Nice problem. Same kind of idea. Let A=(x-3)/(x+1) and B=(x+3)/(1-x). If you repeatably substitute x in A with A you cycle through A ---> B, B ---> x, x ---> A, A---> B,... So, doing the same repeated substitution for this problem you get the 3 equations f(A)+f(B) = X, f(B)+f(X)=A, and f(X)+f(A)=B which quickly reduces to f(x)=(A+B-x)/2=( 8x/(1-x^2) - x)/2
I really liked this question. It was the perfect sort of difficulty for me, as I was able to solve it, but at the same time, it was still difficult enough that I had to think about it for a while in order to work out how to answer it, as I am still very new to functional equations. I referred back to other functional equation videos in order to work this one out. Great video anyway!
@@SyberMath No worries at all! It was awesome 👏, I really love all these sorts of problems you always present to us, and for me at least, most of them are challenging but still doable, but the odd one is impossible for me. Also, when I do manage to solve one, you always seem to have this neat, clever trick that is like 10x quicker than the way I did it. And I love seeing those neat tricks, it’s so amazing seeing how they work, and now I’m finally beginning to use them myself which is even better. So thank you very much for helping me with that Anyway, congratulations on recently hitting 20,000 subscribers, and I, along with plenty of others, am really, really enjoying these videos, so thank you very much for continuously making them and continually stretching my mathematical abilities every day. Sorry for the long essay. Two more quick things: 1. I really admire the fact that you reply to basically everyone’s comments in the comment section, it is very nice to see that you have taken the time to do that, and it is very nice to get a response from you, it’s very nice to see 2. Do you have any book recommendations for questions like the algebraic equation solving/number theory type problems, particularly like the ones that you have done on your channel? I would like to practice them some more Sorry for the really long comment Thanks again and congrats!
Here's a slight reformulation of the solution. If you set g(x) = (x-3) / (x +1), then the original equation can be rewritten as f(g(x)) + f(g(g(x)) = x By substituting x = g(x) and x = g(g(x)), and using the curious fact that g(g(g(x))) = x we also have f(g(g(x)) + f(x) = g(x) f(x) + f(g(x)) = g(g(x)) So we have three linear equations with three unknowns: f(x), f(g(x)), and f(g(g(x)). For example by adding the second and the third, and subtracting the first we get: 2f(x) = g(x) + g(g(x)) - x EDIT: I just realized that several other posters proposed the same idea before. Well, I'm keeping the post in case someone finds this mini-writeup useful.
This is a really elegant approach. Thanks for sharing. When I did the problem, I saw that we had f(g(x)) + f(g⁻¹(x)), and that g(g(x)) = g⁻¹(x), and that g⁻¹(g⁻¹(x)) = g(x), but didn't think to put it all in terms of g(x), g(g(x)), and g(g(g(x))). I think your approach is much tidier.
A less clever and more systematic way to think about it would be to include the original equation in the system of equations. Then there are 3 equations and 3 unknowns, and you can use well known methods like gaussian elimination to solve for f(x)
I've only started watching recently, and this is my favorite so far. Before viewing, I was able to solve the problem after much trial and error, with a similar but less efficient method. In other words, it took more algebra. Anyway, I've learned a lot about solving functional equations. Thanks!
Finally i found technique for equations like this…..thanks i have been looking for this for along time…..this is really really useful for someone like me
Wow. That's the first functional equation I have ever solved and it's correct. What can I say, I did exactly what you do in the video. I even used the same letters for a substitution. : )
@@SyberMath Why not just replace the second expression with something in terms of y so you dont have to introduce a third variable..more streamlined and elegantno? That's how I did it.
Love the problem and plan to share it with my students. You may note that the two arguments are inverses, so there's another way to find f(x)! I'll post a video if anyone is interested.
I spent a good amount of time figuring out how to plug in the right values to get enough equations to cancel out the noisy parts. It’s essentially the same as what you did.
