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Is it possible to map from a parameter to a trajectory?


What is a polynomial and how is it different from a function?Function that is straight and starts oscillating from a certain point.Conditions to guarantee unique limits of trajectories.Is it always possible to algebraically express a function defined by a set of rules?Extension of a linear mapinclusion map between quotient spacesNorm of a continuous linear operatoris it possible to hash a range?Do exist an injective linear map from $mathbb{R}^5rightarrowmathbb{R}^4$Show that there exists a unique continuous linear map













1












$begingroup$


If I have a polynomial trajectory



$$y(t)=at^2+bt$$



The idea I would like to express is that by fixing a and b, I will get a unique trajectory $y(t)$.
Can I say that there exists a map $M$ such that the trajectory $y(t)$ is
$$y(t) = M(a,b)$$
? If not, how can I express the idea in proper math terms?






functional-analysis functions







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    If I have a polynomial trajectory



    $$y(t)=at^2+bt$$



    The idea I would like to express is that by fixing a and b, I will get a unique trajectory $y(t)$.
    Can I say that there exists a map $M$ such that the trajectory $y(t)$ is
    $$y(t) = M(a,b)$$
    ? If not, how can I express the idea in proper math terms?






    functional-analysis functions







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      2



      $begingroup$


      If I have a polynomial trajectory



      $$y(t)=at^2+bt$$



      The idea I would like to express is that by fixing a and b, I will get a unique trajectory $y(t)$.
      Can I say that there exists a map $M$ such that the trajectory $y(t)$ is
      $$y(t) = M(a,b)$$
      ? If not, how can I express the idea in proper math terms?






      functional-analysis functions







      share|cite|improve this question











      $endgroup$




      If I have a polynomial trajectory



      $$y(t)=at^2+bt$$



      The idea I would like to express is that by fixing a and b, I will get a unique trajectory $y(t)$.
      Can I say that there exists a map $M$ such that the trajectory $y(t)$ is
      $$y(t) = M(a,b)$$
      ? If not, how can I express the idea in proper math terms?






      functional-analysis functions




      functional-analysis functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 hours ago









      idriskameni

      723319




      723319










      asked 2 hours ago









      NyarukoNyaruko

      1256




      1256






















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          There is such map, and it is often referred to as a functional, or a mapping from a set of parameters (real or complex numbers or vectors, in your case $a,b$) to a set of functions $y(a,b,;t)$. In other words, a functional maps or "identifies" a function from a single parameter or set of parameters.



          In your case, such mapping would be like:
          $$
          M[y]:quad yto y(a,b,;t)=at^2+bt
          $$



          or, letting $mathbf{p}=(a,b)$, you could rewrite the above as:
          $$
          M[y]:quad yto y(mathbf{p},;t)=p_1t^2+p_2t
          $$



          You could even define a single real number that maps to the above functional, and such number is usually taken as the definite integral of the function between an arbitrary time interval, in this case let it be $[0,1]$:
          $$
          I[y]=int_{0}^{1}y(a,b,t),mathrm{d}t=frac{a}{3}+frac{b}{2}
          $$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            I would express it as:



            Consider $$y(t)=at^2+bt$$ where $a,b$ are fixed elements.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But what if I would like to mean is that, by varying a and b, y(t) also change accordingly?
              $endgroup$
              – Nyaruko
              2 hours ago










            • $begingroup$
              Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
              $endgroup$
              – idriskameni
              2 hours ago



















            1












            $begingroup$

            You can just use $$y_{a,b}(t)=at^2+bt$$






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

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              active

              oldest

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              2












              $begingroup$

              There is such map, and it is often referred to as a functional, or a mapping from a set of parameters (real or complex numbers or vectors, in your case $a,b$) to a set of functions $y(a,b,;t)$. In other words, a functional maps or "identifies" a function from a single parameter or set of parameters.



              In your case, such mapping would be like:
              $$
              M[y]:quad yto y(a,b,;t)=at^2+bt
              $$



              or, letting $mathbf{p}=(a,b)$, you could rewrite the above as:
              $$
              M[y]:quad yto y(mathbf{p},;t)=p_1t^2+p_2t
              $$



              You could even define a single real number that maps to the above functional, and such number is usually taken as the definite integral of the function between an arbitrary time interval, in this case let it be $[0,1]$:
              $$
              I[y]=int_{0}^{1}y(a,b,t),mathrm{d}t=frac{a}{3}+frac{b}{2}
              $$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                There is such map, and it is often referred to as a functional, or a mapping from a set of parameters (real or complex numbers or vectors, in your case $a,b$) to a set of functions $y(a,b,;t)$. In other words, a functional maps or "identifies" a function from a single parameter or set of parameters.



