You are making it too difficult! Let cbr(x)=cube root of x , a=f(-1) , b=f(cbr(2)) and c=f(cbr(1/2)). We replace a, b and c and we obtain: a+b = 2 , b + c =1/2 and a+c =-1 Thus 2(a+b+c)=3/2 => a+b+c=3/4 and a=3/4 - 1/2 = 1/4
The approach in the video gives a more general and mechanical approach. You often see these kinds of functional equations, where you have to use some action of a group (in this case, the group is Z/3Z, with the function 1/(1-u) being a generator acting on u), so you apply all the actions to get as many equations as there are group elements (3 elements in Z/3Z gives 3 equations), and then you can use linear algebra to solve for the function.
Simpler approach: Substitute x = -1; x = 2^(1/3); x = 1/(2^(1/3)) respectively and obtain 2 Equations say 1, 2 and 3 Now Eqn No. 1+2-3 will implies; 2f(-1) = 1/2 and f(-1) = 1/4😀
You wrote the best solution. Concise and simple. I saw the task today . I decided the same way.) (I usually pause the video and don't read the comments while I decide for myself.)
Try this: first replace x=-1; then replace x=3-root of 2; the replace x=1/3-root of 2. Solve the linear problem for f(-1)...that is 1/4.
That’s what I did as well
That is exactly what he did in the video, just with concrete values
You are making it too difficult!
Let cbr(x)=cube root of x , a=f(-1) , b=f(cbr(2)) and c=f(cbr(1/2)).
We replace a, b and c and we obtain:
a+b = 2 , b + c =1/2 and a+c =-1
Thus 2(a+b+c)=3/2 => a+b+c=3/4 and a=3/4 - 1/2 = 1/4
The approach in the video gives a more general and mechanical approach.
You often see these kinds of functional equations, where you have to use some action of a group (in this case, the group is Z/3Z, with the function 1/(1-u) being a generator acting on u), so you apply all the actions to get as many equations as there are group elements (3 elements in Z/3Z gives 3 equations), and then you can use linear algebra to solve for the function.
@@user-jc2lz6jb2e how is the group Z/3Z in this case
If you let h(u) = 1/(1-u), then h^3(u) = h(h(h(u))) = u. The set {id, h, h^2} is a group like Z/3Z with the group action being function composition.
Can you please do JBMO 2012 Prob 4??
Wow thats so cool.
thanks
Simpler approach:
Substitute x = -1; x = 2^(1/3); x = 1/(2^(1/3)) respectively and obtain 2 Equations say 1, 2 and 3
Now Eqn No. 1+2-3 will implies; 2f(-1) = 1/2 and f(-1) = 1/4😀
Sorry obtain 3 Equations
You wrote the best solution. Concise and simple.
I saw the task today . I decided the same way.) (I usually pause the video and don't read the comments while I decide for myself.)
As 3 right side I don't get. Plus the sum of 3 eq. Something uncleared.
I thought first to solve the functional equation
You can do thatz
But still nice process. Just with the right side of as 3 and their sums.
asnwer=-1 isit 🤣🤣🤣🤣