Solved in a similar way to the second method. Set y = 1/x so it forces the first part to be f(1) which basically leads to the same equation in second method, but I like setting x = 1 better. It seems a bit more elegant
I looked at the beginning of the article and saw that there was a special thing 8 = 4 *2 so instead of x=2, y=4, f(8)=f(2)/4 => f(2)=2f(8) , replace x=4 , y= 2, then f(4)=2f(8) and then substitute in equation 2, we have 6f(8)=6 => f(8)=1
I used kind of a mix of 1st and 2nd method. I set x=1 and then y=2 and y=4 respectively. That gave me two equations with expressions for f(2) and f(4) depending on f(1). Adding these 2 equations gave me an equation with f(2)+f(4)=f(1)/4+f(1)/2. Since f(2)+f(4)=6 I could solve for f(1)=8. From here I could easily calculate f(8) with x=1 and y=8 using the original functional equation with f(8)=f(1)/8=1
Another method f(x)=yf(xy) x=2 and y= 4 gives f(2)=4f(8) Switching x and y values gives f(4)=2f(8) Add the two equations f(2)+f(4)=6f(8)=6 f(8)=1 But I always like to find f(x).
considering the original expression, if we multiply everything with xy then we have x*f(x)=y*f(y)=xy*f(xy)=C with the fact that f(8)=1 then x*f(x)=8f(8)=8 which mean f(x)=8/x
I just kind of eyeballed that from: f(xy) = f(x)/y If f := 1/x, we have 1/xy = (1/x)/y, which works. Then based on the initial conditions I knew it had to be some c/x then went from there to solve for c.
Continuing with your thought at the end of method 1: Let xy not be 0. Now f(xy)=f(yx) so f(x)/y = f(y)/x and xf(x) = yf(y). But two expressions depending on different variables must be a constant. Thus f(x) = C/x. Using the second equation C=8, as shown by you. Hence f(8)=1.
@@angelmendez-rivera351 Indeed, by setting x=0. You can also prove it is undefined by letting y=0. In the final form f(x)=C/x, 0 is not in the domain of f(x).
@@Utesfan100 You cannot set y = 0. You can set x = 0. Also, your statement is incorrect. The domain is the set of all real numbers, R, and f(0) = 0, f(x) = 8/x otherwise. This is the correct solution.
I tried some values for x and y that ended up being unnecessary in the end. Setting x = y = 2 sets up the solution very nicely. The moral of the story: always remember 2 x 2 = 4!
I think we only need to fill 0 or 1 into one of the parameters in the functional equation to solve this kind of functional equation problem in most of the case by my experience.
No, this is false. The functional equation as stated can hold for all x in R, and all y in R\{0}. It implies that f(0·y) = f(0)/y = f(0) for all y in R\{0}, and this means that f(0) = 0, but it says nothing about what f is elsewhere.
Consider all f : R -> R such that f(x·y) = f(x)/y for all x in R and all y in R\{0}. This implies that f(0·y) = f(0) = f(0)/y for all y in R\{0}, which means that f(0) = 0. It also implies that f(1·y) = f(y) = f(1)/y for all y in R\{0}. Find all f : R -> R such that the above is true, AND such that f(2) + f(4) = 6. Notice that f(2) = f(1)/2 and f(4) = f(1)/4, so f(2) + f(4) = f(1)/2 + f(1)/4 = f(1)·(1/2 + 1/4) = 3/4·f(1) = 6, hence f(1) = 8. Therefore, f(x) = 8/x for all x in R\{0}.
f(4) = f(2)/2. Done.
With this you found f(2) or f(4). Then f(8) = f(2)/4 = f(4)/2.
So we have 3f(4) = 6. Then f(8) = 1.
Nice one.
Solved in a similar way to the second method. Set y = 1/x so it forces the first part to be f(1) which basically leads to the same equation in second method, but I like setting x = 1 better. It seems a bit more elegant
I looked at the beginning of the article and saw that there was a special thing 8 = 4 *2 so instead of x=2, y=4, f(8)=f(2)/4 => f(2)=2f(8) , replace x=4 , y= 2, then f(4)=2f(8) and then substitute in equation 2, we have 6f(8)=6 => f(8)=1
f(2) = 4f(8), not 2f(8)
@@ಭಾರತೀಯ_ನಾಗರಿಕ oh , thank you i wrote it wrong
Even simpler : f(4) = f(2)/2. Then use fingers.
