- #1

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##\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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- Thread starter tade
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- #1

- 552

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##\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?

- #2

DrClaude

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##f(x) = Ce^{\pm x}##

- #3

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

- #4

DrClaude

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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 )Nice. How did you work that out?

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

- #5

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That's pretty neat. What branch of mathematics is that under?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 )

- #6

DrClaude

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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.That's pretty neat. What branch of mathematics is that under?

- #7

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

- #8

jedishrfu

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http://mathworld.wolfram.com/TaylorSeries.html

- #9

mfb

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It is a typical homework problem to show this.

- #10

Stephen Tashi

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"Functional equations" https://en.wikipedia.org/wiki/Functional_equationNo, I was just wondering what topic or branch this question falls under.

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