. Your enthusiasm is infectious. I wish my teachers had had half your ability.
As x approaches 2 from 2+ or 2- we see that the value is 1, thus allowing us to find f(t) as t approaches both negative and positive infinity. Mind Blown.
Why t can be replaced by x directly in the final？
Because it doesn’t change anything. You’re just changing the variable, which is okay if you do it on both sides. The “x” at the end doesn’t have anything to do with the “x” at the beginning. The problem just asks for f(x).
Careful, when you multiply the function by some common factor with the intention of simplifying the whole expression you must divide as well by the same thing, it means that you are multiplying by 1, otherwise you can not say it is still the same equation.
10:55 The top part was a perfect square, you don't even need to distribute everything
((t + 1) + (t - 1))^2 = (2t + 1 - 1)^2 = (2t)^2 = 4t^2
Haha! Now I see it.
yup, this is the comment im looking for
That is how I handled it, too.
Yup, came for this!
I noticed that and was wondering whether you would use it.
At the end you simply substituted de 't' by 'x', although t = (x+2)/(x-2). It seems to me that this is wrong.
You should specify that x cannot be 0 or 2 in domain of f as those values are not in the domain of original functional equation.
that's not entirely correct the final x is not the same as the first one. t cannot be 1 or -1. and if f(x) = t^2/(t^2-1) then x cannot be 1 or -1.
But t = x+2/(x-2), then whatever x t cannot be 1 so there is no problem here. t = -1 when x = 0 so you have one exception in common x = 0 is the same as t = -1.
when x = 2 t is not defined so there is no problem.
The first equation is not defined on 0 and -2 but the answer is not defined on 1 and -1
I was just looking for f(x).
in the original statement
f((x+2)/(x-2)) = (x+2)^2/8x
you would _not_ input 2 into that function by replacing the x with 2, because x is not the input to the function. You would replace the x with a number such that (x+2)/(x-2) is equal to 2, because (x+2)/(x-2) is the input to the function.
(6+2)/(6-2)=2
(6+2)^2)/8(6)=4/3
thus 2 is in the domain of the original function, you can watch him work out how 0 is in the domain of the original function in the beginning of the video.
Changing the value you put into a function does not change the function or its domain. If we had a separate function g, defined so that
g(x) = f((x+2)/(x-2))
then _that_ function, g, would not be defined at 2, but f still is, because when you feed 2 into f, it returns 4/3.
Yes u are almost right (I see what u were trying to say) - clearly plugging in x = 0 ==> there is a simple pole at t = -1 for f(t)
and taking some limit e.g. let x -> 2+ ==> f(t) -> 1 as t-> +inf
let x -> 2- ==> f(t) -> 1 as t -> -inf
This can be seen all from the initial question (and clearly holds with the final answer!), but all he wanted to do was find the function, which he did - not specify the domain and range of the functional equation (which is an obvious 2 second job anybody can do). Slight mistake in your comment: the domain of f(x+2/x-2) has those problems, not the domain of f itself; domain of f only has a singularity at -1
Every one in the comment going crazy
hey, I'm 65 and just starting to do some math again. I was able to follow that long forgotten algebra so thanks, that is encouraging - subscribed.
Hey I don’t understand at the end…at first you took x+2/x-2 = t…so it gives you X= 2t + 2/ t-1…but at the end why did you implied t = X
It's a month late, but basically since f(t) means that t is a variable of that function, t could be replaced by other variable including x.
However t ≠ x, like you said, t = x+2/x-2, t is not equal to x. And f(t) ≠ f(x). Basically means that t is not implied to be equal to x, the variable that is used in the function is changed, not made equal.
For example if f(x) = x+2
Let's say x = 2 and t = 3
Then f(t) = t+2 = 3+2 = 5
And f(x) = x+2 = 2+2 = 4
the formula of (x+2) is still the same, but x ≠ t because 2 ≠ 3 and f(t) ≠ f(x)
I hope anyone reading this comment understood my explanation.
@@fannyliem3536i got your point but aren’t we supposed to calculate the actual f(x)’s value? It is like saying i found the f(P) but not f(x) but since i can rename the variable let’s replace P by x…..the value which we got at last is not of the actual defined function.
Are you a mathmatics teacher in USA?
