 functional equations. This topic is I would say it doesn't have a you can say unique way or procedure or standard operating procedures to solve questions. They're very unpredictable type of questions. You have to you have to think in different way laterally to find your function out or whatever is the required condition out. So many a times these questions are tested in olympiats. If you look at international mathematical olympiats and all you'll find that there is a chapter of functional equations. So the fact that it is being tested in olympiats clearly suggests the fact that these topics are not that straightforward. They're not conventional that you have certain you know methods and formulas and you can apply it to get the function back. There are a lot of hidden trials you have to do. We have to see the periodicity of the function. We have to see what happens to the function when you change the name of the variable. Many times you need to do differentiation also. So I think you have already been exposed to functional equations before so I don't need to talk more about it. What I'm going to do now is I'm going to give you some commonly seen functional equations. But let me tell you that is just a list which you can just copy it on your notes but might not be useful at all. Okay so let me give you some some instances of functions that satisfy certain functional equation. The most commonly seen one is the one which is this one. f of x plus y is f of x into f of y. Mostly the functions which satisfy these kind of functional equations are of the type a to the power x. Okay so this is a commonly seen solution for this type of functional equation and I don't think so I need to tell you why. Okay next type of functional equation that you will see is f of x plus y f of x plus y sorry not f of x plus y it's f of x plus f of y is f of x y. Such kind of functional equation is satisfied by logarithmic functions. So log x with some base a. Okay where you have to figure out that a by the use of some additional condition given to you in the question. Okay needless to say if you if you have f of x minus f of y is equal to f x by y the same type of function will satisfy that is logarithmic function. If you have something like this f of x plus minus f of y okay a plus just consider plus to be once and then minus and same in the formula also the top part will correspond to the plus and the bottom part will correspond to the minus. So I'm writing it in one go. Can you tell me which function satisfies this function sign inverse x. Yeah which function satisfies this functioning equation. This is an inverse. Okay now the next one is the one which is most commonly asked f of x into f of 1 by x is f of x plus f of 1 by x. Okay now this is satisfied by a polynomial function which is given by plus minus x to the power n plus 1. So this function this is something which we need to know. You can easily verify it also. You can easily verify it also. So this type of functional equation is very commonly seen. Okay try to prove this proof is very simple just take a polynomial function in general and try to find the coefficients of that polynomial function which will meet this criteria. Okay you realize that only plus minus will come as a coefficient. That's all the terms would be zero. Do you want me to do this? Do you want me to prove this? Okay let's let's prove it. Prove for the seventh one. So if you have when we claim that a polynomial let's say the polynomial is this satisfies this. Okay now if you put it in this given functional equation f of x we all know is a0 x to the power n a1 x to the power n minus 1 all the way till let's say a n into f of 1 by x will be a0 by x to the power n a1 by x to the power n minus 1 all all the way till a n. And this is equal to sum of these two functions. Now if you compare the coefficients let's start comparing the coefficient of x to the power n on both the sides. Let's compare the coefficient of x to the power n on both the sides. What will I get on the left side? How will you generate x to the power n? Only when a0 multiplies with a n correct? So if a0 multiplies with a n you will get x to the power n. There's no other way to get it. And on the right side it will only come from a0. Yes or no? Now this means two things a0 could be 0 or a n could be 1 but this is not possible because in a in a polynomial expression the leading coefficient cannot be 0. So if you're considering it to a degree n a0 cannot be 0. Correct? Let's compare the coefficient of x to the power n minus 1. What do we get on the left side? Which term will contribute to a x to the power n minus 1 terms? Yeah sorry I was thrown out of the call. Can you see my screen now? Yes. So if you compare the coefficient of x to the power n minus 1 on the left side you will get a0 a n minus 1 correct? No because the term before this will be a n minus 1 by x and that multiplied to this will give you x to the power n minus 1. Can I say I will also get x to the power n minus 1 from a1 a n correct? And this should be equal to a1 on the right side because only this term will