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.
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. 😅
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 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
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..
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
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.
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
Muy interesante, didáctica y buena clase, a mi hija le servirá mucho esta excelente exposición. Estamos muy agradecidos con su bella persona, bendiciones y éxitos para Usted y su linda familia. ❤
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.
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!
(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.
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
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. 😃
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.
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
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
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)
11:15 When simplifying [(t+1)^2 + 2(t+1)(t-1) + (t-1)^2)], instead of expanding everything and cancelling out you could have used the general formula (a+b)^2 = a^2 + 2ab + b^2, would’ve been neater.
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.
Wonderful manner that conveys such enthusiasm and positivity. I would have understood better if a graph of the function had been included when it was found. That might have helped understand the domain issues that got so many commenters in knots.
So my takeaway is that when given a functional equation call it f(g(x)) in order to determine f(x) we simply find the inverse of g(x) so that when we plug that into f(g(x)) we get f(x). Sounds simple enough! Very good example I just wish he would have mentioned the technique in more general terms at the end. After all as a mathematician we want to be able to generalize results.
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)
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))
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.
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😢
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.
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
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
yeah, this is what we did in ms. the method is that which is called the substitution of variates. make t = (x+2)/(x-2)（x≠2 &），then t= g(x), then integrate g(x) into the function on the right side, we will get a f(t)=t²/t²-1(t≠1, x≠0)。so we have f(x)=x²/x²-1(x≠±1 & x≠0 & x≠2)
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.
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
I love your solution! I mostly just winged it, tried x = 1 and x = -1 for f(5) and f(3), then picked more values of x for f(-3) and f(-5) Then I checked by plugging back in (x+2)/(x-2)
Identity of x is different in 2cases. In the function f(x+2/x-2), x is a VARIABLE and value of f on the RHS has been defined in terms of this VARIABLE x. In the simplified expression of f(x), x is the ARGUMENT and the value of f on the RHS has been defined in terms of its ARGUMENT x, as in the way function is conventionally defined. Once we are clear about these 2 subtle but distinct roles played by x, there is absolutely no confusion. For example, when we say x=2, we must be clear that it means VARIABLE x in the original f, which leads to argument of f tending to infinity and value of f as 1. Equivalent case in the simplified f would mean x tending to infinity, since it is serving the role of ARGUMENT here. and indeed, the corresponding value of f with its argument tending to infinity, comes out to be 1 as a limit.
Man I wish I had found you earlier. You make things so interesting and easy. You are such a charismatic person and teacher which makes it very easy for me to learn. Thank you for your videos.
He is a very patient teacher with a very sympathic voice and charisma.
Sympathetic*
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!
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.
i am 70 retired eng;ineer you got my attention love your teaching style and i love math
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
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.
انا من فلسطين . واحب الرياضيات . انت مذهل و رائع . ساتابعك باستمرار . تحياتي
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.
Functional equations were always very cruel to me. Thanks to you, I'm starting to see the light. Keep on teaching!
Never stop teaching Coach !
Thanks
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🇧🇩🇧🇩
I love the fun you have with maths. Your enthusiasm is infectious. I wish my teachers had had half your ability.
this is one of the most compelling math videos it has been my joy to behold. Nice cap, too.
This is where i learnt how to solve functional equations, thank you so much!!
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..
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
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.
Excellent sir. Loved the way you simplified and great explanation.
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
that's amazing. Never seen functional equations before but solving that looked like a lot of fun.
Muy interesante, didáctica y buena clase, a mi hija le servirá mucho esta excelente exposición. Estamos muy agradecidos con su bella persona, bendiciones y éxitos para Usted y su linda familia. ❤
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.
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!
That quote at the end sent me. Very enjoyable personality.
(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.
You just need to be diligent to solve such a tedious exercise. I like the way you're teaching, thanks Prime!
i really like the syle he talks/teaches here!!
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
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. 😃
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.
In the video, the handwriting on the blackboard is the prettiest I have ever seen on UA-cam.
Wow, thank you!
I’m very happy to have found your channel!!!
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.
12:38 you can simply write the numerator as [ (t+1)+(t-1) ]^2=(2t)^2=4t^2
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
Excellent, very interesting this exercise. Thanks so much!!! Greeting from Perú!
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 !!!!
Never seen a black guy do maths, amazing!
11:15 When simplifying [(t+1)^2 + 2(t+1)(t-1) + (t-1)^2)], instead of expanding everything and cancelling out you could have used the general formula (a+b)^2 = a^2 + 2ab + b^2, would’ve been neater.
Your presentation is awesome.
