Functional Equation

In this video, I showed how tosolve a functional equation

[@WhiteGandalfs]

He is a very patient teacher with a very sympathic voice and charisma.

[@VictorGarcia-gv1ri]

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.

[@stevenwilson5556]

Agreed, he has a perfect attitude to teach!

[@paulw176]

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.

[@sophisticatedplayer]

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

[@PrimeNewtons]

Haha! Now I see it.

[@tiramisu_1th]

yup, this is the comment im looking for

[@Mycroft616]

That is how I handled it, too.

[@LucenProject]

Yup, came for this!

[@Zachary_Roemmich]

I noticed that and was wondering whether you would use it.

[@jacobgoldman5780]

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.

[@glorrin]

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

[@PrimeNewtons]

I was just looking for f(x).

[@JuniperHatesTwitterlikeHandles]

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.

[@adw1z]

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

[@UgyenRangdol-gf8cc]

Every one in the comment going crazy

[@malefetsanekoalane4549]

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. 😅

[@PrimeNewtons]

So nice of you

[@yamada.masahiro]

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 )

[@shaswatadutta4451]

I did exactly the same thing!!

[@PooshanHalder]

Same thing I did

[@metadivergence9523]

Me too

[@shashankks8758]

I tried same thing but missed in expressing 8x in terms of X+2 ad X-2 , thanks for the steps

[@heywrandom8924]

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.

[@johnconrardy8486]

i am 70 retired eng;ineer you got my attention love your teaching style and i love math

[@nihadsaid286]

انا من فلسطين . واحب الرياضيات . انت مذهل و رائع . ساتابعك باستمرار . تحياتي

[@favourtube527]

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

[@mdforhad-wk1zo]

Same bro🇧🇩🇧🇩

[@MervynRThomas]

I love the fun you have with maths. Your enthusiasm is infectious. I wish my teachers had had half your ability.

[@subarnodatta]

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..

[@mrunalhatekar11]

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

[@PrimeNewtons]

Wow! That means a lot to me. Thank you, and God bless.

[@matthewkendall5235]

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!

[@boguslawszostak1784]

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

[@daniel-mircea]

(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.

[@IRanOutOfPhrases]

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.

[@Hardman7]

Never stop teaching Coach ! Thanks

[@zk513]

Functional equations were always very cruel to me. Thanks to you, I'm starting to see the light. Keep on teaching!

[@parthtandel7443]

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

[@fannyliem3536]

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.

[@3v4battler]

@@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.

[@ahpx7553]

Best math teacher i have ever seen, most think i love is your smile 😊

[@sounakkundu6115]

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.

[@fisimath40]

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)

[@brahimaboumohamed1284]

Yes, Cooool !!!!

[@beaumatthews6411]

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

[@labbertubes]

this is one of the most compelling math videos it has been my joy to behold. Nice cap, too.

[@freddyalvaradamaranon304]

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. ❤

[@Rhiannon_Autumn]

that's amazing. Never seen functional equations before but solving that looked like a lot of fun.

[@marxcarton3858]

This is where i learnt how to solve functional equations, thank you so much!!

[@davidchung1697]

In the video, the handwriting on the blackboard is the prettiest I have ever seen on KZbin.

[@PrimeNewtons]

Wow, thank you!

[@gooddude9211]

Excellent sir. Loved the way you simplified and great explanation.

[@clp9823]

That quote at the end sent me. Very enjoyable personality.

[@holmbrg-_-2221]

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.

[@MathsScienceandHinduism]

12:38 you can simply write the numerator as [ (t+1)+(t-1) ]^2=(2t)^2=4t^2

[@gourivigneshgembali3496]

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

[@karryy02]

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.

[@tubetigeerr]

i really like the syle he talks/teaches here!!

[@EDWING6017]

Excellent, very interesting this exercise. Thanks so much!!! Greeting from Perú!

[@john-paulderosa7217]

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.

[@ajamessssss]

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))

[@alangrant5278]

I’m very happy to have found your channel!!!

[@glebannya]

Bro you are great! I'm studying maths profoundly at school and your content is exactly what I'm obsessed with. Thank you!

[@user-qx6br9ds2p]

Your presentation is awesome.

[@benhuralexandredossantos1996]

Sou muito fã de suas aulas! Obrigado muito!

[@BlackPhoton]

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.

[@grandstrategy8987]

easy to understand. you're a great teacher!

[@Sal_da_Terra21]

Nice lesson! Congratulations teacher.

[@Scienceguy0]

The best Math teacher i have ever seen Iam from egypt And iam a new subscriber YOU MAKE MATH FUN🎉 THX❤❤❤❤

[@ritwikgupta3655]

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)

[@mcrow312166]

I really like this level of maths. Thanks.

[@johnstebbins6262]

Wonderfully clear explanation!

[@YakobWakjera]

I love your introduction sir...

[@franklinokafor]

Keep up the good work Sir!❤ From Nigeria😊

[@GargamelTheGreat]

We need more math teachers like this dude

[@kailasnathastro]

The way you explain the steps and logic is really remarkable and I enjoy all your videos.

[@jonathanestrada1064]

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.