On doit vérifier que la fonction f satisfait à l'équation de départ puisque dans les étapes de la résolution on a procédé par implication et non par équivalence. Merci
I verified that the function you found does indeed work, but I like to try to be rigorous in my approach to these problems. I understand the concept of a dummy variable. However I am not convinced that you can always do this in general. x, y and z are all distinct values. In this specific case, one of the inner x expressions was the inverse of the other one. I am not sure this technique would work in a more general case. To understand what I am getting at, make a table of values for x, y and z. When x=0, y = -3 and z = 3. So what you then have is that f(-3) + f(3) = 0. Now pick x=3. When x=3, y=0, and z = -3 So what you get in that case is f(0) + f(-3) = 3. I am not convinced that you can just casually replace y and z with the same dummy variable t and still have the the table of values work out the way I described. I think it works only in very specific cases, like this one, where one inner expression was the inverse of the other inner expression.
Formally there is an omissis. After the expression of X as a function of Y, you need to set out the condition that Y is different from 1. Same applies to Z different from -1.
Very informative....but at @7:00 you say replace z with x but note that it is NOT the same x as in the original equation.. Then you go on to also replace y with x. That is also NOT the same x as in the original equation. So you CANNOT add the equations as you do @8:42. If you are allowed to do that then you could also replace (x-3)/(x-1) in the original equation with y. Replace (x+3)/(1-x) in the original equation with z. And then replace y and z with new x…. noting it it’s not the same as the original x. Am I missing something? Can someone point me to a text book that covers this type of functional equations please. Thanks.
Does it help you in anything to notice the input of the 2 functions are the inverse of each other? In order to generalize the result? I can't find yet the connection if they are inverse functions like this case
that's the first thing I noticed, but it's more than that: f(y(x))+f(z(x)) = x (y and z are functions here) since z(x) = y-1(x) we have: f(y(x))+f(y-1(x)) = x but notice how y(y(x)) = y-1(x) and similarly y-1(y-1(x)) = y(x) This is the crucial property y has, so by doing; f(y(y-1(x))+f(y-1(y-1(x))) = y-1(x) you get: f(x)+f(y(x)) = y-1(x) and by doing: f(y(y(x))) + f(y-1(y(x))) = y(x) you get: f(y-1(x)) + f(x) = y(x) by adding the 2 together you get: 2f(x) + x = y-1(x) + y(x) So the real challenge in this problem was finding this niche function y props to the author of the book once you figure out this trick you can instantly write 2f(x) + x = y-1(x) + y(x) and solve!
Good question! There are some books but they do not start at the basic level unfortunately. I'm planning to come up with a document that explains the basics but who knows when I can write it up
I solved f(2x)=f(x)^2 the Hard way given f(x)≠0 and f'(0)=alpha and I can thank 2^n for that, without 2^n I wouldnt have solved it,2^n is my New Best friend also the equations that I Just solved has e^(xalpha) as solution,also It solves f(x+c)=f(x)f(c) since if c=x,then the solution is the same.
@@resilientcerebrum some tips to get you going: something like f(x+3/1-x) has an expression inside the function argument and takes the general form of f(g(x)) So if you have an equation like f(g(y)) = h(y), you can solve it by defining x = g(y) F(x) = h(g^-1(x))
In general I always get started with this kind of problems by trying plugging in some special values and see if I can find any patterns or get some insights during the course of the calculation. For this problem I tried 0, 3, -3 and found it's always about f(3) f(-3) and f(0) and so I can solve all 3 of them. Then I tried 5, -5 and found f(1/3), f(-1/3) f(2) and f(-2) also showed up, so I added in 1/3, -1/3, 2 and -2, then I found it's all about f(5) f(-5) f(2) f(-2) f(1/3) and f(-1/3) and all these 6 can be solved. This gave me some insights as my gut feeling was that this pattern probably can be generalized for any given values, i.e. if I start with 1 or 2 values and keep adding the new ones showing up in the f(), then with a few iterations I should be able to get back and form a closed domain. So I did the same again but with the letters this time, and guess what, this ended up with essentially the same methodology presented in the video.
This was ok until the 7.00 minute mark, where it turned into a mathematical "Shell Game". When Y was replaced by X, that X was different to the X we started with. Same thing happened later when Z was replaced by another X. So we have three different values of X which are represented by the same X in the following equations. How does that make sense?
Keep in mind that the x here isn't the unknown to be solved for, it's just the placeholder used to describe a rule. The unknown the problem is asking us to solve for in this case is a particular rule, a function, namely, what is the rule for f? What rule describes it? Commonly we write f(x) when describing the rule but we could use f(z) or f(w) or f(whatever). It's how we manipulate thing in the parenthesis that counts.
