I took the following way: f(x) = x - f(1/(1-x)) = x - 1/(1-x) + f(1/(1-1/(1-x))) = x + 1/(x-1) + f((x-1) /x) = x + 1/(x-1) + (x-1)/x - f(1/(1-(x-1)/x)) = x + 1/(x-1) + (x-1)/x - f(x). So, f(x) = x + 1/(x-1) + (x-1)/x - f(x). The conclusion easily follows. Thank you and compliments for your work, Sir. 👍
I remember a lot of the progression of learning when too use substitutions and then combining equations by adding or subtracting them. This problem for me, really reinforced the process for evaluating....thank you for posting tutorial.
Hello, thank you for such a great video. May I ask what motivates you to replace x with 1/(1-x)? Also, is there a unique property of 1/(1-x) such that x replace by 1/(1-x) followed by x replaced by (x-1)/x results in x again in f(x)? Thanks :)
It's called the circle method of solving functional equations. It creates a system of equations that helps us solve for the function f. Best to Google it as there are many answers out there better than mine!
@@VSN1001 No problem. Happy to help. This is a really good video that discusses it further. m.ruclips.net/video/OSUQIF5RkqQ/видео.html&feature=emb_logo
Good afternoon! I drew up a drawing for the development of a mechanical gearbox, betraying comfort in the manipulation of the lever. it needs to be sold. Can you help?
You proved that if f(x) satisfies the functional equation, it has the closed form of 12:40 . But you need prove the opposite as well: that the closed form of 12:40 satisfies the functional equation.
Functions are just shorthand for the expression assigned to them. Wherever you see f(something), replace it with the expression assigned to it but substitute (something) wherever it appears in the equation. So if f(x)=x+1, f(7+y) becomes (7+y)+1
@@ovalteen4404 Very good example to understand. So I checked the solution above in this kind and indeed f(1/(1-x)) gets the value x-f(x), if I replace every x->1/(1-x) in the solution for f(x) found here.
Well, there is actually a good reason for choosing these specific substitutions. Going from equation 1 to equation 2, his idea is that he wants equation 2 to contain at least one of the terms of equation 1, and the way to do that is to make the f(x) term turn into the f(1/(1-x)) term, i.e. substitute x --> 1/(1-x). Going from equation 2 to equation 3 is the exact same substitution (this might not have obvious because he frames it as a substitution in the first equation, but you could just as well acquire equation 3 by substituting x --> 1/(1-x) into equation 2). After doing this procedure twice, we find that the second term in equation 3 is one that we already had earlier (namely, f(x)), so we have a system of three equations with three unknowns (f(x), f(1/(1-x)), f((x-1/x)), so we can solve for f(x). This method can be generalized. Suppose we have some equation f(x) + f(g(x)) = h(x), where g and h are known, and we want to solve for f. Furthermore, suppose that there is a positive integer n such that such that g^n(x) = x (g^n means repeated application of g, in our case: g(x) = 1/(1-x) and n = 3).Then we can also find similar equations f(g(x)) + f(g(g(x)) = h(g(x)), f(g(g(x))) + f(g(g(g(x)))) = h(g(g(x)), and so on, until we find f(g^(n-1)(x)) + f(g^n(x)) = h(g^(n-1)(x)). But we know that g^n(x) = x, so this last equation just comes down to f(g^(n-1)(x)) + f(x) = h(g^(n-1)(x)). Again, we have n equations in n unknowns, and we can solve for f(x). I hope this could somewhat clarify the reasoning behind the solution to this problem.
Are you serious???? Wtf???? Step 3 is totally wrong!! This is definitely not the way you add and subtract equations! I mean, nice for me to figure it out, but it's horrible how many actually believe it...
Thank you
You are very welcome.
Thank you for your feedback! Cheers!
You are awesome 😀
@@PreMath A
I took the following way: f(x) = x - f(1/(1-x)) = x - 1/(1-x) + f(1/(1-1/(1-x))) = x + 1/(x-1) + f((x-1) /x) = x + 1/(x-1) + (x-1)/x - f(1/(1-(x-1)/x)) = x + 1/(x-1) + (x-1)/x - f(x).
So, f(x) = x + 1/(x-1) + (x-1)/x - f(x).
The conclusion easily follows.
Thank you and compliments for your work, Sir. 👍
You always present such a patient, excellent treatment of the puzzles!
I remember a lot of the progression of learning when too use substitutions and then combining equations by adding or subtracting them. This problem for me, really reinforced the process for evaluating....thank you for posting tutorial.
Hello, thank you for such a great video. May I ask what motivates you to replace x with 1/(1-x)? Also, is there a unique property of 1/(1-x) such that x replace by 1/(1-x) followed by x replaced by (x-1)/x results in x again in f(x)? Thanks :)
It's called the circle method of solving functional equations. It creates a system of equations that helps us solve for the function f. Best to Google it as there are many answers out there better than mine!
Thank you so much!
@@VSN1001 No problem. Happy to help. This is a really good video that discusses it further. m.ruclips.net/video/OSUQIF5RkqQ/видео.html&feature=emb_logo
@@markwilliams1555 Thank you
Been a while since solving functional equations!
Keep 'em shooting sir!
Kudos and Thanks!
Would be great if u can upload more calculus questions often👍
Super! Love functional equations.