[@PrimeNewtons]

What you explained is brilliant. That wasn't my strategy in any way. I would try that next time. Thanks

[@grinreaperoftrolls7528]

THIS IS SO COOL! How was this never taught to me? I made it all the way through cal 3 without ever knowing this

[@ChaosPod]

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

[@naumtrandos4191]

A mathematics video has never had a harder plot twist than this 🔥

[@MohammedAli-jt7zr]

honestly i liked your explanation quite a lot dam it was interesting how you explained great respect from India Ali 🖖👍

[@stevenwilson5556]

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.

[@thexavier666]

Your enthusiasm is very nice

[@deventerprises2640]

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

[@tanoukouassialban7501]

Very nice. I like your videos. Just continue

[@andreasretsinas2142]

Very nice video! Students will love it! Keep going!

[@roronoazoro8343]

i like this person man, such a happy intraction

[@dan-florinchereches4892]

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

[@xavieryates9782]

At the end you simply substituted de 't' by 'x', although t = (x+2)/(x-2). It seems to me that this is wrong.

[@rogerfroud300]

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.

[@aadityavikram5030]

One of the reasons I like your videos is because you use black board and chalk......good old days.

[@edmurnico7508]

Very, very nice explanation! Greetings from Brasil

[@mdforhad-wk1zo]

Dream math teacher around the world❤❤❤

[@ZqTi0]

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.

[@MathsScienceandHinduism]

5:36 you could have easily got x in terms of t by applying componendo dividendo. btw nice solution

[@PrimeNewtons]

Seriously I need to google that

[@Jperre2010]

Excellent blackboard techniques.

[@5herwood]

You write neatly and you don't stand in front of your work. You have great delivery--clear and concise. You don''t need long generalizations though. Just dive right into the solution. The 84 second introduction is distracting rather than enlightening. If I'm being presumptuous by making critical comments I apologize. No more to say.

[@PrimeNewtons]

Feedback is appreciated. We never stop learning!

[@puckchibi]

You are a very good teacher!

[@nYEOSUh]

t로 치환하는 방법은 미처 몰랐네요. 멋진 아이디어 감사합니다!

[@vivekrajput..]

You have a Amazing attitude A god's gift

[@francescocastaldi4045]

Why did you not substitute t with (x+2)/(x-2) at the end?

[@ThePROestRedn99]

Did in mind in 2 mins.....by just dividing by x-2 whole square and then manupulating the terms😊

[@jandirpassos5327]

Very good. Greetings from Brazil

[@Navxxn12]

Sir here we use partial derivative method by first using partial derivative wrt x and then simplifying to eliminate x or yand then b putting f'(x)=dy/dx we can use veriable seperable form to integrate then we will finally get our funtion

[@kalakla]

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

[@jeffcava007]

Right

[@tubesteaknyouri]

Thank you. You are like the Bob Ross of math.

[@henry_dschu]

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)

[@randykim9693]

Are you a mathmatics teacher in USA?

[@ridzkiwachyudin5804]

Umm, can i say that f(x+2/x-2) is just the same as f(g(x)) or composite function, so in order to get the function f(x), we need to find the inverse function for g(x)?

[@xyz9250]

The top could be rewritten as (x +2)^2 that will make the substitution a little simpler.

[@dushyantrastogi7342]

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

[@aditishah9469]

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?

[@borisdelaine9797]

This brings back great memories. Do dat math!

[@ayoshyadav-yu2dd]

Sir plz bring challenging problem in functional equation mixed with calculus

[@FinalMiro]

13:00 can someone explain the last part? I don't really understand why t got replaced by x on all sides.

[@glorrin]

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.

[@n_man-im2fv]

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

[@biscuit_6081]

The change of variables from calc 2 at the end is so nice

[@peterc.hayward8067]

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

[@nanamacapagal8342]

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)

[@skwbusaidi]

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

[@r04it]

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

[@simeon7450]

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?

[@amoswittenbergsmusings]

We can simplify the algebra if we first factor the numerator, convert thedenominator toa difference of squares and then replace x by x-2. Step by step: f((x+2)/(x-2)) = (x^2+4*x+4)/(8*x) f((x+2)/(x-2)) = (x+2)^2 / ( (x+2)^2 - (x-2)^2 ) f(x / (x-4)) = x^2 / (x^2 - (x-4)^ Now set t=x / (x-4) and solve for x: x = 4*t / (t-1) As an intermediate step we first express x^2 and (x-4)^2 in terms of t: x*2 (4*t)^ / (t-1)^2 (x-4)^2 = 4^2 / (t-1)^2 We can now cancel 4^2 / (t-1)^2 and get f(t) = t^2 / (t^2-1) without further hassle. Alternatively, after recognising the denominator as a difference of squares we can divide numerator and denominator immediately: f((x+2)/(x-2)) = (x+2)^2 / (x+2)^2 - (x-2)^2) = ((x+2)/(x-2))^2 / ( ((x+2)/(x-2))^2 - 1) Now we have the function in terms of the argument and can write without further ado: f(x) = x^2 / (x^2 - 1)

[@adamoksiuta4715]

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.