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












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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    I have edited your question to include the tags notation and parametric (which, I think, are appropriate tags for this question; please feel free to revert if you disagree), and removed the tag functional-analysis, as this question does not seem to be about Banach spaces and their relatives.
    $endgroup$
    – Xander Henderson
    Feb 24 at 19:11
















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?










share|cite|improve this question











$endgroup$












  • $begingroup$
    I have edited your question to include the tags notation and parametric (which, I think, are appropriate tags for this question; please feel free to revert if you disagree), and removed the tag functional-analysis, as this question does not seem to be about Banach spaces and their relatives.
    $endgroup$
    – Xander Henderson
    Feb 24 at 19:11














2












2








2


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?










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?







functions notation parametric






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 24 at 19:07









Xander Henderson

14.8k103555




14.8k103555










asked Feb 24 at 12:07









NyarukoNyaruko

1326




1326












  • $begingroup$
    I have edited your question to include the tags notation and parametric (which, I think, are appropriate tags for this question; please feel free to revert if you disagree), and removed the tag functional-analysis, as this question does not seem to be about Banach spaces and their relatives.
    $endgroup$
    – Xander Henderson
    Feb 24 at 19:11


















  • $begingroup$
    I have edited your question to include the tags notation and parametric (which, I think, are appropriate tags for this question; please feel free to revert if you disagree), and removed the tag functional-analysis, as this question does not seem to be about Banach spaces and their relatives.
    $endgroup$
    – Xander Henderson
    Feb 24 at 19:11
















$begingroup$
I have edited your question to include the tags notation and parametric (which, I think, are appropriate tags for this question; please feel free to revert if you disagree), and removed the tag functional-analysis, as this question does not seem to be about Banach spaces and their relatives.
$endgroup$
– Xander Henderson
Feb 24 at 19:11




$begingroup$
I have edited your question to include the tags notation and parametric (which, I think, are appropriate tags for this question; please feel free to revert if you disagree), and removed the tag functional-analysis, as this question does not seem to be about Banach spaces and their relatives.
$endgroup$
– Xander Henderson
Feb 24 at 19:11










3 Answers
3






active

oldest

votes


















2












$begingroup$

Depending on what you are doing with these polynomials, there are a number of ways that you could think about them. Depending on how you think about them, you might adopt different notation. A non-exhaustive list of possibilities follows:




  • One possibility is that you might be interested in some kind of algebraic structure. In that case, what you are "really" doing is associating a quadratic polynomial to a point in $mathbb{R}^2$. If this is the kind of structure that you are trying to emphasize, then you might define
    $$ M: mathbb{R}^2 to mathbb{R}[t] : (a,b) mapsto at^2 + bt. $$
    Here, $mathbb{R}[t]$ denotes the ring of polynomials with real coefficients. Hence $M$ is a function which eats an ordered pair of parameters and spits out a polynomial. For example,
    $$M(3,-2) = 3t^2 - 2t. $$

  • Another possibility is that you are not so much interested in the abstract polynomials with a "formal" symbol $t$, but that you are actually interested in the actual functions defined this way. Notationally, I suppose that we could write
    $$ M: mathbb{R}^2 to C(mathbb{R}) : varphi(a,b) mapsto y_{a,b}, $$
    where $y_{(a,b)}$ is the function defined by the formula
    $$ y_{a,b}(t) = at^2 + bt. $$
    Here, $C(mathbb{R})$ denotes the set of continuous functions on $mathbb{R}$. You could replace this with some other target space (say, the space of locally integrable functions, or the space of smooth functions, or whatever). The point is that $M$ is a function which eats an ordered pair of real numbers, and spits out another function. You needn't necessarily carry the indexing around on the $y$ (though I think that Garmekain totally has the right idea on that); you could instead write
    $$ M(a,b) = (t mapsto at^2 + bt). $$
    In some environments, this might even be the "right" thing to do. For example, this is (more or less) how the software package Maple would handle this.

  • If you don't really care about how the parameters are getting mapped from one space to another, you might simply consider the family of functions indexed by the parameters. Notationally, this can look basically identical to the above: here, we consider the family of functions
    $$ { y_{a,b}}_{a,binmathbb{R}}, $$
    where $y_{a,b}$ is the function defined by
    $$ y_{a,b}(t) = at^2 + bt. $$
    On some level, this is completely equivalent to the previous characterization (since an indexing of a set is really just a map from the index set to the set), but the emphasis here is on the actual objects (i.e. the functions $y_{a,b}$) rather than on the process which gives you those functions.