have x to the power n minus 1. Now remember a n is 1. a n is 1 means this term will be a1 again. So a1 a1 gets cancelled off that means a0 a n minus 1 is equal to 0. a0 cannot be 0. What does it mean? a n minus 1 has to be 0. Am I right? Similarly you realize that a n minus 2 would also be 0. a n minus 3 would also be 0 and so on. Now what about the last term? What about the last term? I think we had already figured out the last term is going to be 1. What about a 0? Sorry what about a 0? Let us compare the constant terms on both the sides. If you compare the constant terms on both the sides you would realize on this side you will get a n square. Let me write comparing constants on both the sides. Comparing constants on both the sides you will get a n square and on this side you will end up getting constant term as as as as as how much this into this sorry a n square and also a not sorry my bad a not square that will also give you constant term correct? Rest other terms they would be there but they would be 0 because a 1 square a 2 square etc they will all be 0 we have already seen that over here so those terms you need not write. On the right hand side on the right hand side the constant terms would be a n right? So this gives you a not square plus 1 is equal to 2 so a not square is equal to 1 so a not could be plus or minus 1. In other words you realize that a not is plus minus 1 rest a 1 a 2 all the way till a n minus 1 is equal to 0 right and your last term is 1 that means your given polynomial equation was plus minus x to the power n plus 1 so this becomes your solution for that functional equation okay other than this I would like you also to note down actually do you want to copy this or should I move wide you want to copy this please do so okay few things I would like you to add on to that list number one was was there any numbering for the same did I give you any numbering for the same okay just I would write that important points here if your function satisfies this functional equation that is f of a minus x is equal to f of a plus x remember the graph of such a function will be symmetrical about x equal to a line okay so if any function graph if any functional equation is satisfying this functional equation then it's graph would be symmetrical about x equal to a line please make a note of this very important concept several questions have been framed on this in past and if you realize your function f of a minus x is negative of f of a plus x then the graph of this function is symmetrical about the point a comma zero even function and odd functions are special cases for this if you're a is zero if you're a is zero then it becomes f of x becomes a even function okay here if you're a is zero then your f of x will become an odd function because a comma zero will become origin in that case and x equal to zero will become the y-axis in that case okay but these are the properties in general that we need to know these are the properties in general that we need to know okay so what I'll do is I'll take a quick break over here I'm sure most of you would be hungry now so let's take a break right now the time I'm assuming it's 11 17 am will meet at 11 32 am with more questions on functional equations I think I won't be able to do much in different definite integrals but I'll see if I can start the topic okay see you after the break so let us begin with some questions on functional equation again as I told you there is no you can say conventional way to solve questions you have to try a lot of things you have to try special values you have to try putting one value as the other and see whether you can solve it I think most more clear more clarity will come when we take questions let us begin with this question of x is x plus f of x minus 1 if f 0 is given to you as 1 find f 100 easy question to begin with okay that's not correct by the way correct richer but I'm here oh no he is he has made a mistake richer party is correct but I'm is correct that's deep again small mistake okay let's discuss this I think you would have got an idea how to do it hey do you realize that your f of 1 is 1 plus f of 0 that's nothing but 2 okay f of 2 is nothing but 2 plus f of 1 which is 4 right f of 3 is 3 plus f of 2 which is 7 right so the trend that is going here so your first term sorry your 0th term is 1 right this is your 0th term first term is 2 then 4 then 7 I think you can make out that the difference between these terms are in AP so 1 2 3 the next difference will be 4 so it will be 11 okay so this is your first term this is your second term and like that you need to go till your hundredth term which is your hundredth expression okay now how do I solve this question let's say I call the sum of these terms as let me write f 100 as s okay write the same as 1 shifted okay subtract it what you end up getting 0 as 1 plus 1 plus 2 plus 3 all the way f of 100 minus f of 99 will give you 100 if I'm not mistaken correct f of 100 f of 100 minus f of 99 will give you a hundred correct and this finally is minus f of 100 so take it on the other side so f of 100 will