Solving the Functional Equation f(x+y)=f(x)f(y)

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#ChallengingMathProblems #FunctionalEquations

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[govindam_adi_purusham]

Me: sees the thumbnail, my subconscious mind: a^x!

[davidblauyoutube]

Once you reach f(x) = m e^(c x), you can use f(0) = 1 to see immediately that m = 1 is the only (non-zero) choice. :)

[BCS-IshtiyakAhmadKhan]

Directly differentiate both sides taking y as constant you will get f'(x+y)=f'(x)f(y) Now put x=0 and y=x you will get f'(x)=f'(0)f(x) now proceed with the video

[Darkev]

This style of problems is my favorite, not insanely complex but just tricky enough to be intriguing. Great job!

[idavid]

It's like re-watching your favorite movie: I know where it's going but it's still fun to watch it get there. Great video!

[bscutajar]

I never knew about these types of problems. Functional equations are my new favourite thing. You could easily see that e^Cx is a solution for f(x), but doing it rigorously revealed that it is the only solution (apart from the trivial case).

[dennisdesormier]

I enjoy this problem because it can even be offered to strong algebra students, who can be challenged to think deeply about the operations.
Set y = 0... f(x+0) = f(x) • f(0). So either f(x) = 0 or you can divide by f(x) to get f(0) = 1.
Set y = 1... f(x+1) = f(x) • f(1). So f has a constant factor of change, whatever value f(1) is.
A function with a constant factor of change — that's what exponential functions are all about!
So f(x) = 1 • f(1)^x.
(One can rewrite the value f(1) other ways, say, b or e^k.)

[walterufsc]

f(x + y) = f(x) * f(y)

First, by exchanging x by y, we get exactly the same expression. This allows us to conclude that the expressions for f(x) and f(y) have exactly the same form. Also, making x = 0, it follows that f(y) = f(0) * f(y), that is, f(0) = 1.
Differentiating both sides with respect to x (using the product rule on the right side of the equality):

df(x+y)/dx = f(x) * df(y)/dx + f(y) * df(x)/dx

Using the starting equation:

df(x+y)/dx = [ f(x + y)/f(y)] * df(y)/dx + [f(x + y)/f(x)] * df(x)/dx

[1/f(x + y)] * df(x+y)/dx = [1/f(y)] * df(y)/dx + [1/f(x)] * df(x)/dx

Since f(y) is not a function of x, df(y)/dx = 0. Therefore:

[1/f(x + y)] * df(x+y)/dx = [1/f(x)] * df(x)/dx

Since x and y are independent, the only way for equality to hold is if both sides of the equation are a constant (k). So:

[1/f(x)] * df(x)/dx = k

df(x)/f(x) = k * dx

Integrating both sides, we get:

f(x) = A e^kx

Since f(0) = 1, we have A = 1. Therefore, f(x) = e^kx.

[tonyhaddad]

Wow how a functional equation ends up with the definition of derivative 😍

[josephsun]

The highlight is f(0)=1, which is used to express f’(0). SO COOL👍

[Happy_Abe]

We can let e^c=b for some b>0 and then e^cx=(e^c)^x=b^x
Which we can combine with the solution f(x)=0 to say the solution is
f(x)=b^x for b greater than or equal to 0

[SyberMath]

After seeing the live chat, I realize that I messed up and did not follow the plan. Today should be a radical equation so we'll have to switch today and tomorrow and we'll do the radical equation tomorrow! Sorry for the confusion
EDIT: The post has been edited to reflect the changes!

[titassamanta]

Hello, @SyberMath I deduced that f(x)=0 for all real x. This, being a constant function, is differentiable for all real x.

[fatihsinanesen]

We don't need e, indeed. f(x)=c^x is the solution for any arbitrary constant c>0. Let's check: f(x+y)=f(x)f(y) -> c^(x+y)=(c^x)(c^y) -> c^(x+y)=c^(x+y)

[WahranRai]

classical properties of exponential function as for logarithm (reciprocal function of exponential) f(xy) = f(x) + f(y)

[TarunRanga]

Thankyou so much I had been searching for same time for about 30 min.😊

[skylardeslypere]

Wow that's a really cool solution :0
I've only known this problem with the given that is continuous, not differentiable. It's a much longer solution if we don't assume differentiability.

[prasham_shah]

Thank you I actually was trying to find how to solve such problems.

[ryanburkett]

f(x) = 1 for all x, also solves this equation. So there are two trivial solutions.

[Hobbitangle]

In fact the solution of this functional equation is obvious: f(x) = exp(x), what can be easily verified by substitution exp(x) to the equation. But the result may be achieved by more strong approach.
1) f(x) never can be negative, what follows from partial case y = x, so f(2x) = f(x)^2 > or = 0
2) f(x) is not 0 for either value of argument, except of the trivial case f(x)=0 for all x.
This follows from modified equation
f(z) = f(x) • f(z - x)
So if f(x) = 0 for some x, then f(z)=0 for all z
3) From anothe partial case y=0 we can find f(0):
f(x) = f(x) • f(0)
so f(0) = 1 except the trivial case
4) let's logarithm both sides of equation. We may do this because f(x) > 0
ln(f(x+y) = ln(f(x)) + ln(f(y))
Denoting g(x) = ln(f(x)) we'll get
g(x+y) = g(x) + g(y)
and the only solution of this equation is a linear function f(x)=x