We need more math teachers like this dude
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.
Wonderful manner that conveys such enthusiasm and positivity. I would have understood better if a graph of the function had been included when it was found. That might have helped understand the domain issues that got so many commenters in knots.
I really like this level of maths. Thanks.
I hated maths at school, yet here I am watching this and enjoying it now I'm retired. I guess we just didn't have very good teachers.
I love your introduction sir...
11:18 The numerator is (a+b)² identity. But Absolutely beautiful question and solution!!
The best Math teacher i have ever seen
Iam from egypt
And iam a new subscriber
YOU MAKE MATH FUN🎉
THX❤❤❤❤
easy to understand. you're a great teacher!
A mathematics video has never had a harder plot twist than this 🔥
Dream math teacher around the world❤❤❤
So my takeaway is that when given a functional equation call it f(g(x)) in order to determine f(x) we simply find the inverse of g(x) so that when we plug that into f(g(x)) we get f(x). Sounds simple enough!
Very good example I just wish he would have mentioned the technique in more general terms at the end. After all as a mathematician we want to be able to generalize results.
What you explained is brilliant. That wasn't my strategy in any way. I would try that next time. Thanks
Nice lesson! Congratulations teacher.
Bro you are great! I'm studying maths profoundly at school and your content is exactly what I'm obsessed with. Thank you!
Your enthusiasm is very nice
Wonderfully clear explanation!
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)
You are a very good teacher!
Sou muito fã de suas aulas, obrigado!
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))
Thank you. You are like the Bob Ross of math.
Very nice video! Students will love it! Keep going!
You have a Amazing attitude
A god's gift
Keep up the good work Sir!❤ From Nigeria😊
Did in mind in 2 mins.....by just dividing by x-2 whole square and then manupulating the terms😊
i like this person man, such a happy intraction
t로 치환하는 방법은 미처 몰랐네요. 멋진 아이디어 감사합니다!
Very nice. I like your videos. Just continue
Very, very nice explanation!
Greetings from Brasil
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.
One of the reasons I like your videos is because you use black board and chalk......good old days.
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😢
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.
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
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
The way you explain the steps and logic is really remarkable and I enjoy all your videos.
Very good. Greetings from Brazil
yeah, this is what we did in ms. the method is that which is called the substitution of variates. make t = (x+2)/(x-2)（x≠2 &），then t= g(x), then integrate g(x) into the function on the right side, we will get a f(t)=t²/t²-1(t≠1, x≠0)。so we have f(x)=x²/x²-1(x≠±1 & x≠0 & x≠2)
Можно свернуть в полный квадрат числитель исходной функии и тогда подставлять t. Решается гораздо проще! Максимум 4 минуты!
honestly i liked your explanation quite a lot dam it was interesting how you explained great respect from India Ali 🖖👍
FELICIDADES ERES MUY BUENO
Best math teacher i have ever seen, most think i love is your smile 😊
Bob Ross of algebra
Excellent blackboard techniques.
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.
Thank you, awesome training.
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
This brings back great memories. Do dat math!
The change of variables from calc 2 at the end is so nice
This is the part I didn't understand! Why can you arbitrarily decide to call it x again? I thought x was defined in a specific way
I love your solution! I mostly just winged it, tried x = 1 and x = -1 for f(5) and f(3), then picked more values of x for f(-3) and f(-5)
Then I checked by plugging back in (x+2)/(x-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
Identity of x is different in 2cases. In the function f(x+2/x-2), x is a VARIABLE and value of f on the RHS has been defined in terms of this VARIABLE x. In the simplified expression of f(x), x is the ARGUMENT and the value of f on the RHS has been defined in terms of its ARGUMENT x, as in the way function is conventionally defined. Once we are clear about these 2 subtle but distinct roles played by x, there is absolutely no confusion. For example, when we say x=2, we must be clear that it means VARIABLE x in the original f, which leads to argument of f tending to infinity and value of f as 1. Equivalent case in the simplified f would mean x tending to infinity, since it is serving the role of ARGUMENT here. and indeed, the corresponding value of f with its argument tending to infinity, comes out to be 1 as a limit.
I am inspired by you my Brother
Man I wish I had found you earlier. You make things so interesting and easy. You are such a charismatic person and teacher which makes it very easy for me to learn. Thank you for your videos.
your a great teacher
This is we call assumption but i really like your step it make sense and looks easier.😊
HAA, I did it correct. And....very nice handwriting!
You are awesome, subscribed immediately!
The top could be rewritten as (x +2)^2 that will make the substitution a little simpler.
Very nice video !