Holy crap I found the answer by trial and error. I reasoned through and evaluated f at 0 and plus or minus 1/3, 2, 3, and 5. Can prove that f is odd. Plotted the points. Noted that there's probably vertical asymptotes at plus and minus 1, so put an (x-1)^2 in the denominator. Fiddled around with cubic equations in the numerator. Presto. Okay, now to figure out how to actually do it.
At 7:50 by using X to replace both Y and Z isn't he supposedly assuming that y equals Z? What justification do we have to do that? They were solved against different X expressions.
This mathematics is for the middle school student in Moroccan before you get to the secondary school so it's so easy even for me unfortunately I was very very bad at math
Занимательное решение! Как так удачно получилось, что из суммы двух уравнений с «y» и «z» вышла часть исходного уравнения, дающая в сумме икс? *** An interesting solution! How did it happen so successfully that a part of the original equation came out of the sum of two equations with "y" and "z", giving the sum of x?
Explain me please. At deriving 2nd set of equations (time 8:01 minutes) how can you replace y with x. You understand x is not equal to y. As we assumed y=(x-3)/(x+1), so they are not equal and cannot be replaced with same notation variable used during earlier assumption. Isn't it ??? e.g for x=2, y will be equal to -1/3 hence x not= y.. same with z.. ??? So how does the f(x) value is correct at the end??
Is this only correct for when z=y? Because that’s the only way you can replace both with the same variable. If the domains of z and y never overlap can this method still produce a solution?
I don’t see how at one step you used: “z = (x+3)/(1-x)” then when you arrived at a simplified version you were able to plug in “x” for “z” implying “x = z”. I think perhaps a 4th variable would make more sense?
It is a nice lecture. But is there a general solution if 3 and 1 are replaced by arbitrary a and b? Or it works just accidentally for this choice of numbers?
Fuck, I spent an hour solving this. Just forgot to write some necessary step and covered the entire A4 page starting over and over. So stupid. I lost my mind in these scribbles but it came back on the next blank page. In an hour. Damn it, I am a furious anger.
Oh summation im late,but still ur awesome and entertaining us. Also here are some variants of me. Ok Now Im Ready Ok Now Your Ready Ok Now Hes Ready Ok Now -Shes- Ready Ok now were ready Ok now Theyre ready Ok Now Its Ready
This is an interesting approach, but I don't think this method works in general, in other words, the Putnam problem is "rigged". I mean I tried my own problem: Find f(x) if f(4x+3)+f(2x-7)=x+3. Going through exactly the same process I arrive (after addition) 2f(x)+f(0.5x-8.5)+f(2x+17)=0.75x+15.75 which leads me nowhere
It works here because the two terms are inverse function of each other, so in general: f(g(x)) + f(g-1(x)) =x -> f(x) = g(x) + g-1(x) - x It would be interesting to learn how to solve equations like your example.
The substitution you do at around 7:00 seems dubious. y and z have already defined in terms of x, and x equals neither y nor z. How can you make this direct substitution?
no. you can replace variable with other as you wish you need just to respect the domaine. in other term, x and z and y and whatever are unkown variables so they can take all possible values
@around 10:17, you multiplied the right side of the equation by (1-x) and (x+1). Don't you have to multiply the left-hand side by the same? 2f(x) times (1-x) and (x+1)? If so, why or why not?
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It was so easy, tho so difficult to think about that idea
I agree
It is so easy for our Chinese students
It may be easy like an egg of Columbus.
Nice problem. Same kind of idea. Let A=(x-3)/(x+1) and B=(x+3)/(1-x). If you repeatably substitute x in A with A you cycle through A ---> B, B ---> x, x ---> A, A---> B,... So, doing the same repeated substitution for this problem you get the 3 equations f(A)+f(B) = X, f(B)+f(X)=A, and f(X)+f(A)=B which quickly reduces to f(x)=(A+B-x)/2=( 8x/(1-x^2) - x)/2
I'm Japanese. this video is very helpful because I can improve my English listening skills as well as learn math.
That's great!
I really liked this question. It was the perfect sort of difficulty for me, as I was able to solve it, but at the same time, it was still difficult enough that I had to think about it for a while in order to work out how to answer it, as I am still very new to functional equations. I referred back to other functional equation videos in order to work this one out. Great video anyway!
