Solving Two Functional Equations

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#ChallengingMathProblems #FunctionalEquations

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[kushaldey]

f(xy)=xf(y)+yf(x)
Put x=y=-1, f(-1)=½f(1)=0
Put y=-1, f(-x)=-f(x)
Thus, f(x) is odd, thus f(0)=0.
So f(x)=kxln(x) is kind of incomplete,
f(x)=kxln(|x|), with f(0)=0, f(x) is continuous at x=0, ofcourse in the previous definition if we take limit as x approaches 0 from right, it will be 0, but still we would assume that the function will be discontinuous at x=0 due to ln(x).

[geoffreytrang]

The second equation is similar to the product rule for derivatives, which states that (fg)'=fg'+gf'.

[pkmath]

I somehow get to check your second method first over the first method haha good job!

[ahmadmazbouh]

2 problems in 1 video? this is a thing we don't see everyday!! respect

[jclzjbe]

You can also have solutions to the second equation that are continuous NOWHERE.

As another commenter showed, f is odd, so we can focus on just the positive values of x and y, and just reflect the rest of the graph about the origin.

Since x and y are positive, let x = e^a, y = e^b, where a = ln(x), b = ln(y). Divide both sides of the functional equation by xy, then plug in e^a and e^b. You get that h(a+b) = h(a) + h(b), where h(z) = f(e^z)/e^z. There are continuous-nowhere and everywhere dense solutions to this in R (that involve the axiom of choice and R as a vector space over Q), so you can pick any of them for h(z), then work back what the original function is.

[peterchan]

7:48 'We know that the solutions are logarithmic'
How do you justify this?

[stmmniko]

66, 6 thousands subscribers. Evil number

[RisetotheEquation]

Greetings, Syber - and Happy New Year! Just sent you a little message.

[faridganouche]

في الجواب الأخير هناك خطأ بسيط لأن الدوال f معرفة على *R إذن الدالة لوغارتم تكتب بالقيمة المطلقة أي (/×/)k ln

[chaosredefined]

For the second problem, for a method which doesn't use division:

We want to turn the product into a sum, somehow. So, let x = 2^a and y = 2^b
f(2^(a+b)) = 2^a f(2^b) + 2^b f(2^a)
Denote g(x) = f(2^x)
Hence g(a+b) = 2^a g(b) + 2^b g(a)

Set a = 1
g(b+1) = 2 g(b) + 2^b g(1)
We'll call g(1) = c.
g(b+1) = 2 g(b) + c 2^b

Let's generate some terms for this, just so we can see what's happening. g(1) = c. g(2) = 4c. g(3) = 12c. g(4) = 32c. g(5) = 80c. That looks like g(n) = c*n*2^(n-1). We have 5 base cases for a proof by induction, so we'll test the induction hypothesis:
g(n+1) = c*(n+1)*2^n = c*n*2^n + c*2^n = 2(c*n*2^(n-1)) + c*2^n = 2g(n) + c^2^n.

Confirmed that that is the series.


So, g(n) = c*n*2^(n-1)
f(2^n) = c*n*2^(n-1)
Let x = 2^n
f(x) = c * log_2(x) * x/2
f(x) = k x ln(x), where k = log_2(e)/2.

[JohnRandomness]

The first problem: f(x+0) = x*f(0) + 0*f(x) = x*f(0) = f(x). Now, take x=0. 0*f(0) = f(0). The conclusion is f(x) = 0.

[rajeevk]

In question no 1
Can we div both side by xy
F(x+y)/xy =f(x)/x+f(y)/y
G(x+y)=G(x)+G(y)

G(x) = mx
F(x)=m (X^2)
Later putting f(0)and f2 we will get m=0

F(x)=0

It would be 3rd way

[shmuelzehavi]

We know that the function: g(x) = ln⁡x satisfies the equation:
g(xy) = g(x) + g(y),
but where is the proof that it's the single one which does it ?

[rachidtanan]

Please Can you do this:. f is from IN to IN. with f(n^2+m^2)=f^2(n)+f^2(m)
& f(1)≠0

[fakelove]

What about some topics about discrete mathematics? I am totally new to the graph theory world

[marconniaugusto]

Hi, congratulations on the math videos, very good!! Do you use Android tablet / Apple ipad or graphics tablet on windows? And what program do you use for the videos? I'm going to start recording math videos for my students.

[angshukNag]

What happens if the function is not defined at x = 0 ?

[sevencube]

Is there something like NoteAbility for Android?

[govindam_adi_purusham]

Hey, can you solve this : f(x²+y)=f²(x)+f(y) ? This was on my test yesterday

[ahmadmazbouh]

syber you're at 666 hundreds subscribers which is cool if you know what i mean, you deserve some more zeros at the end tho