be 1 plus summation of 1 to 100 which is 100 into 101 by 2 so answer is 1 plus 5050 which is 5051 easy question you should have got it okay any questions here anybody okay let's move on to the next a function f of x is defined for all real x if f of a plus b is f of ab for all a and b are as inputs to the function f of minus half is minus half then find f 1005 okay Richard both the Richard's okay good okay Keaton okay let's check this out see let's say I put a as 0 in this let's say I put a as 0 so I'll get f b is equal to f 0 correct even if I put b as 0 I get f a is equal to f 0 okay what does it imply it implies f a is equal to f b for any a and b for all a and b right correct that means any two values you put there equal what does it comment on the nature of the function it comments that this function is actually a constant function if it is a constant function f of 1005 will be same as f of minus half the answer will be minus half only okay this is no clear cut defined way to solve it it's up to you how smart you are in lateral thinking that you can solve it okay the real fun in maths is when there is no method known to solve it and you're trying to solve a question okay else maths becomes physics and chemistry which is driven by laws and rules math is supposed to be an abstract concept that is why these topics are favorites in Olympiads is that fine can I move on to the next question any question that you would like to ask I'm just giving an exposure it's not like you know I can take all the different types of functional equation question no I'm just giving an exposure how to deal with such situations f of x satisfies f of x plus f 2x plus f of 2 minus x plus f of 1 minus sorry f of 1 plus x is x for all x belonging to real number find f 0 find f 0 correct here yeah I can read your message because yeah I can read your answers I can see which is things answer because answer he has answer which is thing we'll discuss yeah we'll discuss it others please respond okay Chaitanya I got your answer see sometimes what happens when the power goes off and I'm thrown out of the call it doesn't allow you to send messages to me so what is the approach here correct Ruchir Parik so what is the approach here you put x is 0 so if you put x is 0 you get f 0 plus f 0 plus f 2 plus f 1 is equal to 0 okay that means to f 0 plus f 1 plus f 2 is equal to 0 then what do you do you put x as 1 because you want to know f 1 and f 2 also so let's try putting x as 1 so f 1 plus f 2 plus again f 1 plus f 2 is equal to 1 that means 2 times f 1 plus f 2 is going to give you 1 which means f 1 plus f 2 is equal to half okay and from here I know that f 0 is negative half of f 1 plus f 2 so clearly your answer will become minus half into half which is minus 1 4th so if you have got minus 1 4th your answer is absolutely correct those who got a half or a minus half please check where you went wrong okay easy okay next problem if a function f satisfying f x y is equal to f x by y for all positive real all positive all positive real x and y if f 30 is given to you f 30 is given to us 20 find f 40 absolutely correct here because kdg adhrik ruchar parek padyat so how do I get this very simple if you want to get f 40 try writing 40 as such a combination which will give you 30 in its result so I think I can write it as 40 is 40 can be written as 30 into 4 by 3 right so if I write it like this I can write it as f of 30 divided by 4 by 3 that means 20 into 3 by 4 that's 15 15 is the right answer for this okay easy let's move on to the next okay this is our old function that we had derived which had proved a little while ago f of x is a polynomial function satisfying this condition f 4 is 65 find f 6 excellent okay again very easy question now you know that a function that satisfies this type of equation is of the nature x to the power n plus minus plus 1 now if f 4 is 65 that means you're claiming that this is 65 that means 64 is plus minus 4 cube oh sorry I already taken yeah yeah this can only happen when you have taken your sign as a plus okay sign should be plus and n value should be 3 that means your function is revised as x to the power 3 plus 1 so f 6 will be f 6 will be 6 cube plus 1 which is 270 absolutely correct any questions here easy next f is the real valued function says that f of x plus 2 f of 2000 by x is equal to 3x find f of x yes any success anybody f of x in terms of x what else what type of question f of x will be in terms of x only no or without x also it can be can be constant also no that's not correct which is saying but good you tried others correct sorry through I'm habituated to say Dhruva because of one student that we have correct Dhruva ah no because KTG also no what is the full form of KTG that it's Keethan Gopal Rishnan sir KTG I've heard of somewhere bike is it a bike those bikes so that's KTM those stupid of me sorry KTM okay see guys when these kind of questions are there the simplest way to solve this is one equation is this other equation is you replace your x with 2002 by x okay you can play with your inputs so this side will become f