I love you're enthusiasm. It makes me feel like I'm not crazy or left alone because sometimes I find math or science fascinating and when I try to talk to people about it they look at me weird. We need more teachers like you.
Agreed, he has a perfect attitude to teach!
After substitution of t := (x+2)/(x-2), I found f(t) = 1 + 1/[(t - 1)(t + 1)] which can be reasonably defined for all real (or complex) values of t except for t = ±1.
It's an even function with a double zero at t=0, two poles of order 1 at t=-1 and t=+1, and a horizontal asymptote y=1. 😃
Sir sorry to abrupt you
But in the end of solution
You just replaced x in place of t
And did not put the value of t taken x+2/x-2
Your content is so good that i think you deserve atleast a million subs. I am from India and i love watching your content. If for any reason you get depressed or think that you should stop making your videos, there's always me and my group of friends watching your vdos. Your teaching skills are fabulous. The way you make maths interesting. Thanks a lot my man. Love from india
Wow! That means a lot to me. Thank you, and God bless.
Neat algebra - you might wish to explain how the original function won't given an answer at x = 2, whereas the revised function won't give an answer at x = 1 or -1 and how that works okay - as you have shifted the points where the function doesn't converge because of a divide by zero and why that would be allowed!
Your way of solving it is universal. Great!
I found the numerator of RHS equals ( x + 2 )^2, and then I tried to express the denominator with ( x + 2) and ( x - 2 ).
8 x = ( x + 2 )^2 - ( x - 2 )^2
∴ RHS = ( x + 2 )^2 ÷ { ( x + 2 )^2 - ( x - 2 )^2 }
= { ( x + 2 ) / ( x - 2 ) }^2 ÷ [ { ( x + 2 ) / ( x - 2 ) }^2 - 1 ]
Replace ( x + 2 ) / ( x - 2 ) with x, you can get x^2 / ( x^2 - 1 )
I did exactly the same thing!!
Same thing I did
Me too
I tried same thing but missed in expressing 8x in terms of X+2 ad X-2 , thanks for the steps
I also did this but the identity for the denominator might not be known by many so I will try make a story that might help finding solutions in the future.
We will try to guess the function. First notice the x square in the numerator which means that there is some squaring involved. So try f(t) = t^2. You get the numerator of the Right Hand Side (RHS) but not the denominator. You can multiply and divide the denominator by the thing you want which is (x-2)^2. Then you have 8x/(x-2)^2 in the denominator.
Issue is that there is no evident simplification unless you saw the relevant identity in the past and remembered it.
So we will have to write 8x in some way that involves (x-2)^2.
If in the end we want to write a function of (x+2)/(x-2) we will probably need to write 8x in terms of (x-2)^2 and (x+2).
If we want to get rid of the x^2 in (x-2)^2 when that term is expanded then it might be interesting to look at (x-2)^2-(x+2)^2.
I am a pensioner and I alternate between doing math and the garden.Your presentation is just so captivating. I just can't imagine what I would be doing if I couldn't do math .Kudos from Johannesburg. Been thinking that functional equations were reserved for IMOs. 😅
So nice of you
انا من فلسطين . واحب الرياضيات . انت مذهل و رائع . ساتابعك باستمرار . تحياتي
I prefer a clear and simple formulation to avoid any confusion
In the first and second lines, the letter 'x' is used in different ways. We're used to writing y=f(x), so it's easier to change the 'x' to 't' in the first line.
This gives us the equations:
f(x)=y
x=(t+2)/(t-2)
y=(t^2+4t+4)/(8t)
Our task is to eliminate the variable 't' from these equations.
(t-2)*x=(t+2)
x*t-2x=t+2
x*t-t=2x+2
t=2(x+1)/(x-1)
y=(t+2)^2/(8t)= ... etc
12:38 you can simply write the numerator as [ (t+1)+(t-1) ]^2=(2t)^2=4t^2
Well i actully solved this in my mind with a different solution.
(X+2)²=X²+4X+4
(X+2+X-2)(X+2-X+2)=(2X)(4) = 8X
so we can say:
f(a/b) = (a²)/(a+b)(a-b)
-> f(X/1) = X²/(X+1)(X-1)
-> f(X) = X²/X²-1 😊
Pls like until he see this😢
The solution is actually easy. On the first sight, we can already see that 8x = (x+2)² - (x-2)², let u = (x+2)/(x-2), the eq becomes f(u)=1/(1-u⁻²)
And that's the function we need to find.
