Solving f(xy)=xf(y) in Two Ways

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#ChallengingMathProblems #FunctionalEquations

PLAYLISTS 🎵 :

[vdivoqp]

I spent precious time in my life evaluating f(1)

[angelmendez-rivera]

Let f : M —> M, where M is any monoid, it could even be noncommutative. f(x·y) = x·f(y) for all x, y in M implies f(x) = x·f(1) for all x in M, and substituting into the functional equation means (x·y)·f(1) = x·[y·f(1)] for all x, y in M. Since M is a monoid, multiplication is associative, x·[y·f(1)] = (x·y)·f(1). Thus, f(1) can be any arbitrary member of the monoid. Notice, though, that one cannot say f(x) = f(1)·x solves the equation, unless you have commutativity.

[varionmori]

Assuming that we're looking at something that's not unusual, then we know that y = f(x). Making that assumption, our second step should likely be to do a simple test using the identity function. Considering, no matter what type of function(s) we're being questioned about, doing so would likely give us some sort of idea as to what we're looking at.

So, putting the two together, we would basically be saying to replace and simplify any values with x, of the following:
y = f(x) = x

This test would give us a LHS of:
f(xy) = f(xx) = f(x^2)

Okay, cool. Looks good. This same test would give us a RHS of:
xf(y) = xf(x) = xy = xx = x^2

Putting the two together, we get the following:
f(x^2) = x^2

Since, this actually tells us that we're putting x^2 into our function named f and that we're getting back the same value, putting in x instead actually gives us the final answer to the problem:
y = f(x) = x

So, there's honestly no such thing as c = f(1). The only way this could actually exist, is if we first figured out that, y = f(x) = x. In which case, we could then make the claim that, c = f(1) = 1, which simplifies to, f(x) = cx = 1x = x, but only after we state that we also know what f(x) actually is without writing it in terms of itself.

This is an example of Circular Dependancy, among other things, and doesn't fully answer the question, as it just rephrased the question in a way that basically says, "The person asking the question already know the answer, so ask them what f(x) is and we'd also then know what f(x) is, but until then, f(x) = xf(c)."

It never gave an actual answer to what f(c) or f(q) or f(a^2 + b^2) is equal to since f(x) = xf(c), which is just as good as the question being asked, honestly. The question is supposed to be, "what is f(x) equal to, in terms of x, " and not "what is f(x) equal to, in terms of anything you want it to be?" This would honestly be marked completely wrong. Since, f(xy) = xf(y) just as much as f(x) = xf(c).

Despite this bluper, great work on all your videos including this one, and keep up the great job. You do a very great job, and doing a great job isn't always easy, nor does it always come out as something perfect. I think even this video is a perfect amount of effort, though. 👏

[dlnkbfv]

Basic problem. A linear operator f : ℝ^n → ℝ^m is expressed by matrix.
This is an important fact of linear algebra.

[minhdoantuan]

Let y=1, f(x)=x f(1)
f(1) is a number, so let f(1)=k
So f(x)=kx, check the 2 sides of the equation:
kxy = x × ky, which is true
Therefore, f(x)=kx for any real number k.

[threstytorres]

F(x) = x and F(y) = (cy)/x, where c is an arbitrary paramter

If F(x) = x
F(xy) = xF(y)
xy = xy, Which is True for all numbers x and y

If F(y) = (cy)/x, therefore F(x) = c
F(xy) = xF(y)
c(xy)/x = x((cy)/x)
cy = cy, Which us True for all numbers y because c is a arbitrary parameter

[laokratis]

And a third one: Take the partial derivative in respect to x the first time. This gives y*(df(xy)/d(xy))=f(y). Doing this again for the partial derivative in respect to y we get x*(df(xy)/d(xy))=x*df(y)/dy. Eliminating df(xy)/d(xy) we get df/f=dy/y ->lnf=lny +C or f(y)=C'*y.

[Shadi]

Initially, i started defining:
X=0 => f(0) = 0
X=1, y=0 => f(0) = x
I found 2 values of f(0) not sure why!
Then i came to the point matches your 2nd method when y=1 f(x) = x.f(1)

But if f is continuous as been mentioned at the begining, how i got 2 values of f(0)?

[kanankazimzada]

I appreciate these methods as well. These can work when I solve math problems like this. 👍🏻

[maccollo]

Well I know from my engineering track at college that this is the most linear of linear things.

[rossellagiovanardi]

What do you have to solve...
That's perfectly the same thing.

Furthermore you don't know what you should have to be looking for.

[angshukNag]

We have f(x) = 0 for x = 0 now considering x != 0, f(x*(1/x)) = x*f(1/x) i.e f(1) = x*f(1/x), taking y = 1 f(x) = x*f(1) putting f(1) we get f(x) = x*x*f(1/x) hence f(x) * f(1/x) = x^2, now taking x = 1 we get f(1) = (+ or -) 1 hence f(x) = (+ or -) x .. I am not understanding why i am getting a specific answer?

[vectoria]

Thank you sir for this explanation 🙂. But i have something i want to add i don't know if it's true

If we place y by x we find

if we place x by y we find
f(y^2)=xf(y)
from 1and2we find
f(y^2)=f(xy)
So xy=y^2
x=y

[SuperYoonHo]

thank you your solutions are so amazing it allways makes me click the like button

[jmart]

Very nice second method.
I used derivatives and solved the corresponding differential equation which leads to the same result.
As we say here, I used a canon to kill flies.😊

[gavintillman]

I don’t think you need to assume continuity?

[dublistoeo]

hahaha, this is the shortest 2nd method ever seen in the channel.