of 2002 by x and this side will become back f of x and of course this will become 2006 by x right correct now think as if you have a simultaneous equation x plus 2 y equal to something and y plus 2 x equal to something and you want to solve for x correct so what I will do is I'll multiply this with a 2 okay I'll multiply this with a 2 so I'll make it as 4 f of x plus 2 times f of 2002 by x plus I mean 12,012 whatever is it okay I think I put more zeros than required okay anyways we're now going to subtract the two results we're now going to subtract this result with this result okay so 3 f of x will give me this x minus 3x divided by 3 so your f of x will become 4004 by x minus x so this is your answer so your function is 4004 by x minus x this is a very commonly seen question okay it's not a difficult one is that fine any questions here anybody can I move on to the next one let me know once you're done with this done copied yes okay done I think Ruchir Singh is correct because that's not correct Ruchir Parik is correct oh I hope you have regained the internet now Baram yes it's back okay let's discuss this see here you can play with your x and y right you can play with your x and y now what I'll do is I will choose my x and y relation in such a way that this input here becomes same as this input that is I'll choose my x and y relation in such a way that 2x plus y is equal to 3x minus y in other words I'll choose y as x by 2 if I do that see what will happen this will become f of x this will become f of x square by 2 this will get already cancelled with this so I'm not writing it and this will remain 2x square plus 1 so directly you can get your f of x as x square by 2 in fact minus x square by 2 plus 1 this is your x so ultimately the trick was to substitute this in place of y let's take a slightly more difficult question on this then we can wrap this up so it is known that there is a function from natural numbers to natural numbers in fact I'll include 0 also so from whole numbers to whole numbers okay and 0 means including 0 natural numbers including 0 okay better I like whole numbers only else people will get confused so there's a function whose domain and co-domain are whole numbers and this function satisfies this criteria f of x square plus f of y y is x f of x plus y f of x square plus f y is x f of x plus y for all x and y belonging to whole numbers find the function correct Ruchir did you guess it use y as equal to 0 otherwise orders sound something like a because no yeah kind of guess is that yes okay correct Chaitanya correct KTG okay anybody wants more time or should I discuss it correct param let's discuss this I'll do it in a very rigorous way I think some of you would have also guessed it let me do it in a very rigorous way okay let's put x as 0 and y as x let's let's put x as 0 and y as x in the given in the given functional equation so that will give you f of f of x is equal to x f of f of x is equal to x let's say I call this as one of our findings now let me put x as 1 and y as 0 let's put x as 1 and y as 0 so it'll give you f of 1 plus f of 0 is equal to f of 1 right okay now let me do one thing let me take f of f of 1 plus f of 0 as f of f of 1 that is known to be 1 and here it really known to be 1 plus f of 0 because of the property number 1 see f of f of x is giving you x so here f of f of this guy should be this guy correct and here f of f of 1 should be 1 what does it mean it means f of 0 is 0 that means f of 0 is 0 no problem so far okay now what I'm going to do is I'm going to put x as 1 and y as f of x in the very same equation in the very same functional equation I'm going to put let me let me take it on the upside as I'll lose a track of it let's take x as 1 and y as f of x so this will give me 1 plus f of f of x it will become isn't it that is actually x and this will become f of x plus f of x oh my bad x is 1 right so f of 1 1 f of 1 f of 1 plus f of x so far so good okay now what does it say it says f of 1 plus x minus f of x is f of 1 always let's say I call f of 1 value as a okay let's start putting the values when you put 0 f of 1 will be a plus f of 0 f of 0 is known to be 0 so f of a is a f of 2 will be f of 2 will be put x as 1 here so it'll be 2a f of 3 will be 3a and so on okay so it gives me a general feeling that f of x would be xa correct now what is this a I have to figure out what is my a for that I can put this in the original equation I have so in the original equation in the parent equation that has been given to me I'll put these inputs in ax and find it out so my f of x is known to be xa so if I use here it becomes x so a times x square plus a times f of y f of y itself is a y so I can write this as a square y correct is equal to ax square plus y so if I see this condition I'll get a square is equal to 1 right so my a has to be plus minus 1 but it cannot be minus 1 because if it is minus 1 f of 2 would have been minus 2 and minus 2 is not allowed in the core domain so it can only be 1 right if it is only 1 your function here would actually be x this is your answer is that clear a very