Thank you! I'm glad to hear that! 😊
@@SyberMath No worries at all! It was awesome 👏, I really love all these sorts of problems you always present to us, and for me at least, most of them are challenging but still doable, but the odd one is impossible for me. Also, when I do manage to solve one, you always seem to have this neat, clever trick that is like 10x quicker than the way I did it. And I love seeing those neat tricks, it’s so amazing seeing how they work, and now I’m finally beginning to use them myself which is even better. So thank you very much for helping me with that
Anyway, congratulations on recently hitting 20,000 subscribers, and I, along with plenty of others, am really, really enjoying these videos, so thank you very much for continuously making them and continually stretching my mathematical abilities every day. Sorry for the long essay.
Two more quick things:
1. I really admire the fact that you reply to basically everyone’s comments in the comment section, it is very nice to see that you have taken the time to do that, and it is very nice to get a response from you, it’s very nice to see
2. Do you have any book recommendations for questions like the algebraic equation solving/number theory type problems, particularly like the ones that you have done on your channel? I would like to practice them some more
Sorry for the really long comment
Thanks again and congrats!
Same here, I really had to fight (and cheat a little with graphing) but it was worth it!
@@txikitofandango Yeah exactly! It’s so satisfying once you get the answer!
Here's a slight reformulation of the solution. If you set g(x) = (x-3) / (x +1), then the original equation can be rewritten as
f(g(x)) + f(g(g(x)) = x
By substituting x = g(x) and x = g(g(x)), and using the curious fact that g(g(g(x))) = x we also have
f(g(g(x)) + f(x) = g(x)
f(x) + f(g(x)) = g(g(x))
So we have three linear equations with three unknowns: f(x), f(g(x)), and f(g(g(x)). For example by adding the second and the third, and subtracting the first we get:
2f(x) = g(x) + g(g(x)) - x
EDIT: I just realized that several other posters proposed the same idea before. Well, I'm keeping the post in case someone finds this mini-writeup useful.
This is a really elegant approach. Thanks for sharing. When I did the problem, I saw that we had f(g(x)) + f(g⁻¹(x)), and that g(g(x)) = g⁻¹(x), and that g⁻¹(g⁻¹(x)) = g(x), but didn't think to put it all in terms of g(x), g(g(x)), and g(g(g(x))). I think your approach is much tidier.
Adding the 2 equations was a brilliant idea ,I did not think that it wud simplify to x, great work
Glad you liked it
A less clever and more systematic way to think about it would be to include the original equation in the system of equations. Then there are 3 equations and 3 unknowns, and you can use well known methods like gaussian elimination to solve for f(x)
Woah, that's a really interesting one!
Wow ! My heart is filled with joy .Marvellous . Your method of solving is unique .God bless you !
I've only started watching recently, and this is my favorite so far. Before viewing, I was able to solve the problem after much trial and error, with a similar but less efficient method. In other words, it took more algebra. Anyway, I've learned a lot about solving functional equations. Thanks!
Finally i found technique for equations like this…..thanks i have been looking for this for along time…..this is really really useful for someone like me
Glad it was helpful!
Hahaha lol....i'm trying to study English now thanks for your help...
Wow. That's the first functional equation I have ever solved and it's correct. What can I say, I did exactly what you do in the video. I even used the same letters for a substitution. : )
Excellent!
I used s and t, so I must be on a different wavelength. :)
@@SyberMath Why not just replace the second expression with something in terms of y so you dont have to introduce a third variable..more streamlined and elegantno? That's how I did it.
you' re a genius
Bonito ejercicio, me hizo recordar a mi primer año en la universidad. Saludos desde Perú.
Greetings from the United States! 💖
Love the problem and plan to share it with my students. You may note that the two arguments are inverses, so there's another way to find f(x)! I'll post a video if anyone is interested.
Just posted a video illustrating the use of inverses. ruclips.net/video/nr97kn6Uidw/видео.htmlsi=A8sPlJv6CU9rVZ8Y
This video made my day. Every good math video makes my day. Thank you for bringing this nice problem. Good luck.
Happy to help! 😊
I spent a good amount of time figuring out how to plug in the right values to get enough equations to cancel out the noisy parts. It’s essentially the same as what you did.
^^^ Great use of word 'noisy'! 👍
So nice problem! I solved it by my self and i am happy i got the same answer:)
Great!