You have very good content and scenic mastery.
The form presented shows the equivalence with the change of variable
It could also have been done like this
x²+4x+4=(x+2)²
(x²+4x+4)/8x=(x+2)²/8x, dividing numerator and denominator by (x-2)²
=((x+2)²/(x-2)²)/(8x/(x-2)²), adding and subtracting 1 from the denominator
=((x+2)/(x-2))²/(8x/(x-2)²+1-1)
=((x+2)/(x-2))²/(((x+2)/(x-2))²-1) then the change
f(x)=x²/(x²-1)
Yes, Cooool !!!!
I have a very easy solution.
in the RHS, the numerator can be written as (x+2) ^2 and denominator can be written as ((x+2) ^2 - (x-2) ^2) and then divide the numerator and denominator with (x-2) ^2. Then replace x + 2/x - 2 with x. The solution is x^2/x^2 - 1
Hey, really nice. I noticed something though, before 12:36 but at that time it's the step above the one you're pointing at. The top is the form a^2 + 2ab + b^2, so it equals (a+b)^2, which is ((t+1)+(t-1))^2 which evaluates to (2t)^2 then 4t^2, which is what you end up with as well. Just thought it was interesting, I immediately noticed it when I saw it
12:21 i just noticed that (t+1)^2+2(t+1)(t-1)+(t-1)^2 wich is equal to (t+1+t-1)^2 wich is also equal to (2t)^2 or 4t^2
Bob Ross of algebra
In school I learned this approach was called the dummy variable approach.
Sir I am an Indian student studying in class 12th (high school)..
i substituted t = x+2/x-2, and then directly used COMPONENDO-DIVIDENDO to get t+1/t-1 = x/2.. so x = 2(t+1/(t-1))..
then I directly considered (x^2 + 4x + 4)/8x as (x+2)^2/8x and substituted x as 2(t+1/(t-1)) on both the sides to get the desired answer.
Thanks a lot for this question sir..
x^2+4x+4 = (x+2)^2
if you use this it will be easier.
i really like the syle he talks/teaches here!!
Excellent video sir, i thoroughly enjoyed it.
just by looking at the thumbnail. I guessed we would have to plug in another variable,
But I made the mistake of substituting a directly into the equation.
like, f(a) = (((x+1)/(x-1))^2 + 4(x+1)/(x-1) + 4 )/ 8((x+1)/(x-1))
Been WAAAAY too long since I looked at this stuff. I was always pretty good and keen on math, but once this stuff started to turn up, it made the subject loads more interesting. It's hard to describe, but the way these functions relate to one another, it almost feels like you're peeling away at the layers of how the universe as a whole operates.
Some of the discoveries end up being more exciting than others, of course. Very similar vibes with how taking the derivative of a function, and then taking that derivative, and then taking that derivative, and all these functions you end up with all relate to one another. It's like the numbers behind the numbers behind the numbers.
You're introduced to things like parabolas and other common graph shapes well before learning derivatives, so it just felt like a huge plot twist when you first learn that these derivatives were there 'driving' the shapes of the graph all along. I don't know, just always seemed very cool to me.
I confused, t=(x+2)/(x-2), but end of the solution t = x
You are just transforming the function to be write as f(x).
X is just a variable you can use any variable you want.
i am 70 retired eng;ineer you got my attention love your teaching style and i love math
You forgot to plug in
(x +2)/(x - 2) for t to get the equation in relation to x. You can't just change t into x.
His answer is wrong imo. Subbing in to check fails miserably too.1/8 (x+2)(x-2) or something I got. Did it in my head and forgot.while checking his answer in my head.
Might be wrong, dont know, dont care.
that's amazing. Never seen functional equations before but solving that looked like a lot of fun.
Never stop teaching Coach !
Thanks
Excellent sir. Loved the way you simplified and great explanation.
Excellent, very interesting this exercise. Thanks so much!!! Greeting from Perú!