rigorous way to do it most of you would have taken a guess am I right how many of you did it rigorously very few I did like half rigorously half rigorously okay let's take another one again these are not easy questions let me tell you probably if you get them them in j advance the only option that you have is to take a clever guess so there's a function from integers to integers there's a function from integers to integers this satisfies we satisfies this condition f of f of n plus 2 is equal to n okay and it is also known that f 1 is 0 find f of n correct Mahit Richer Singh Richer Parik Vikas how did you guys do it correct pattern coming up okay yeah anybody who did it rigorously so it's my thing correct because I did it rigorously I don't know if it's correct yeah it's correct okay f of n is actually n minus 1 how do you do it actually now that I look at it and say it's kind of rigorous I don't think yeah if you put value in just start getting the previous summer okay so from there you figured out it's n minus 1 pattern yes okay anybody who did it without pattern I'm not saying that by knowing the pattern it's not correct absolutely correct anybody who did it without doing the pattern let's look into this in a rigorous way a lot of analysis is involved over here if you call this inside function as let's say g of n okay that means you are writing something like this now this is something which we call as f o g function isn't it so f o g function is equal to n right we know that n or you can say f of x is equal to x kind of a thing this is a 1 1 and on to function correct we all know that f of x equal to x even if your x is only allowed to take integers that will be a 1 1 and on to function okay no doubt about it okay now if composition of a function is one one and on to then tell me which one of them should necessarily be one one and which one of them should be necessarily be on to if f o g is one one which one of them should be necessarily be one one f no g yeah g is necessarily one one g must be one one g must be one one f may or may not be one one g must be one one because if g is not one one for two different values of the input g will show the same answer and f and hence f will show the same answer okay that will make the entire function become non one one or you can say uh non-injective right similarly if f o g is on to then which of them must be necessarily be on to f must be on to i think i had given this as a property to you when we did the types of functions please revisit those notes i think i had discussed with you uh in the chapter inverse functions because i was i needed this one one and on to concept to talk about inverse of the function okay so g must be one one okay and f must be on to now here one more thing that you'll see since g of n let me write over here since g of n is f of n plus two and if this is one one then can i say f of n must be one one yes sir and if f is if f of n is on to can i say g of n will also be on to in this short what i'm trying to say is that both these functions are one one and on to that means f n and g n both are one one and on to that means they are bijective and if they are bijective both are inverses of each other that means f and n are inverses of each other because they meet this criteria because they meet this criteria that means f o g is giving you an identity function right and they are both 1 and 2 that means f and g functions are inverses of each other I think till here nobody has any problem let's go further now it's given to me in the question that f1 is 0 f1 is 0 means what f0 is 1 sorry g0 is 1 isn't it because both the functions are inverses of each other am I right now g0 is what g0 is f0 plus 2 that means f0 is minus 1 is that fine agreed okay now since f of n plus 2 is g of n let us take let us take a composition with f okay so we have already seen that this isn't is already given to us right so what I'm going to do is let's not do this step this step is of no use let's start with f of n plus 2 is equal to gn let us replace n with f of n let us replace f with f of n so this will become f of f of n plus 2 is equal to g of f of n okay and g of f of n is known to be n yes sir that means f of f of n plus 2 is equal to n now replace your n here replace your n here by f of n plus 2 replace your n here by f of n plus 2 so what will happen f of n plus 2 I'm just writing the right side first on the left side you'll get f of f of f of n plus 2 plus 2 yes or no now remember f of f of n is given to us as n minus 2 from this equation so can I write can I write this term as can I write this term as the one on the right hand side as f of n plus 2 minus 2 plus 2 correct me if I'm wrong that is nothing but f of n plus 2 so what does it mean it means I have gone out of space it means f of n plus 2 is equal to f of n plus 2 now start putting some values let's start let's start our value with some let's say I start with a zero okay f of 2 f of 2 is f of 0 plus 2 f of 3 is f of 1 plus 2 f of 4 is equal to f of 2 plus 2 f of 5 is equal to f of 3 plus 2 and so on if I go on all the way till f of n