Jojo
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Member for 3 years, 11 months
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Integral calculus with Gamma function
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One observation of special type of square matrix exponentiation
definitions for u and v were mixed up
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Priming for the primes
In your example with the forks and spoons, what does addition correspond to?
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Mapping exponentiation onto addition
Thankyou. This is much simpler than trying to look at $a=b$.
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Mapping exponentiation onto addition
Thanks for the link, this was very interesting and counterintuitive to me. What method would you use to show that the solution you gave is unique? My default is to try and rewrite functional equations as ODEs, but it doesn't always work very well
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Mapping exponentiation onto addition
Could you explain why that is? Might it be unique if $f$ is continuous?
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Mapping exponentiation onto addition
Usual usage for 'function' in mathematical physics is probably something like 'defined on the reals minus some subset of points, differentiable or probably $C^\infty$'. I appreciate that I'm posting on a mathematics forum now though
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Mapping exponentiation onto addition
This is a very nice and simple solution. Do you have thoughts on whether it's unique as a solution to my $(3)$? I think the logarithm is the unique solution to $f(ab) = f(a) + f(b)$, up to a scaling maybe. I edited my question to specify more clearly
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Mapping exponentiation onto addition
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Mapping exponentiation onto addition
Yes OK, I'm coming from mathematical physics and I'm just getting used to being more specific about these kinds of things. I added a restriction to remove 1 from the domain of $f$
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Mapping exponentiation onto addition
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Mapping exponentiation onto addition
I added the restriction $a>0$ to the question. I think if $f$ diverges at $1$ then it's not necessary to have $h$ constant
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Mapping exponentiation onto addition
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Recursion for generating functions
@Gupta I don't quite follow your argument about the polynomial and $\int dz F/z$, but I think if you made it clearer and put it in the question it might help you get a better answer. (Also I realised that my previous comment is wrong, but what you has was correct)