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May 25 Information

Functions from convolutions

For idle curiosity....if I have some unknown functions f,g,h but I know their convolutions f*g, g*h, and f*h, is this enough information to figure out what f,g,h are? If not, what additional information is needed to pin them down? -- HappyCamper 12:53, 25 May 2019 (UTC) reply

After Fourier transform, the convolutions become products, so you get 3 algebraic equations with 3 unknowns. Count Iblis ( talk) 14:21, 25 May 2019 (UTC) reply

Periods and subgroups of the real numbers

Given any function f from the real numbers R to itself, a period of f is any real number p for which f(x + p) = f(x) for all real numbers x. Clearly, the periods of any function (including zero) always form a subgroup of the additive group of real numbers. For which subgroups G of (R, +) does there exist a continuous function f for which the subgroup of periods of f is exactly equal to G? Note that if continuity is not required, then such a function always exists: just let f be the indicator function of G. GeoffreyT2000 ( talk) 14:24, 25 May 2019 (UTC) reply

For subgroup {0, 2π, 4π, 6π, ...} there exists such a function: . Ruslik_ Zero 20:57, 25 May 2019 (UTC) reply
In fact, thanks to this ProofWiki result, the answers are exactly just R itself and its cyclic subgroups. Indeed, the subgroup of periods of a continuous function f must be a closed subgroup because it is the intersection , where, for any real number x, fx is the function sending a real number y to f(x + y). If it is a proper subgroup, then it cannot be dense, and so must be cyclic per the linked ProofWiki page. Clearly, any aperiodic function will work for the trivial subgroup containing just zero, while any constant function will work for the improper subgroup. For the cyclic subgroup aZ, where a is a nonzero real number, either or will work. GeoffreyT2000 ( talk) 03:46, 27 May 2019 (UTC) reply
And you may like to consider the analogous question for a continuous function on Rn. The set of p in Rn for which f(x + p) = f(x) for all x in Rn is a closed subgroup of Rn, and any closed subgroup of Rn is the set of periods of some continuous function. In fact, a closed subgroup of Rn is the direct sum of a discrete group (isomorphic to some Zk) and a linear subspace (isomorphic to some Rh). pm a 21:54, 28 May 2019 (UTC) reply
From Wikipedia, the free encyclopedia
Mathematics desk
< May 24 << Apr | May | Jun >> Current desk >
Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.


May 25 Information

Functions from convolutions

For idle curiosity....if I have some unknown functions f,g,h but I know their convolutions f*g, g*h, and f*h, is this enough information to figure out what f,g,h are? If not, what additional information is needed to pin them down? -- HappyCamper 12:53, 25 May 2019 (UTC) reply

After Fourier transform, the convolutions become products, so you get 3 algebraic equations with 3 unknowns. Count Iblis ( talk) 14:21, 25 May 2019 (UTC) reply

Periods and subgroups of the real numbers

Given any function f from the real numbers R to itself, a period of f is any real number p for which f(x + p) = f(x) for all real numbers x. Clearly, the periods of any function (including zero) always form a subgroup of the additive group of real numbers. For which subgroups G of (R, +) does there exist a continuous function f for which the subgroup of periods of f is exactly equal to G? Note that if continuity is not required, then such a function always exists: just let f be the indicator function of G. GeoffreyT2000 ( talk) 14:24, 25 May 2019 (UTC) reply

For subgroup {0, 2π, 4π, 6π, ...} there exists such a function: . Ruslik_ Zero 20:57, 25 May 2019 (UTC) reply
In fact, thanks to this ProofWiki result, the answers are exactly just R itself and its cyclic subgroups. Indeed, the subgroup of periods of a continuous function f must be a closed subgroup because it is the intersection , where, for any real number x, fx is the function sending a real number y to f(x + y). If it is a proper subgroup, then it cannot be dense, and so must be cyclic per the linked ProofWiki page. Clearly, any aperiodic function will work for the trivial subgroup containing just zero, while any constant function will work for the improper subgroup. For the cyclic subgroup aZ, where a is a nonzero real number, either or will work. GeoffreyT2000 ( talk) 03:46, 27 May 2019 (UTC) reply
And you may like to consider the analogous question for a continuous function on Rn. The set of p in Rn for which f(x + p) = f(x) for all x in Rn is a closed subgroup of Rn, and any closed subgroup of Rn is the set of periods of some continuous function. In fact, a closed subgroup of Rn is the direct sum of a discrete group (isomorphic to some Zk) and a linear subspace (isomorphic to some Rh). pm a 21:54, 28 May 2019 (UTC) reply

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