This channel is really very interesting and informative.
Glad you think so! 💖
One of wonderful Functional Equation. Congratulations on being methodical.
Thanks a lot.
You are most welcome
You really make a good recap and recreational stuff ,really enjoy solving and learning.
Thanks! I'm glad to hear that! 🥰
Nice solution.Thank you @syberMath.
You're welcome!
On doit vérifier que la fonction f satisfait à l'équation de départ puisque dans les étapes de la résolution on a procédé par implication et non par équivalence. Merci
Can we directly replace x by x+3/1-x then x by x-3/x+1.
I verified that the function you found does indeed work, but I like to try to be rigorous in my approach to these problems. I understand the concept of a dummy variable. However I am not convinced that you can always do this in general. x, y and z are all distinct values. In this specific case, one of the inner x expressions was the inverse of the other one. I am not sure this technique would work in a more general case. To understand what I am getting at, make a table of values for x, y and z. When x=0, y = -3 and z = 3. So what you then have is that f(-3) + f(3) = 0. Now pick x=3. When x=3, y=0, and z = -3 So what you get in that case is f(0) + f(-3) = 3. I am not convinced that you can just casually replace y and z with the same dummy variable t and still have the the table of values work out the way I described. I think it works only in very specific cases, like this one, where one inner expression was the inverse of the other inner expression.
replacing is ok coz after those are functional eqn and it doesn't matters if all variables are x or x1 or y
YOU ARE SO AWESOME
you are the best pro i have seen!
Thanks 😅
SO much algebra could be dispensed with by using matrix multiplication and inversion on the linear fractional transformations.
yep, mobius transformations
SyberMath you never fail to impress me and astound me with the math problem videos you come up with. Keep up the awesome content👏👏👏👏!!!!
Thank you!!! 💖
Formally there is an omissis. After the expression of X as a function of Y, you need to set out the condition that Y is different from 1. Same applies to Z different from -1.
Very informative....but at @7:00 you say replace z with x but note that it is NOT the same x as in the original equation.. Then you go on to also replace y with x. That is also NOT the same x as in the original equation. So you CANNOT add the equations as you do @8:42.
If you are allowed to do that then you could also replace (x-3)/(x-1) in the original equation with y. Replace (x+3)/(1-x) in the original equation with z. And then replace y and z with new x…. noting it it’s not the same as the original x.
Am I missing something? Can someone point me to a text book that covers this type of functional equations please. Thanks.
I solved problem in the same way, assuming g(x) = (x-3)/(x+1). Then g^{-1}(x)=(x+3)/(1-x), g(g(x))=g^{-1}(x).
Nice! I like that!
@@SyberMath it's just the same thing just with a better tip of understanding..
i hope you get 1 million subscribers at the end of this year :)
you are awesome :)
take care :)
Thank you so much 😀💖
I think so! I’m waiting for to see there!! Your content is the best!
@@MathZoneKH Thanks! I appreciate it! 💖
@@SyberMath you are probably the most underrated YT Channel.
Does it help you in anything to notice the input of the 2 functions are the inverse of each other? In order to generalize the result? I can't find yet the connection if they are inverse functions like this case
that's the first thing I noticed, but it's more than that:
f(y(x))+f(z(x)) = x (y and z are functions here)
since z(x) = y-1(x) we have:
f(y(x))+f(y-1(x)) = x
but notice how y(y(x)) = y-1(x)
and similarly y-1(y-1(x)) = y(x)
This is the crucial property y has, so by doing;
f(y(y-1(x))+f(y-1(y-1(x))) = y-1(x)
you get: f(x)+f(y(x)) = y-1(x)
and by doing: f(y(y(x))) + f(y-1(y(x))) = y(x)
you get: f(y-1(x)) + f(x) = y(x)
by adding the 2 together you get: 2f(x) + x = y-1(x) + y(x)
So the real challenge in this problem was finding this niche function y props to the author of the book once you figure out this trick you can instantly write 2f(x) + x = y-1(x) + y(x) and solve!
Wow idea is interesting.
Cool problem and solution
This was a very easy question from putnam.
thanks a lot for this beautifull work , i enjoy it .maths make me feel good best wishes from Algeria
Hi! Thank you for the kind words! 💖
Very tricky, brilliant solution 👍
Beautiful!
