I got the three equations 3f(0) = f(f(0)) f(2a) + 2f(-a) = f(f(0)) f(-2b) + 2(b) = f(f(0)), realised they were all the same as the middle one, and gave up
Nice video. However, I have a complaint! I think you missed some possible solutions with your solution. I solved the problem with derivation of both sides with respect to x. The chain rule gives us: f’*d(2a)/dx+2f’*db/dx = f’*f’*d(a+b)/dx Next i divided both sides with f’ and multiplied with dx: d(2a)+2db = f’*d(a+b) Now I complete the individual integrations: ∫d(2a) + ∫2db = ∫(f’)d(a+b) -> 2(a+b) + C = ∫(f’)d(a+b) Now we can either use intuition or derivate both sides with respect to a+b to get f’(x) = 2 -> f(x) = 2x+C My answer says that C can be any constant, including rational and irrational numbers, and as far as I checked, this is true! Your solution only includes C= n where n is an integer. What do you think of this?
I’m so sorry, just realized now that you said in the beginning that f is defined to the set of integers. My bad My question then is if it is okey to let f(x) temporarly be extended to all values so that it becomes continuous and thereby possible to derivate it? And then later go back to the original domain of f(x) - only integers
Similar to @moorsyjam I got the following but here's how I approached it: One can see that f(x) = 0 for all x is the trivial solution. Now let's consider any nontrivial solutions: Let a = b = 0 --> f(0) + 2f(0) = f(f(0)) Let f(0) = k --> 3k = f(k) This is not sufficient for f(x) to be linear, but it gave me that suspicion. Then, let a = -b --> f(-2b) + 2f(b) = f(f(0)) = 3k Let b = -2b --> f(4b) + 2f(-2b) = 3k --> f(-2b) + 2f(b) = f(4b) + 2f(-2b) --> f(-2b) = 2f(b) - f(4b) Another suspicion of linear f(x). And then I said perhaps it is the case then that f(x) is linear, following the same approach as @MathAPlus to finish the problem off.
Great video, thanks for sharing ❤😊
I got the three equations
3f(0) = f(f(0))
f(2a) + 2f(-a) = f(f(0))
f(-2b) + 2(b) = f(f(0)),
realised they were all the same as the middle one, and gave up
👌🏻👌🏻
Nice video. However, I have a complaint! I think you missed some possible solutions with your solution.
I solved the problem with derivation of both sides with respect to x. The chain rule gives us:
f’*d(2a)/dx+2f’*db/dx = f’*f’*d(a+b)/dx
Next i divided both sides with f’ and multiplied with dx:
d(2a)+2db = f’*d(a+b)
Now I complete the individual integrations:
∫d(2a) + ∫2db = ∫(f’)d(a+b) ->
2(a+b) + C = ∫(f’)d(a+b)
Now we can either use intuition or derivate both sides with respect to a+b to get f’(x) = 2 -> f(x) = 2x+C
My answer says that C can be any constant, including rational and irrational numbers, and as far as I checked, this is true! Your solution only includes C= n where n is an integer.
What do you think of this?
I’m so sorry, just realized now that you said in the beginning that f is defined to the set of integers. My bad
My question then is if it is okey to let f(x) temporarly be extended to all values so that it becomes continuous and thereby possible to derivate it? And then later go back to the original domain of f(x) - only integers
😆
@@MathAPlus haha naïveté at its finest! Do you think that solution would be possible?
Similar to @moorsyjam I got the following but here's how I approached it:
One can see that f(x) = 0 for all x is the trivial solution. Now let's consider any nontrivial solutions:
Let a = b = 0
--> f(0) + 2f(0) = f(f(0))
Let f(0) = k
--> 3k = f(k)
This is not sufficient for f(x) to be linear, but it gave me that suspicion.
Then, let a = -b
--> f(-2b) + 2f(b) = f(f(0))
= 3k
Let b = -2b
--> f(4b) + 2f(-2b) = 3k
--> f(-2b) + 2f(b) = f(4b) + 2f(-2b)
--> f(-2b) = 2f(b) - f(4b)
Another suspicion of linear f(x).
And then I said perhaps it is the case then that f(x) is linear, following the same approach as @MathAPlus to finish the problem off.