10:49 You could have factorised the numerator (t+1)^2 + 2(t+1)(t-1) + (t-1)^2 = ((t+1) + (t-1))^2 = (2t)^2 = 4t^2 since it is of the form (a + b)^2 = a^2 +2ab +b^2
5:36 you could have easily got x in terms of t by applying componendo dividendo. btw nice solution
Seriously I need to google that
If you had observed that x²+4x+4 = (x+2)² then it would not have been so cumbersome. That's how I solved it.
The numerator x^2+4x+4 is (x+2)^2. Should you have used this, it would have saved you a lot of calculations.
One of the reasons I like your videos is because you use black board and chalk......good old days.
(x^2+4x+4)/8x=(x+2)^2/((x+2)^2-(x-2)^2). After dividing both numerator and denominator of the fraction by(x-2)^2, the result is: f(z)=z^2/(z^2-1), where z=(x+2)/(x-2). It is always a pleasure to watch your enthusiastic presentations.
Great videos you make, they are super useful. For me personally i have, in the last couple of days, learned a bunch of new techniques from your videos.
You have great "board-side" manner. Cool...But sometimes shorter methods are easier to follow.
Put x+2=a, x-2=b and a/b=c,
then, f((x+2)/(x-2)) is f(a/b) or f(c) and RHS
= a^2/(a^2-b^2) = 1/(1-(b/a)^2)
= 1/(1-(1/c)^2) = c^2/(c^2-1)
Now, as f(c)=c^2/(c^2-1)
Substituting x for c, gives
f(x)= x^2/(x^2-1)
That quote at the end sent me. Very enjoyable personality.
there is a simple way altogether, even we can do it on fly and I have told on fly , that f(x) = x^2/(x^2-1)
as follows.
Let y/1=x+2 by x-2 then x/2=y+1 by y-1 -> x = 2 times y+1 by y-1.
Right side, numerator = (x+2) ^2. so. first find out x+2.
x+2 = 2 times ( y+1 by y-1 plus 1) = 2 ties 2y by y-1 = 4y by y-1.
so numerator = 16 y^2 by (y-1) ^2.
Dr = 8x = 8 * 2 * y+1 by y-1.
Nr/Dr -> 16 cancels. y-1 , one cancels .. so, remains y^2 / (y+1) *y-1) which is y^2/ (y^2-1).
so f(y) = y^2/ (y^2-1)
so, f(x) = x^2 / (x^2-1)
That big niominator can be simplify without expanding by noticing that it is a prefect square
(t+1)^2 +2(t+1)(t-1) + (t-1)^2
=(t+1+t-1)^2
=(2t)^2
=4t^2
I think it would have been easier to use full square formula instead of opening braces: (t-1)^2 +2(t-1)(t+1)+(t+1)^2=((t-1)+(t+1))^2=(2t)^2=4t^2
I am from Bangladesh. And mymother langyage is not english. But your lecture is incredible. Despite being a bangali i can understand your solution so easily.your way of teaching is not boring at all. You are a really great teacher
Same bro🇧🇩🇧🇩
In the last step you substituted t=x but in fact at the start we assumed t=x+2/x-2
The f(x) should be
(X+2)^2/8x
Right
My thought was to consider the expression as f(g(x))= E(X) and then see if g(x)=(x+2)/(x-2) is bijective to see if and inverse function exists. I see you just jumped to obtaining the inverse function and plugging it in.
It means that we need to check if any value of the codomain will have a value in the domain of g(x) which is not specified directly but by sketching a rough graph we get the horizontal asymptote y=1 and vertical asymptote for x=2 with g(x) taking values from -inf to +inf for x real.
Secondarily we need to prove that if we have g(a)=g(b) then it means a=b. Or alternatively just show g'(x) is not zero for any x or the function admits no minimum/maximum.
Just then can we go trough and get the inverse to plug in. Not sure why we just glossed over the problem of an inverse existing
I would argue that this problem is ill defined. (x+2)/(x-2) can never be one so the functional equation does not restrict f at one. You have solved for f(x) at all x different from one. f(1) can be arbitrary. A functional equation like this will only be solvable on the range of the argument of the LHS.