If I want to solve a functional equation, I always come to your channel :)
From first equation follows that
f(8) = f(2*4) = f(2)/4 and
f(8) = f(4*2) = f(4)/2
So
f(2)/4 = f(4)/2
Multiply by 4:
f(2) = 2 * f(4)
Plug that into second equation:
2 * f(4) + f(4) = 6
3 * f(4) = 6
Divide by 3:
f(4) = 2
And since f(2) is twice f(4),
f(2) = 2 * 2 = 4
Plug this into above:
f(8) = f(2)/4 = 4/4 = 1
f(8) = f(4)/2 = 2/2 = 1.
Correct.
I did it applyying method #1. Thanks for the analysis of method #2.
I used kind of a mix of 1st and 2nd method. I set x=1 and then y=2 and y=4 respectively. That gave me two equations with expressions for f(2) and f(4) depending on f(1). Adding these 2 equations gave me an equation with f(2)+f(4)=f(1)/4+f(1)/2. Since f(2)+f(4)=6 I could solve for f(1)=8. From here I could easily calculate f(8) with x=1 and y=8 using the original functional equation with f(8)=f(1)/8=1
Another method
f(x)=yf(xy)
x=2 and y= 4 gives f(2)=4f(8)
Switching x and y values gives f(4)=2f(8)
Add the two equations f(2)+f(4)=6f(8)=6
f(8)=1
But I always like to find f(x).
considering the original expression, if we multiply everything with xy then we have x*f(x)=y*f(y)=xy*f(xy)=C
with the fact that f(8)=1 then x*f(x)=8f(8)=8 which mean f(x)=8/x
f(y)=8/y
Replace y with xy then f(xy)=8/xy
Replace xy with yx then f(yx)=8/yx=8/xy=f(xy)
I just kind of eyeballed that from:
f(xy) = f(x)/y
If f := 1/x, we have
1/xy = (1/x)/y, which works.
Then based on the initial conditions I knew it had to be some c/x then went from there to solve for c.
Continuing with your thought at the end of method 1:
Let xy not be 0. Now f(xy)=f(yx) so f(x)/y = f(y)/x and xf(x) = yf(y).
But two expressions depending on different variables must be a constant.
Thus f(x) = C/x.
Using the second equation C=8, as shown by you.
Hence f(8)=1.
Wow! That’s neat
Letting x·y not equal 0 is unnecessary, since one can prove anyway that f(0) = 0.
@@angelmendez-rivera351 Indeed, by setting x=0. You can also prove it is undefined by letting y=0.
In the final form f(x)=C/x, 0 is not in the domain of f(x).
@@Utesfan100 You cannot set y = 0. You can set x = 0. Also, your statement is incorrect. The domain is the set of all real numbers, R, and f(0) = 0, f(x) = 8/x otherwise. This is the correct solution.
Hey dude! What do you mean about f(xy)=xyf(x)f(y)
i thought it was f(2)+f(4)=f(6), but luckily its still solvable. f(x)=0 in this situation
I tried some values for x and y that ended up being unnecessary in the end. Setting x = y = 2 sets up the solution very nicely.
The moral of the story: always remember 2 x 2 = 4!
Actually, 2x2 = 4! is not true 🤓
@@DeJay7 It’s true in Mod 4, lol.
I think we only need to fill 0 or 1 into one of the parameters in the functional equation to solve this kind of functional equation problem in most of the case by my experience.
Actually x also cannot be 0 because our function ends up being c/y therefore the function has no solution at 0
No, this is false. The functional equation as stated can hold for all x in R, and all y in R\{0}. It implies that f(0·y) = f(0)/y = f(0) for all y in R\{0}, and this means that f(0) = 0, but it says nothing about what f is elsewhere.
Awesome
I am struggling with proving f(xy) = f(yx), any hints?
For all real x, y, [x·y = y·x]. Therefore, for all real x, y, and for all f : R -> R, [f(x·y) = f(y·x)]. Q. E. D.
I think it was an easy one ❤
Easy one
Answer = 1
Consider all f : R -> R such that f(x·y) = f(x)/y for all x in R and all y in R\{0}. This implies that f(0·y) = f(0) = f(0)/y for all y in R\{0}, which means that f(0) = 0. It also implies that f(1·y) = f(y) = f(1)/y for all y in R\{0}.
Find all f : R -> R such that the above is true, AND such that f(2) + f(4) = 6. Notice that f(2) = f(1)/2 and f(4) = f(1)/4, so f(2) + f(4) = f(1)/2 + f(1)/4 = f(1)·(1/2 + 1/4) = 3/4·f(1) = 6, hence f(1) = 8. Therefore, f(x) = 8/x for all x in R\{0}.
Very nice! 🤩
First method I like better.
1
Nice problem man
❤
🧡
1 minut to solve
15 seconds to solve. Fucking 2 minutes to write it down.