For every differentiable function f : R → R , there is a function g : R × R → R such that g ( f ( x ) , f ′ ( x ) ) = 0 for every x and for every differentiable function h : R → R holds thatbeing true that for every x ∈ R , g ( h ( x ) , h ′ ( x ) ) = 0 and h ( 0 ) = f ( 0 ) implies that h ( x ) = f ( x ) for every x ∈ R .i.e every differentiable function f is a solution to some first order differential equation that has translation symmetry.

Question

For every differentiable function   f  :      R    →      R  , there is a function   g  :      R    ×      R    →      R   such that   g  (  f  (  x  )  ,      f    ′    (  x  )  )  =  0 for every   x and for every differentiable function   h  :      R    →      R   holds thatbeing true that for every   x  ∈      R  ,   g  (  h  (  x  )  ,      h    ′    (  x  )  )  =  0 and   h  (  0  )  =  f  (  0  ) implies that   h  (  x  )  =  f  (  x  ) for every   x  ∈      R  .i.e every differentiable function   f is a solution to some first order differential equation that has translation symmetry.

Carleigh Shaffer · Accepted Answer

It&#039;s easy to construct a differentiable function   f  (  x  ) such that   f  (  0  )  =  f  (  1  ) and       f    ′    (  0  )  =      f    ′    (  1  ) but   f  (  1  )  ≠  f  (  2  ). If   f  (  x  ) satisfies the differential equation   g  (  y  (  x  )  ,      y    ′    (  x  )  )  =  0, then so does   h  (  x  )  =  f  (  x  +  1  ), and   h  (  0  )  =  f  (  0  ) but   h  (  1  )  ≠  f  (  1  ).EDIT: In response to your comment, here&#039;s another example. Consider a differentiable function   f  (  x  ) such that   f  (  0  )  =  f  (  1  )  =  f  (  2  ),       f    ′    (  0  )  =      f    ′    (  1  )  =      f    ′    (  2  )  =  0, and   f  (  1      /    2  )  ≠  f  (  3      /    2  ), and let  h  (  x  )  =      {                            f          (          x          +          1          )                                       if                           x              ∈              [              0              ,              1              )                                                            f          (          x          −          1          )                                       if                           x              ∈              [              1              ,              2              )                                                            f          (          x          )                                       otherwise

motorinum6fh9v · Accepted Answer

If you restrict yourself to algebraic differential equations, Hölder&#039;s theorem says that the   Γ function (or equivalently its reciprocal if you want it to be differentiable everywhere) does not satisfy one.

For every differentiable function f : <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="dou

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