  • Yet another possibility (which might arise in, for example, optimization problems) is to consider a function of three variables, i.e.
    $$ M : mathbb{R}^2 times [0,infty) to mathbb{R} : (a,b,t) mapsto at^2 + bt. $$
    I have chosen to assume that $t$ is nonnegative, as we often work in settings where $t = text{time}$ is assumed to be nonnegative. This assumption is totally arbitrary, but makes sense from the point of view that $a$ and $b$ are some kind of initial condition (e.g. $a$ might be half of an initial acceleration, and $b$ might be an initial velocity).



    Given that your stated goal has something to do with observing changes in the function as the parameters are perturbed, this last notation might be the most reasonable. It allows you to easily discuss, for example, partial derivatives with resect to the parameters $a$ and $b$, while keeping $t$ fixed.








share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

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



    This implies that $y$ is a function taking some $tin T$ parametrized by $a$ and $b$, i.e. $a$ and $b$ are fixed.






    share|cite|improve this answer











    $endgroup$





















      2












      $begingroup$

      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
        Feb 24 at 12:20










      • $begingroup$
        Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
        $endgroup$
        – idriskameni
        Feb 24 at 12:25










      • $begingroup$
        @Nyaruko If you change $a$ and $b$ the same way you change $t$ then $y$ becomes a function taking a 3-tuple as an argument.
        $endgroup$
        – Garmekain
        Feb 24 at 16:04











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Depending on what you are doing with these polynomials, there are a number of ways that you could think about them. Depending on how you think about them, you might adopt different notation. A non-exhaustive list of possibilities follows:




      • One possibility is that you might be interested in some kind of algebraic structure. In that case, what you are "really" doing is associating a quadratic polynomial to a point in $mathbb{R}^2$. If this is the kind of structure that you are trying to emphasize, then you might define
        $$ M: mathbb{R}^2 to mathbb{R}[t] : (a,b) mapsto at^2 + bt. $$
        Here, $mathbb{R}[t]$ denotes the ring of polynomials with real coefficients. Hence $M$ is a function which eats an ordered pair of parameters and spits out a polynomial. For example,
        $$M(3,-2) = 3t^2 - 2t. $$

      • Another possibility is that you are not so much interested in the abstract polynomials with a "formal" symbol $t$, but that you are actually interested in the actual functions defined this way. Notationally, I suppose that we could write
        $$ M: mathbb{R}^2 to C(mathbb{R}) : varphi(a,b) mapsto y_{a,b}, $$
        where $y_{(a,b)}$ is the function defined by the formula
        $$ y_{a,b}(t) = at^2 + bt. $$
        Here, $C(mathbb{R})$ denotes the set of continuous functions on $mathbb{R}$. You could replace this with some other target space (say, the space of locally integrable functions, or the space of smooth functions, or whatever). The point is that $M$ is a function which eats an ordered pair of real numbers, and spits out another function. You needn't necessarily carry the indexing around on the $y$ (though I think that Garmekain totally has the right idea on that); you could instead write
        $$ M(a,b) = (t mapsto at^2 + bt). $$
        In some environments, this might even be the "right" thing to do. For example, this is (more or less) how the software package Maple would handle this.

      • If you don't really care about how the parameters are getting mapped from one space to another, you might simply consider the family of functions indexed by the parameters. Notationally, this can look basically identical to the above: here, we consider the family of functions
        $$ { y_{a,b}}_{a,binmathbb{R}}, $$
        where $y_{a,b}$ is the function defined by
        $$ y_{a,b}(t) = at^2 + bt. $$
        On some level, this is completely equivalent to the previous characterization (since an indexing of a set is really just a map from the index set to the set), but the emphasis here is on the actual objects (i.e. the functions $y_{a,b}$) rather than on the process which gives you those functions.


      • Yet another possibility (which might arise in, for example, optimization problems) is to consider a function of three variables, i.e.
        $$ M : mathbb{R}^2 times [0,infty) to mathbb{R} : (a,b,t) mapsto at^2 + bt. $$
        I have chosen to assume that $t$ is nonnegative, as we often work in settings where $t = text{time}$ is assumed to be nonnegative. This assumption is totally arbitrary, but makes sense from the point of view that $a$ and $b$ are some kind of initial condition (e.g. $a$ might be half of an initial acceleration, and $b$ might be an initial velocity).



