A Functional Equation | Math Olympiads

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[TecknoVicking]

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.

[chahn]

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

[thietsniper]

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

[MathOrient]

If I want to solve a functional equation, I always come to your channel :)

[goldfing]

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.

[SG]

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

[KennethChile]

I did it applyying method #1. Thanks for the analysis of method #2.

[ManjulaMathew-wbzn]

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)

[thomaspickin]

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.

[李承恩-bb]

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.

[ManjulaMathew-wbzn]

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).

[namanhnguyen]

i thought it was f(2)+f(4)=f(6), but luckily its still solvable. f(x)=0 in this situation

[Paul-]

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!

[sampersonguy]

Actually x also cannot be 0 because our function ends up being c/y therefore the function has no solution at 0

[KurenkovDanya]

Hey dude! What do you mean about f(xy)=xyf(x)f(y)

[Utesfan]

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.

[journeytotheparallel]

I am struggling with proving f(xy) = f(yx), any hints?

[angelmendez-rivera]

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}.