Function as a Solution to Specific Conditions

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  • Thread starter tade
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  • #1
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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?
 

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  • #2
DrClaude
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##f(x) = Ce^{\pm x}##
 
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  • #3
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##f(x) = Ce^{\pm x}##
Nice. How did you work that out? Is an exponential function the only solution to this problem?
 
  • #4
DrClaude
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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 :smile:)

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.
 
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  • #5
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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 :smile:)
That's pretty neat. What branch of mathematics is that under?
 
  • #6
DrClaude
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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.
 
  • #7
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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.
 
  • #9
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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.
 

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