But how did you get the insight of where to start?
very well done, great job solving this function equation, thanks for sharing
Thank you sir! Because of you, I'm starting to find math interesting and sure am improving!
I have trouble understanding your change of variable since you kept the same variable substitution.
Thank you for sharing.
Can use polynomial division to show x^3+14=(x^2+7x-2)(x-7) +51x. Since x^2+7x-2=0, x^3+14=51x. The answer 51 follows by division by x.
Really helpful i want such more questions of function i want to practice them
Excellent Pranav.
Thank you for your feedback! Cheers!
You are awesome 😀
Great video! I really enjoy the way u explain it! Keep it up!
Good afternoon! I drew up a drawing for the development of a mechanical gearbox, betraying comfort in the manipulation of the lever. it needs to be sold. Can you help?
Please can all functional equations be solved using this approach?
Its very defficult sir
You are great sir
I love you
As my master of maths
Nice and helpful video👏👏👏👏👌👌👌👌
Thanks for video. Good luck!!!!
In which grade it has come?
Excellent. It doesn't occur to me to bring in an Extraneous
value to be assigned to ' x'.
👍👍
Very nicely done. Good job. Keep it up.
Very interesting method.Thank you
Thanks you teacher for video 🙏✏️
Wohhaaa!! Solved by me... Within 4 minutes 🙌❤
Date :- 8th May 2022 1:23 AM
nice solution
Excellent explaination any average student can understand
#binomial #polynomial #linearequation
cảm ơn, giải phương trình hàm bằng biến đổi hàm số.
You proved that if f(x) satisfies the functional equation, it has the closed form of 12:40 . But you need prove the opposite as well: that the closed form of 12:40 satisfies the functional equation.
Wow sir thanku 🙏👍 but i puzzle into these questions
You are very welcome.
Thank you for your feedback! Cheers!
You are awesome 😀
I am awesome 😎
Very nice question
Nice sir
Nice
Would be nice to add that x=/0 and x=/1.
Superb!
Hey 👋👋 thanks sir
Very well done.
Good
very nice
Great
Extremely beautiful ❤️
Nice 🙏🙏🙏🌹
Thanks a mil
#LCD #lcddenominator #denominator #LeastCommonDenominator
#crossmultiply #crossmultiplication #crossmultiplying
It's all way beyond me - I just don't understand f or function; I still enjoy it, though!
Functions are just shorthand for the expression assigned to them. Wherever you see f(something), replace it with the expression assigned to it but substitute (something) wherever it appears in the equation. So if f(x)=x+1, f(7+y) becomes (7+y)+1
@@ovalteen4404 Many thanks for that - I think my addled 72 year old brain just about understands it! You have been most helpful
@@ovalteen4404 Very good example to understand. So I checked the solution above in this kind and indeed f(1/(1-x)) gets the value x-f(x), if I replace every x->1/(1-x) in the solution for f(x) found here.
toán học rất hay.
How on earth somebody knows to take all those steps to solve the problem?
Проще сначала продиференцировать уравнение, а потом проинтегрировать.
I understand the solution but I dunno how could anyone solve it. The need to substitute x as so many other things is just bizarre.
Well, there is actually a good reason for choosing these specific substitutions. Going from equation 1 to equation 2, his idea is that he wants equation 2 to contain at least one of the terms of equation 1, and the way to do that is to make the f(x) term turn into the f(1/(1-x)) term, i.e. substitute x --> 1/(1-x). Going from equation 2 to equation 3 is the exact same substitution (this might not have obvious because he frames it as a substitution in the first equation, but you could just as well acquire equation 3 by substituting x --> 1/(1-x) into equation 2). After doing this procedure twice, we find that the second term in equation 3 is one that we already had earlier (namely, f(x)), so we have a system of three equations with three unknowns (f(x), f(1/(1-x)), f((x-1/x)), so we can solve for f(x).
This method can be generalized. Suppose we have some equation f(x) + f(g(x)) = h(x), where g and h are known, and we want to solve for f. Furthermore, suppose that there is a positive integer n such that such that g^n(x) = x (g^n means repeated application of g, in our case: g(x) = 1/(1-x) and n = 3).Then we can also find similar equations f(g(x)) + f(g(g(x)) = h(g(x)), f(g(g(x))) + f(g(g(g(x)))) = h(g(g(x)), and so on, until we find f(g^(n-1)(x)) + f(g^n(x)) = h(g^(n-1)(x)). But we know that g^n(x) = x, so this last equation just comes down to f(g^(n-1)(x)) + f(x) = h(g^(n-1)(x)). Again, we have n equations in n unknowns, and we can solve for f(x).
I hope this could somewhat clarify the reasoning behind the solution to this problem.
i said 1
🥰🤩🥰💫💫💫🙏🙏🙏
What is Premath Gmail???
I want to send you my homework math help
First
Thank you for your feedback! Cheers!
You are awesome 😀
But (x^3 - x + 1) = (x - 1)(x^2 + x - 1) so the answer is not fully reduced
Are you serious???? Wtf???? Step 3 is totally wrong!! This is definitely not the way you add and subtract equations! I mean, nice for me to figure it out, but it's horrible how many actually believe it...
As long as LHS and RHS symmetry is maintained, you can do any number of combinations