#functionalequation — Public Fediverse posts
Live and recent posts from across the Fediverse tagged #functionalequation, aggregated by home.social.
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@mittelwertsatz @chris_sangwin To express that idea more formally, define functions \[ L_n(x) = \int_1^x \xi^{n-1} d\xi. \]
Then \( L_0(x) = \ln x \) and \( L_n(x) = \frac{x^n - 1}n \) otherwise. But all \( L_n \) obey the functional equations
\[ L_n(x y) = y^n L_n(x) + L_n(y), \quad 𝐿_n(1)=0. \]
My question now is: Do these "generalized logarithms" occur somewhere naturally? -
@mittelwertsatz @chris_sangwin To express that idea more formally, define functions \[ L_n(x) = \int_1^x \xi^{n-1} d\xi. \]
Then \( L_0(x) = \ln x \) and \( L_n(x) = \frac{x^n - 1}n \) otherwise. But all \( L_n \) obey the functional equations
\[ L_n(x y) = y^n L_n(x) + L_n(y), \quad 𝐿_n(1)=0. \]
My question now is: Do these "generalized logarithms" occur somewhere naturally? -
@mittelwertsatz @chris_sangwin To express that idea more formally, define functions \[ L_n(x) = \int_1^x \xi^{n-1} d\xi. \]
Then \( L_0(x) = \ln x \) and \( L_n(x) = \frac{x^n - 1}n \) otherwise. But all \( L_n \) obey the functional equations
\[ L_n(x y) = y^n L_n(x) + L_n(y), \quad 𝐿_n(1)=0. \]
My question now is: Do these "generalized logarithms" occur somewhere naturally? -
@mittelwertsatz @chris_sangwin To express that idea more formally, define functions \[ L_n(x) = \int_1^x \xi^{n-1} d\xi. \]
Then \( L_0(x) = \ln x \) and \( L_n(x) = \frac{x^n - 1}n \) otherwise. But all \( L_n \) obey the functional equations
\[ L_n(x y) = y^n L_n(x) + L_n(y), \quad 𝐿_n(1)=0. \]
My question now is: Do these "generalized logarithms" occur somewhere naturally? -
@mittelwertsatz @chris_sangwin To express that idea more formally, define functions \[ L_n(x) = \int_1^x \xi^{n-1} d\xi. \]
Then \( L_0(x) = \ln x \) and \( L_n(x) = \frac{x^n - 1}n \) otherwise. But all \( L_n \) obey the functional equations
\[ L_n(x y) = y^n L_n(x) + L_n(y), \quad 𝐿_n(1)=0. \]
My question now is: Do these "generalized logarithms" occur somewhere naturally?