it becomes f of n minus 2 plus 2 add them all you would realize that when you're adding terms starting from here will start going off this will go off with this this will go off with this okay yes or no so what will be left ultimately what will be left ultimately can I say f of n minus 1 will be left here and f of 0 and f of 1 will be left here so what do I get from there what do I get from there f of n minus 1 plus f of n will be equal to f 0 which is already 0 f 0 is minus 1 because there only sir oh f 0 is minus 1 and f 1 is 0 right so this term will go off completely what was f 0 value minus 1 yeah f 1 was 1 and how many 2s will you get so f 1 is 0 oh f 1 is 0 f 1 is 0 and how many 2s will you get n minus 2 n minus 2 plus 1 so okay minus 1 but does it help me to get f of n I'm still left with f of n minus 1 right yes so post this I think we need to check the pattern I think I mean we did all this thing to get back to the pattern checking n plus so f 0 is known to be minus 1 f 1 is known to be 0 f 2 would be 1 1 yeah I think is there any better way to reach the result after having got this relation it's a recursive relation can can you try x square minus x minus 2 is it factorizable no no this factorizable no yes so x minus 1 and we get x minus 2 right no sorry you take x x minus 2 plus 1 x minus 2 yeah x plus 1 and x minus 2 is equal to 0 so I think I got it from here so once we added those things if we add everything we get f of n plus f of n minus 1 is equal to 2 n minus 3 and then we add we added those things that you got right all of those they add up to f of n plus f of n minus 1 is equal to 2 n minus 3 you're talking about yes when we added all these equations we got f of n plus f of n minus 1 is equal to 2 n minus 3 okay and then when we substitute n as n plus 1 we get f of n plus 1 plus f of n is equal to n minus 1 and then we subtract these two and we get a constant value so we'll get it's a linear function f of n plus f of n minus 1 is 2 n minus 3 f of n f of n minus 1 is was it coming yeah it's n minus 1 times okay 2 n minus 3 correct yes so now we take n as n plus 1 so then we get f of n plus 1 plus f of n is equal to 2 n minus 1 okay then we subtract the 2 so we get f of n plus 1 minus f of n minus 1 as or 2 and that's a constant value so it doesn't depend on n so that is a linear function if you get f of n plus 1 minus f of n minus 1 yes that'll be equal to 2 so it's a linear function then anyways we got here no how do you see it's a linear function oh yeah sorry sir I'm not excited okay now see one thing that I can say here is it is alpha minus 1 to the power n beta 2 to the power n this is your f of n f f 0 was minus 1 right f 0 was minus 1 so alpha plus beta is minus 1 and f 1 was 0 f 1 was 0 okay so if you added beta becomes minus 1 third okay and and and and and alpha becomes minus 2 third so minus 2 third minus 1 to the power n minus 1 3rd 2 to the power n is your f of n okay let's check what are you what were you getting for f f 1 was 0 right yes is it coming out to be 0 yeah f 2 what were you getting for f 2 on f 2 was 4 by 3 so then it's not correct no no no it starts becoming corrupted so why did you take this to be f of n though actually I was trying a recursive relation so this is something called the auxiliary equation but you need not know all that thing anyways see in this case in this case in this case I think you have to take a guess I have to you have to take a call with the pattern so there's no other route to that so if you use that telescopic sum you'll only end up getting stuck with this relation so yes f 0 is minus 1 f 1 is 0 f 2 is 1 that will only leach to f n is equal to n minus 1 can't say anything but yes you can say from here that can you say from here that it's going to be a linear function can you say from this that is a linear function from here because quadratic term will generate a function of n right so as soon as you can prove linear then we already have enough because we have f of n is equal to y if you can prove it's linear then I think we can we can directly take in a function as a x plus b and proceed but is the question here is if a function satisfies a condition like this is it always linear let's say if you take a quadratic and if you subtract this will we get a n square minus b n and c will get cancelled off so this will become a yeah this will leave you with 2 n plus 2 into 2 plus 2 b which will leave a term of n at least yeah it has to be linear so Richard was not completely wrong yeah this has to be linear if it is linear then yes you can go ahead and prove that it is n minus 1 so if it is a linear term then a n plus 2 plus b is equal to a n plus b plus 2 so 2 a plus b is equal to b plus 2 so a is going to be 1 okay and if a is 1 that means your function your function is n plus b kind of a thing and you know f 0 is minus 1 right that means 0 plus b is minus 1 so b is minus 1 so this function was n minus okay so these are the type of questions you can expect in advanced version it will not come in j e main for sure fine so with this we close the chapter of functions for you