        Given that your stated goal has something to do with observing changes in the function as the parameters are perturbed, this last notation might be the most reasonable. It allows you to easily discuss, for example, partial derivatives with resect to the parameters $a$ and $b$, while keeping $t$ fixed.








      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Depending on what you are doing with these polynomials, there are a number of ways that you could think about them. Depending on how you think about them, you might adopt different notation. A non-exhaustive list of possibilities follows:




        • One possibility is that you might be interested in some kind of algebraic structure. In that case, what you are "really" doing is associating a quadratic polynomial to a point in $mathbb{R}^2$. If this is the kind of structure that you are trying to emphasize, then you might define
          $$ M: mathbb{R}^2 to mathbb{R}[t] : (a,b) mapsto at^2 + bt. $$
          Here, $mathbb{R}[t]$ denotes the ring of polynomials with real coefficients. Hence $M$ is a function which eats an ordered pair of parameters and spits out a polynomial. For example,
          $$M(3,-2) = 3t^2 - 2t. $$

        • Another possibility is that you are not so much interested in the abstract polynomials with a "formal" symbol $t$, but that you are actually interested in the actual functions defined this way. Notationally, I suppose that we could write
          $$ M: mathbb{R}^2 to C(mathbb{R}) : varphi(a,b) mapsto y_{a,b}, $$
          where $y_{(a,b)}$ is the function defined by the formula
          $$ y_{a,b}(t) = at^2 + bt. $$
          Here, $C(mathbb{R})$ denotes the set of continuous functions on $mathbb{R}$. You could replace this with some other target space (say, the space of locally integrable functions, or the space of smooth functions, or whatever). The point is that $M$ is a function which eats an ordered pair of real numbers, and spits out another function. You needn't necessarily carry the indexing around on the $y$ (though I think that Garmekain totally has the right idea on that); you could instead write
          $$ M(a,b) = (t mapsto at^2 + bt). $$
          In some environments, this might even be the "right" thing to do. For example, this is (more or less) how the software package Maple would handle this.

        • If you don't really care about how the parameters are getting mapped from one space to another, you might simply consider the family of functions indexed by the parameters. Notationally, this can look basically identical to the above: here, we consider the family of functions
          $$ { y_{a,b}}_{a,binmathbb{R}}, $$
          where $y_{a,b}$ is the function defined by
          $$ y_{a,b}(t) = at^2 + bt. $$
          On some level, this is completely equivalent to the previous characterization (since an indexing of a set is really just a map from the index set to the set), but the emphasis here is on the actual objects (i.e. the functions $y_{a,b}$) rather than on the process which gives you those functions.


        • Yet another possibility (which might arise in, for example, optimization problems) is to consider a function of three variables, i.e.
          $$ M : mathbb{R}^2 times [0,infty) to mathbb{R} : (a,b,t) mapsto at^2 + bt. $$
          I have chosen to assume that $t$ is nonnegative, as we often work in settings where $t = text{time}$ is assumed to be nonnegative. This assumption is totally arbitrary, but makes sense from the point of view that $a$ and $b$ are some kind of initial condition (e.g. $a$ might be half of an initial acceleration, and $b$ might be an initial velocity).



          Given that your stated goal has something to do with observing changes in the function as the parameters are perturbed, this last notation might be the most reasonable. It allows you to easily discuss, for example, partial derivatives with resect to the parameters $a$ and $b$, while keeping $t$ fixed.








        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Depending on what you are doing with these polynomials, there are a number of ways that you could think about them. Depending on how you think about them, you might adopt different notation. A non-exhaustive list of possibilities follows:




          • One possibility is that you might be interested in some kind of algebraic structure. In that case, what you are "really" doing is associating a quadratic polynomial to a point in $mathbb{R}^2$. If this is the kind of structure that you are trying to emphasize, then you might define
            $$ M: mathbb{R}^2 to mathbb{R}[t] : (a,b) mapsto at^2 + bt. $$
            Here, $mathbb{R}[t]$ denotes the ring of polynomials with real coefficients. Hence $M$ is a function which eats an ordered pair of parameters and spits out a polynomial. For example,
            $$M(3,-2) = 3t^2 - 2t. $$

