# I Function as a Solution to Specific Conditions

#### tade

I'm trying to solve some statistical mechanics. This problem appeared.

$\frac{f\left(0\right)}{f\left(a\right)}f\left(x+a\right)=f\left(x\right)$

Any idea as to which function will satisfy this equation?

#### DrClaude

Mentor
$f(x) = Ce^{\pm x}$

• tade

#### tade

$f(x) = Ce^{\pm x}$
Nice. How did you work that out? Is an exponential function the only solution to this problem?

#### DrClaude

Mentor
Nice. How did you work that out?
From the observation that $f(x) \propto f(x+a) / f(a)$ requires that an addition in the argument becomes an (inverse) multiplication of the function, and I recognized the exponential. (I actually thought about log first, but then realized I had it backwards )

Is an exponential function the only solution to this problem?
No idea. Apart from the answer I gave and $f(x) = \mathrm{const.}$, I don't see any.

• tade

#### tade

From the observation that $f(x) \propto f(x+a) / f(a)$ requires that an addition in the argument becomes an (inverse) multiplication of the function, and I recognized the exponential. (I actually thought about log first, but then realized I had it backwards )
That's pretty neat. What branch of mathematics is that under?

#### DrClaude

Mentor
That's pretty neat. What branch of mathematics is that under?
I'm not a mathematician. It just comes from years of working with, and getting a feeling for, those kind of mathematical functions.

#### tade

I'm not a mathematician. It just comes from years of working with, and getting a feeling for, those kind of mathematical functions.
No, I was just wondering what topic or branch this question falls under.

Mentor

#### mfb

Mentor
The general solution is $f(x) = Ce^{d x}$, C and d arbitrary constants. C and d can be complex if you work with complex numbers.
It is a typical homework problem to show this.

#### Stephen Tashi

Science Advisor
• DrClaude

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