Sybermath can you suggest me how can I start learning functional equations, techniques and approaches from the very basic?
Good question! There are some books but they do not start at the basic level unfortunately. I'm planning to come up with a document that explains the basics but who knows when I can write it up
I solved f(2x)=f(x)^2 the Hard way given f(x)≠0 and f'(0)=alpha and I can thank 2^n for that, without 2^n I wouldnt have solved it,2^n is my New Best friend also the equations that I Just solved has e^(xalpha) as solution,also It solves f(x+c)=f(x)f(c) since if c=x,then the solution is the same.
@@SyberMath : (
@@resilientcerebrum some tips to get you going: something like f(x+3/1-x) has an expression inside the function argument and takes the general form of f(g(x))
So if you have an equation like f(g(y)) = h(y), you can solve it by defining x = g(y)
F(x) = h(g^-1(x))
In general I always get started with this kind of problems by trying plugging in some special values and see if I can find any patterns or get some insights during the course of the calculation. For this problem I tried 0, 3, -3 and found it's always about f(3) f(-3) and f(0) and so I can solve all 3 of them. Then I tried 5, -5 and found f(1/3), f(-1/3) f(2) and f(-2) also showed up, so I added in 1/3, -1/3, 2 and -2, then I found it's all about f(5) f(-5) f(2) f(-2) f(1/3) and f(-1/3) and all these 6 can be solved. This gave me some insights as my gut feeling was that this pattern probably can be generalized for any given values, i.e. if I start with 1 or 2 values and keep adding the new ones showing up in the f(), then with a few iterations I should be able to get back and form a closed domain. So I did the same again but with the letters this time, and guess what, this ended up with essentially the same methodology presented in the video.
I did it, phew!! Although I'll admit that wasn't sure I was going anywhere with it initially.
Nice!
This was ok until the 7.00 minute mark, where it turned into a mathematical "Shell Game". When Y was replaced by X, that X was different to the X we started with. Same thing happened later when Z was replaced by another X. So we have three different values of X which are represented by the same X in the following equations. How does that make sense?
Exactly! I don't get why those replacements are allowed.
They are not the same x. They don't have to be. You can choose x to be whatever you want for that particular occasion
@@SyberMath Why did you use X and not another letter such as W?
Keep in mind that the x here isn't the unknown to be solved for, it's just the placeholder used to describe a rule. The unknown the problem is asking us to solve for in this case is a particular rule, a function, namely, what is the rule for f? What rule describes it? Commonly we write f(x) when describing the rule but we could use f(z) or f(w) or f(whatever). It's how we manipulate thing in the parenthesis that counts.
great solution sir thanks
Most welcome
Holy crap I found the answer by trial and error. I reasoned through and evaluated f at 0 and plus or minus 1/3, 2, 3, and 5. Can prove that f is odd. Plotted the points. Noted that there's probably vertical asymptotes at plus and minus 1, so put an (x-1)^2 in the denominator. Fiddled around with cubic equations in the numerator. Presto. Okay, now to figure out how to actually do it.
🤩
It's f(g(x))+f(g-1(x))=x with g(x)=(x-3)/(1+x) maybe more simple like that ?
I thought about it
Beautiful & clever! Love it. 😍
Thank you! 😊
Calculus is so beautiful.
The way the rigorous problem is solved is quite effective
Thank you!
This channal is growing fast .. I hope the best for you ❤
Thank you! 💖
At 7:50 by using X to replace both Y and Z isn't he supposedly assuming that y equals Z? What justification do we have to do that? They were solved against different X expressions.
You can make any replacement you want. No justification needed
Nice one
We can use that way if and only if y and z are the inverses of eachother.
This mathematics is for the middle school student in Moroccan before you get to the secondary school so it's so easy even for me unfortunately I was very very bad at math
This is really nice
Thank you! 💖
Awesome !!! Good job
Thank you so much 😀
@@SyberMath 😍😍
I like functional equations👍
😊
That’s great! It’s what I want functional equation!❤️❤️❤️pretty cool 😎 solution
Glad you like it! 💖
Wonderful I love this question
Clean solution
I love it
please do videos with matrices!
Як завжди вподобайка.
Super
Занимательное решение! Как так удачно получилось, что из суммы двух уравнений с «y» и «z» вышла часть исходного уравнения, дающая в сумме икс?