          • Another possibility is that you are not so much interested in the abstract polynomials with a "formal" symbol $t$, but that you are actually interested in the actual functions defined this way. Notationally, I suppose that we could write
            $$ M: mathbb{R}^2 to C(mathbb{R}) : varphi(a,b) mapsto y_{a,b}, $$
            where $y_{(a,b)}$ is the function defined by the formula
            $$ y_{a,b}(t) = at^2 + bt. $$
            Here, $C(mathbb{R})$ denotes the set of continuous functions on $mathbb{R}$. You could replace this with some other target space (say, the space of locally integrable functions, or the space of smooth functions, or whatever). The point is that $M$ is a function which eats an ordered pair of real numbers, and spits out another function. You needn't necessarily carry the indexing around on the $y$ (though I think that Garmekain totally has the right idea on that); you could instead write
            $$ M(a,b) = (t mapsto at^2 + bt). $$
            In some environments, this might even be the "right" thing to do. For example, this is (more or less) how the software package Maple would handle this.

          • If you don't really care about how the parameters are getting mapped from one space to another, you might simply consider the family of functions indexed by the parameters. Notationally, this can look basically identical to the above: here, we consider the family of functions
            $$ { y_{a,b}}_{a,binmathbb{R}}, $$
            where $y_{a,b}$ is the function defined by
            $$ y_{a,b}(t) = at^2 + bt. $$
            On some level, this is completely equivalent to the previous characterization (since an indexing of a set is really just a map from the index set to the set), but the emphasis here is on the actual objects (i.e. the functions $y_{a,b}$) rather than on the process which gives you those functions.


          • Yet another possibility (which might arise in, for example, optimization problems) is to consider a function of three variables, i.e.
            $$ M : mathbb{R}^2 times [0,infty) to mathbb{R} : (a,b,t) mapsto at^2 + bt. $$
            I have chosen to assume that $t$ is nonnegative, as we often work in settings where $t = text{time}$ is assumed to be nonnegative. This assumption is totally arbitrary, but makes sense from the point of view that $a$ and $b$ are some kind of initial condition (e.g. $a$ might be half of an initial acceleration, and $b$ might be an initial velocity).



            Given that your stated goal has something to do with observing changes in the function as the parameters are perturbed, this last notation might be the most reasonable. It allows you to easily discuss, for example, partial derivatives with resect to the parameters $a$ and $b$, while keeping $t$ fixed.








          share|cite|improve this answer









          $endgroup$



          Depending on what you are doing with these polynomials, there are a number of ways that you could think about them. Depending on how you think about them, you might adopt different notation. A non-exhaustive list of possibilities follows:




          • One possibility is that you might be interested in some kind of algebraic structure. In that case, what you are "really" doing is associating a quadratic polynomial to a point in $mathbb{R}^2$. If this is the kind of structure that you are trying to emphasize, then you might define
            $$ M: mathbb{R}^2 to mathbb{R}[t] : (a,b) mapsto at^2 + bt. $$
            Here, $mathbb{R}[t]$ denotes the ring of polynomials with real coefficients. Hence $M$ is a function which eats an ordered pair of parameters and spits out a polynomial. For example,
            $$M(3,-2) = 3t^2 - 2t. $$

          • Another possibility is that you are not so much interested in the abstract polynomials with a "formal" symbol $t$, but that you are actually interested in the actual functions defined this way. Notationally, I suppose that we could write
            $$ M: mathbb{R}^2 to C(mathbb{R}) : varphi(a,b) mapsto y_{a,b}, $$
            where $y_{(a,b)}$ is the function defined by the formula
            $$ y_{a,b}(t) = at^2 + bt. $$
            Here, $C(mathbb{R})$ denotes the set of continuous functions on $mathbb{R}$. You could replace this with some other target space (say, the space of locally integrable functions, or the space of smooth functions, or whatever). The point is that $M$ is a function which eats an ordered pair of real numbers, and spits out another function. You needn't necessarily carry the indexing around on the $y$ (though I think that Garmekain totally has the right idea on that); you could instead write
            $$ M(a,b) = (t mapsto at^2 + bt). $$
            In some environments, this might even be the "right" thing to do. For example, this is (more or less) how the software package Maple would handle this.

