I think it'll be good to highlight that the domain was restricted to integers because the linear functions are the only solutions to the Cauchy equation over the rationals. If the domain is over the reals, there can be highly pathological functions which are not linear that satisfy the Cauchy equation.
Using the method from 10:15 already at 8:50 , we see much earlier that the coefficients c(c-2) and (c-2)f(0) must be =0, so ( c=0 and f(0)=0 ) or ( c=2 and f(0) arbitrary )
Good afternoom! Very intersting solution. Most of time , i am not able to solve theese kind o f problems. I liked too much. I hope you post more problems of function like that. May be seeing more problems I get the point. Congratulations
f(2a)-2f(a)=f(2b)-2f(b) implies f(2x)-2f(x) is invariant. This is only possible when f is of degree one. We can assume f(x)=mx+n and solve by comparing coefficients of x in the original equation.
I have a theory of problem in the process. ;) But, I have resolved already. Problems in combinations are too tough but somehow when you got the solutions with the right process, all the given problems of any functions that are exactly evel (2n, as I would implied) will lead to 0. ;)
He showed that. By 9:05 he has derived that f must be of the form f(x)=cx+f(0), where c and f(0) are unknown but must satisfy the equation (c-2)(c(a+b)+f(0))=0 for all integers a and b. It follows from this that there are no other solutions.
He have already shown. As he found that g(x + y)= g(x) + g(y), then he showed that f(x) = g(x) + c. You have to wath on the video again and pausing in order to get the steps that he prooved.
I have been following your channel and other math channels for a long time. I want to participate in the IMO from Bangladesh and ensure my best performance. How can I fulfill my dream?
Good explanation sir😍😍 sir you can help me solve this following exercises: Find the integer tribles (x, y, z) / x + y-z = 12 and x ^ 2 + y ^ 2-z ^ 2 = 12. It is very difficult for me😔😔.
Bro i have a very short solution for the problem , you just put b = 0 and take inverse both side of the eqn formed. You will get f(x) = 2x which satisfy the realation.👍 Try this
One line solution.....put a=0, u will get f(0)+2f(b)=f(f(b)). Now replace f(b) by x as the mapping is Z to Z. So u will get f(x)=2x+f(0). For the 2nd solution, put f(x)=c. And proceed to get c=0. Hence two solutions
@@Charliethephysicist it is not wrong because as per definition of function, all elements of domain should have UNIQUE value in range, this has nothing to do with image...because that image is subset of Z
I think it'll be good to highlight that the domain was restricted to integers because the linear functions are the only solutions to the Cauchy equation over the rationals. If the domain is over the reals, there can be highly pathological functions which are not linear that satisfy the Cauchy equation.
Using the method from 10:15 already at 8:50 , we see much earlier that the coefficients c(c-2) and (c-2)f(0) must be =0, so ( c=0 and f(0)=0 ) or ( c=2 and f(0) arbitrary )
Great solution, upload combinatorics problems pls :)
Good afternoom! Very intersting solution. Most of time , i am not able to solve theese kind o f problems. I liked too much. I hope you post more problems of function like that. May be seeing more problems I get the point. Congratulations
f(2a)-2f(a)=f(2b)-2f(b) implies f(2x)-2f(x) is invariant. This is only possible when f is of degree one. We can assume f(x)=mx+n and solve by comparing coefficients of x in the original equation.
What about e.g. f(2^k*l)=(l+2024)*2^k, where l is odd and k is thus the highest power of two that divides the argument?
This one is quite easy for an IMO. I solved it quickly so I thought I made a mistake 😅
Problem 1 of IMO is always the easiest problem
@@advaykumar9726 Yes but this is quite easy
I was not abble to solve it.
@@pedrojose392 Me neither :(
Since it's only integers, calculate f(f(n+1)) and get f(n+1)-f(n)...
Who knows what you could possibly get... something recognisable, by any chance ?
Idk if this works, but for any c: f(f(x)) = f(f(c + (x - c))) = f(2c) + 2f(x - c). Hence if c = 0, f(f(x)) = f(0) + 2f(x), i.e. f(x) = f(0) + 2x
I have a theory of problem in the process. ;) But, I have resolved already. Problems in combinations are too tough but somehow when you got the solutions with the right process, all the given problems of any functions that are exactly evel (2n, as I would implied) will lead to 0. ;)
Which competition problems are the hardest???
is it IMO or Putnam?????
i think putnam..
how can we show that there are not any more functions satisfying the expression
He showed that. By 9:05 he has derived that f must be of the form f(x)=cx+f(0), where c and f(0) are unknown but must satisfy the equation (c-2)(c(a+b)+f(0))=0 for all integers a and b. It follows from this that there are no other solutions.
He have already shown. As he found that g(x + y)= g(x) + g(y), then he showed that f(x) = g(x) + c. You have to wath on the video again and pausing in order to get the steps that he prooved.
I have been following your channel and other math channels for a long time. I want to participate in the IMO from Bangladesh and ensure my best performance. How can I fulfill my dream?
Thanks for the great video!
Good explanation sir😍😍 sir you can help me solve this following exercises: Find the integer tribles (x, y, z) / x + y-z = 12 and x ^ 2 + y ^ 2-z ^ 2 = 12. It is very difficult for me😔😔.
This was probably the easiest recent IMO problem ever!
Easy if you know the Cauchy equation’s solutions lol
How long...i just wrote g(u) as derivative of f(x) and showed it is constant only.
Physics principals or theories where published
Hello, you can just replace a=0, than we will have, f(f(b))=2f(b)+f(0), for all b.
So f(x)=2x+f(0). Or f(x)=0
How can we show f(x) can also be zero
f(b) may not be any whole number, so the equation f(x)=2x+f(0) may not be right for all integers
Bro i have a very short solution for the problem , you just put b = 0 and take inverse both side of the eqn formed. You will get f(x) = 2x which satisfy the realation.👍 Try this
To take inverse of the function you must prove that f is bijective first.
One line solution.....put a=0, u will get f(0)+2f(b)=f(f(b)). Now replace f(b) by x as the mapping is Z to Z. So u will get f(x)=2x+f(0).
For the 2nd solution, put f(x)=c. And proceed to get c=0. Hence two solutions
Perfect
Wrong. It is only true for when x is in the image of f. You can that for x outside of the image.
@@Charliethephysicist it is not wrong because as per definition of function, all elements of domain should have UNIQUE value in range, this has nothing to do with image...because that image is subset of Z