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An interesting solution! How did it happen so successfully that a part of the original equation came out of the sum of two equations with "y" and "z", giving the sum of x?
my compi can't do it. no surprise!
Yay! Finally!!! 😜😁
@@SyberMath lol!! haha
Can someone tell me why we can replace z with x after getting the second expression
good
Thank for sybermath
You’re welcome, Taha!
great。give more limit，function questions please。thanks。
I have a very nice proof for this problem, but unfortunately, this comment space is not enough for it. Great job.
Thanks! 😁😁😁
Amazing work sir! 😇🙏
Thank you!
@@SyberMath most welcome 😇🙏
The only part I do not understand is how you can return from the variables y and z to x just like that...
You can use any variable you want
@@SyberMath and we find the solution in that variable?
Excellent!
Many thanks!
Explain me please. At deriving 2nd set of equations (time 8:01 minutes) how can you replace y with x. You understand x is not equal to y. As we assumed y=(x-3)/(x+1), so they are not equal and cannot be replaced with same notation variable used during earlier assumption. Isn't it ???
e.g for x=2, y will be equal to -1/3 hence x not= y.. same with z.. ???
So how does the f(x) value is correct at the end??
Is this only correct for when z=y? Because that’s the only way you can replace both with the same variable. If the domains of z and y never overlap can this method still produce a solution?
Thank you
Please keep it up.
Thank you, I will
Great Solutions
Thanks
I don’t see how at one step you used:
“z = (x+3)/(1-x)” then when you arrived at a simplified version you were able to plug in “x” for “z” implying “x = z”. I think perhaps a 4th variable would make more sense?
No you can use any variable you want
Nice problem, thanks for sharing!
No problem 👍 Thanks for watching! 😊
Thank you very good
thanks
Beautifull solution!
Thank you! 🤩
Found you today 😄... Looks very interesting !!
Welcome aboard! 😁
Namaste.
Namaste
Excelente. Gracias.
Thank you! 💕
@@SyberMath Todos los que miran tus videos, aprenden un montón. Yo estoy en ese grupo.
I enjoyed
Glad to hear that
It is a nice lecture. But is there a general solution if 3 and 1 are replaced by arbitrary a and b? Or it works just accidentally for this choice of numbers?
Good method!
Glad you think so!
Good math tricks.
It becomes easier if you notice and use g(g(g(x)))=x.
Fuck, I spent an hour solving this. Just forgot to write some necessary step and covered the entire A4 page starting over and over. So stupid. I lost my mind in these scribbles but it came back on the next blank page. In an hour. Damn it, I am a furious anger.
Beautiful..
Thank you! 😊
Oh summation im late,but still ur awesome and entertaining us.
Also here are some variants of me.
Ok Now Im Ready
Ok Now Your Ready
Ok Now Hes Ready
Ok Now -Shes- Ready
Ok now were ready
Ok now Theyre ready
Ok Now Its Ready
😂
This is an interesting approach, but I don't think this method works in general, in other words, the Putnam problem is "rigged". I mean I tried my own problem: Find f(x) if f(4x+3)+f(2x-7)=x+3. Going through exactly the same process I arrive (after addition) 2f(x)+f(0.5x-8.5)+f(2x+17)=0.75x+15.75 which leads me nowhere
It works here because the two terms are inverse function of each other, so in general:
f(g(x)) + f(g-1(x)) =x ->
f(x) = g(x) + g-1(x) - x
It would be interesting to learn how to solve equations like your example.
beautiful
Thank you!
The substitution you do at around 7:00 seems dubious. y and z have already defined in terms of x, and x equals neither y nor z. How can you make this direct substitution?
no. you can replace variable with other as you wish you need just to respect the domaine. in other term, x and z and y and whatever are unkown variables so they can take all possible values
Good problem.
Good video!
Glad you enjoyed it
Really pretty!
Thank you! 😊
Havent done any calc equations for about half a decade now. Why does it seem they were all this length though.
You're very good! Thanks for sharing :)
Thank you too!
Watching from india . ❤️❤️
Hello! ☺️
@around 10:17, you multiplied the right side of the equation by (1-x) and (x+1). Don't you have to multiply the left-hand side by the same? 2f(x) times (1-x) and (x+1)? If so, why or why not?