                In your case, such mapping would be like:
                $$
                M[y]:quad yto y(a,b,;t)=at^2+bt
                $$



                or, letting $mathbf{p}=(a,b)$, you could rewrite the above as:
                $$
                M[y]:quad yto y(mathbf{p},;t)=p_1t^2+p_2t
                $$



                You could even define a single real number that maps to the above functional, and such number is usually taken as the definite integral of the function between an arbitrary time interval, in this case let it be $[0,1]$:
                $$
                I[y]=int_{0}^{1}y(a,b,t),mathrm{d}t=frac{a}{3}+frac{b}{2}
                $$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  There is such map, and it is often referred to as a functional, or a mapping from a set of parameters (real or complex numbers or vectors, in your case $a,b$) to a set of functions $y(a,b,;t)$. In other words, a functional maps or "identifies" a function from a single parameter or set of parameters.



                  In your case, such mapping would be like:
                  $$
                  M[y]:quad yto y(a,b,;t)=at^2+bt
                  $$



                  or, letting $mathbf{p}=(a,b)$, you could rewrite the above as:
                  $$
                  M[y]:quad yto y(mathbf{p},;t)=p_1t^2+p_2t
                  $$



                  You could even define a single real number that maps to the above functional, and such number is usually taken as the definite integral of the function between an arbitrary time interval, in this case let it be $[0,1]$:
                  $$
                  I[y]=int_{0}^{1}y(a,b,t),mathrm{d}t=frac{a}{3}+frac{b}{2}
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  There is such map, and it is often referred to as a functional, or a mapping from a set of parameters (real or complex numbers or vectors, in your case $a,b$) to a set of functions $y(a,b,;t)$. In other words, a functional maps or "identifies" a function from a single parameter or set of parameters.



                  In your case, such mapping would be like:
                  $$
                  M[y]:quad yto y(a,b,;t)=at^2+bt
                  $$



                  or, letting $mathbf{p}=(a,b)$, you could rewrite the above as:
                  $$
                  M[y]:quad yto y(mathbf{p},;t)=p_1t^2+p_2t
                  $$



                  You could even define a single real number that maps to the above functional, and such number is usually taken as the definite integral of the function between an arbitrary time interval, in this case let it be $[0,1]$:
                  $$
                  I[y]=int_{0}^{1}y(a,b,t),mathrm{d}t=frac{a}{3}+frac{b}{2}
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  alandellaalandella

                  359514




                  359514























                      1












                      $begingroup$

                      I would express it as:



                      Consider $$y(t)=at^2+bt$$ where $a,b$ are fixed elements.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        But what if I would like to mean is that, by varying a and b, y(t) also change accordingly?
                        $endgroup$
                        – Nyaruko
                        2 hours ago










                      • $begingroup$
                        Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
                        $endgroup$
                        – idriskameni
                        2 hours ago
















                      1












                      $begingroup$

                      I would express it as:



                      Consider $$y(t)=at^2+bt$$ where $a,b$ are fixed elements.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        But what if I would like to mean is that, by varying a and b, y(t) also change accordingly?
                        $endgroup$
                        – Nyaruko
                        2 hours ago










                      • $begingroup$
                        Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
                        $endgroup$
                        – idriskameni
                        2 hours ago














                      1












                      1








                      1





                      $begingroup$

                      I would express it as:



                      Consider $$y(t)=at^2+bt$$ where $a,b$ are fixed elements.






                      share|cite|improve this answer









                      $endgroup$



                      I would express it as:



                      Consider $$y(t)=at^2+bt$$ where $a,b$ are fixed elements.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 hours ago









                      idriskameniidriskameni

                      723319




                      723319












                      • $begingroup$
                        But what if I would like to mean is that, by varying a and b, y(t) also change accordingly?
                        $endgroup$
                        – Nyaruko
                        2 hours ago










                      • $begingroup$
                        Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
                        $endgroup$
                        – idriskameni
                        2 hours ago


















                      • $begingroup$
                        But what if I would like to mean is that, by varying a and b, y(t) also change accordingly?
                        $endgroup$
                        – Nyaruko
                        2 hours ago










                      • $begingroup$
                        Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
                        $endgroup$
                        – idriskameni
                        2 hours ago
















                      $begingroup$
                      But what if I would like to mean is that, by varying a and b, y(t) also change accordingly?
                      $endgroup$
                      – Nyaruko
                      2 hours ago




                      $begingroup$
                      But what if I would like to mean is that, by varying a and b, y(t) also change accordingly?
                      $endgroup$
                      – Nyaruko
                      2 hours ago












                      $begingroup$
                      Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
                      $endgroup$
                      – idriskameni
                      2 hours ago




                      $begingroup$
                      Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
                      $endgroup$
                      – idriskameni
                      2 hours ago











                      1












                      $begingroup$

                      You can just use $$y_{a,b}(t)=at^2+bt$$






                      share|cite|improve this answer









                      $endgroup$


















                        1












                        $begingroup$

                        You can just use $$y_{a,b}(t)=at^2+bt$$






                        share|cite|improve this answer









                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          You can just use $$y_{a,b}(t)=at^2+bt$$






                          share|cite|improve this answer









                          $endgroup$



                          You can just use $$y_{a,b}(t)=at^2+bt$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 1 hour ago









                          GarmekainGarmekain

                          1,461720




                          1,461720






























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