У меня получилось F(x)= (x^2-1)/4. Если (x+2)/(x-2)=y, то x=2(y+1)/(y-1). После подстановки и замены переменной я получила свой ответ.
you should simply the right side of original equation first,(x^2+4x+4）/8x=x/8+1/2+1/(2x), this way, the calculation should be easier
شكرا لكم على المجهودات
يمكن استعمال
x réel différent de 2 et de 0
On pose y=x+2/x-2 ,y différent de 1 et de -1
x=2(y+1)/y-1
.....
f(y)=y^2/y^2-1
Pour x différent de 1 et de -1
f(x)=x^2/x^2 -1 😂
When you get (t+1)^2 + 2(t+1)(t-1) + (t-1)^2 in numerator you can use the formula for (a+b)^2, where a= t+1 and b= t-1. If you do that you will get (t+1 + t-1)^2. This is equal (2t)^2 and this is 4t^2.
Could someone please explain the concept of functional equations? What I don't understand is: say one has a function F(x) = y. X is a independent variable, So one chooses an X and the function gives a Y. straightforward. Say, If one wants to know what f(x+4)is? ,then one adds 4 to x and plugs it into the same function, right? I really don't know why the function needs to be reworked? ie F(x) = x^2 + 3, if x= 2, then F(2) = 7. then F(x+2) = F(4) = 19. What am i missing?
There is a mistake. You have arrived at the equality
x = (2t + 2)/(t - 1)
that is correct.
But when you substitute x in the functional equation, you had a mistake when you put the factor 2 multiplying the whole fraction, i.e., multiplying both numerator and denominator. Really, factor 2 applies to the numerator only.
In the process shown, t=x+2/x-2,
Using COMPONENDO-DIVIDENDO,
X=2(t+1) /t-1
For this specific case I found It easier to write It as f(g(x))=p(x) find the inverser of (x+2)/(x-2) and them plug It into the equation.
Ok the result I got was wrong but thats only because among all the simplifications I missed a negative sign.
Why can you replace t with x at the end? I understand replacing t with another variable like z or y. But t had already been defined in terms of x (t=(x+2)(x-2)); so replacing t with x directly (t=x) seems weird to me because you are overwriting the previous assumption?
General concept, if you have f(g(x)) = h(x) and g^{-1} exists, then just do f(g(g^{-1}(x))) = h(g^{-1}(x)) = f(x)
What The logic!!! First Substitution
T = (X+2) /(X-2) at first and the End U substitution
X= T
Plz any body can explain this discrepancy ??
can we just let g(x)=(x+2)/(x-2)? then by the property of function f(x) = f[gg-1(x)], just subtitute g inverse in that function.
For the step where you distribute, you can actually use the formula a² + 2ab + b² = (a + b)², so the numerator will simplify to ((t + 1) + (t - 1))² which further simplifies to (t + 1 + t - 1)² = (2t)² = 4t²
13:18 Why can you switch the f(t) = (t^2)/( (t^2) - 1) to f(x) = (x^2)/( (x^2) - 1) when we defined t as (x + 2) / (x - 1)? It doesn't make sense to me because I think that is like saying t = x, when at the start it did not. Was the goal just to get the input to be one letter?
Easier solution give x the value x+2 and you will have 1+4/x. Then give x the value 4/x-1 and you will end up with x. Dont forget the numerator on the solution side is (x+2) square whick makes things easier
f(0/0)=f(0)?? , We know that 0/0=undefined(it has three possibilities 0, 1, infinity), doesn't it??
Question must have limitation
11:24 you should make it a whole square (t+1)^2 + 2(t+1)(t-1) + (t-1)^2 = (t+1+t-1)^2 which will give you t^2 as a result and make it easier
I’m very happy to have found your channel!!!
Sorry if this answer chosed you with No prior decision
Who are thé countries strong inMaths :Algèbra and Geometry ?
Level of industrialisation??
Tell me once Studies done
In the end why can I not substitute the value of t=(x+2)/(x-2) in the f(t)=t²/(t²-1), I am curious as ti what I would get if I did do that ☠️
Why didn't you write( t ) as( x+2/x-2)
at the end
This is, in my opinion, the most complexe part of this exercise, if you do that you would go back to the begining and would have undo every thing. The final x is not the same as the first one.
@@glorrinThat's the only possible explanation for this in my understanding too. "Final X is not the same as the first one". But this method works for other functions too which has same X. If anyone knows any other explanation , please enlighten me.
If you multiply a fraction with a fraction, you'll get numeratorxnumerator over denominatorxdenominator.
In contrast to that, you multiplied numerator x fraction by denominator x fraction which ist oviously incorrect.
Nevermind, we're all fallible Humans.
Aren’t you supposed to put the value of “t” i.e. (x+2)/(x-2) in the last step cause just by simply replacing the the letter t by x won’t give you the actual f(x) do let me know if I’m incorrect.
How did you subtitute t=x directly in the last step? t was substituted as x+2/x-2. and finally x was 2t+2/t-1 so how did you directly substitute t=x?
Great channel, I really appreciate what you're doing and how you explain math concepts. Regarding this algebra the only thing I miss is to determine the function domain which is also part of the solution.
Ребята здесь я не вижу математики. Это демагогия функция f(x) уже есть зачем ее находить))))
Comparing the last two steps, we can say that t = x ,but we have assumed t = x+2/x-2....Can someone explain plz😅
I try like this
Put x=m+2 , then take 4 common from lhs both in numerator and denominator, then put 4/m =t , and then put 1+t =p we get f(p) , then put p=x to get the same result
Kolay bir fonksiyon sorusu, sadece uğraştırıyor.
I have a doubt, Can exist multivariable functional equations? like f(x+5,y-2)=xy??
Sou muito fã de suas aulas! Obrigado muito!
Would have been more clean and simple if you have solved RHS in form of (x+2)sq/8x
salam alaykom, your expression is ok fo x not equal to 1 or -1 but the right expression that you should write all possibility then you can discuss the continuity of function f(x)
There is a problem : f(1) is not defined in the final form whereas we can compute the value of f while replacing x by one in the first equation ?!?!
11:22 Don't distribute and factorise instead: it is of the form A^2 + 2AB + B^2 = (A+B)^2 = 4t^2 :)
(t+1)^2 + 2(t+1)(t-1) +( t-1)^2 is same thing like x^2 + 2x + 1 , so too much calculation I think.
This is easy given f(x+2/x-2) = (x+2)^2/(x+2) ^2- (x-2) ^2
So f(x) = f(x/1) = x^2/x^2-1
Non je ne suis pas d'accord vous posez un changement de variable t=x+2/x-2 après vous vous dites tt simplement que t=x parce qu'il la solution vient de satisfaire une condition limite
Water pump A needs 15 hours to fill a Pool, water pump B needs 10 hours, how much they need together ?
x^2+4x+4=(x+2)^2. then replace x with x(t), i think this way is more simple.
Did in mind in 2 mins.....by just dividing by x-2 whole square and then manupulating the terms😊
13:00 can someone explain the last part? I don't really understand why t got replaced by x on all sides.
let's say you have g(y) = y. You can also say g(z) = z, without much of an explanation right ?
You can also say g(t) = t or g(x) = x.
y z t and x are defined localy, they only exis fort this function and do not depend on anything else around.
It is the same thing at the end, you just replace the variable inside without any trouble. f(t) = t^2 / (t^2 -1) is the same as f(x) = x^2/(x^2-1)
The last x is not the same as the first one.
Moreover the first x has not the same domain as the last one.
Hello! At the beginning of the video we are given a function f. We know that if the input is x+2/x-2 then the output is x²+4x+4/8x. We want to know what will be the output if the input is just x. And of course, x is just some number. Sio if we know the output od the function for some number, we can subatitute x as input value and find the result!
If we create a new variable t such that t=x+2/x-2, we can say that f(x+2/x-2)=f(t)=x²+4x+4/8x. As you can see, we now have one single number as an input, instead of expression. However, function of t is equal to the x²+4x+4/8x, how do we calculate that? Well, t is a number that we defined as t=x+2/x-2, and that implies that x=2t+2/t-1. So now we can replace every x in x²+4x+4/8x with 2t+2/t-1, because that's what x is equal to, if x+2/x-2=t.
After simplifying we get f(t)=t²/t²-1. I remind you that t is just some number. So if we plug some number into the function, we get number²/number²-1. Just plug x as the number and the result is f(x)=x²/x²-1
I never studied functional equations and I have a degree in math with 4 semesters of calculus under my belt. I did not focus on algebra, more on probability and statistics and this sort of mathematics does not come up much in that area of math.