          • If you don't really care about how the parameters are getting mapped from one space to another, you might simply consider the family of functions indexed by the parameters. Notationally, this can look basically identical to the above: here, we consider the family of functions
            $$ { y_{a,b}}_{a,binmathbb{R}}, $$
            where $y_{a,b}$ is the function defined by
            $$ y_{a,b}(t) = at^2 + bt. $$
            On some level, this is completely equivalent to the previous characterization (since an indexing of a set is really just a map from the index set to the set), but the emphasis here is on the actual objects (i.e. the functions $y_{a,b}$) rather than on the process which gives you those functions.


          • Yet another possibility (which might arise in, for example, optimization problems) is to consider a function of three variables, i.e.
            $$ M : mathbb{R}^2 times [0,infty) to mathbb{R} : (a,b,t) mapsto at^2 + bt. $$
            I have chosen to assume that $t$ is nonnegative, as we often work in settings where $t = text{time}$ is assumed to be nonnegative. This assumption is totally arbitrary, but makes sense from the point of view that $a$ and $b$ are some kind of initial condition (e.g. $a$ might be half of an initial acceleration, and $b$ might be an initial velocity).



            Given that your stated goal has something to do with observing changes in the function as the parameters are perturbed, this last notation might be the most reasonable. It allows you to easily discuss, for example, partial derivatives with resect to the parameters $a$ and $b$, while keeping $t$ fixed.









          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 24 at 19:57









          Xander HendersonXander Henderson

          14.8k103555




          14.8k103555























              3












              $begingroup$

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



              This implies that $y$ is a function taking some $tin T$ parametrized by $a$ and $b$, i.e. $a$ and $b$ are fixed.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

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



                This implies that $y$ is a function taking some $tin T$ parametrized by $a$ and $b$, i.e. $a$ and $b$ are fixed.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

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



                  This implies that $y$ is a function taking some $tin T$ parametrized by $a$ and $b$, i.e. $a$ and $b$ are fixed.






                  share|cite|improve this answer











                  $endgroup$



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



                  This implies that $y$ is a function taking some $tin T$ parametrized by $a$ and $b$, i.e. $a$ and $b$ are fixed.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Feb 24 at 15:19

























                  answered Feb 24 at 13:12









                  GarmekainGarmekain

                  1,495720




                  1,495720























                      2












                      $begingroup$

                      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
                        Feb 24 at 12:20










                      • $begingroup$
                        Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
                        $endgroup$
                        – idriskameni
                        Feb 24 at 12:25










                      • $begingroup$
                        @Nyaruko If you change $a$ and $b$ the same way you change $t$ then $y$ becomes a function taking a 3-tuple as an argument.
                        $endgroup$
                        – Garmekain
                        Feb 24 at 16:04
















                      2












                      $begingroup$

                      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
                        Feb 24 at 12:20










                      • $begingroup$
                        Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
                        $endgroup$
                        – idriskameni
                        Feb 24 at 12:25










                      • $begingroup$
                        @Nyaruko If you change $a$ and $b$ the same way you change $t$ then $y$ becomes a function taking a 3-tuple as an argument.
                        $endgroup$
                        – Garmekain
                        Feb 24 at 16:04














                      2












                      2








                      2





                      $begingroup$

                      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 Feb 24 at 12:18









                      idriskameniidriskameni

                      755321




                      755321












                      • $begingroup$
                        But what if I would like to mean is that, by varying a and b, y(t) also change accordingly?
                        $endgroup$
                        – Nyaruko
                        Feb 24 at 12:20










                      • $begingroup$
                        Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
                        $endgroup$
                        – idriskameni
                        Feb 24 at 12:25










                      • $begingroup$
                        @Nyaruko If you change $a$ and $b$ the same way you change $t$ then $y$ becomes a function taking a 3-tuple as an argument.
                        $endgroup$
                        – Garmekain
                        Feb 24 at 16:04


















                      • $begingroup$
                        But what if I would like to mean is that, by varying a and b, y(t) also change accordingly?
                        $endgroup$
                        – Nyaruko
                        Feb 24 at 12:20










                      • $begingroup$
                        Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
                        $endgroup$
                        – idriskameni
                        Feb 24 at 12:25










                      • $begingroup$
                        @Nyaruko If you change $a$ and $b$ the same way you change $t$ then $y$ becomes a function taking a 3-tuple as an argument.
                        $endgroup$
                        – Garmekain
                        Feb 24 at 16:04
















                      $begingroup$
                      But what if I would like to mean is that, by varying a and b, y(t) also change accordingly?
                      $endgroup$
                      – Nyaruko
                      Feb 24 at 12:20




                      $begingroup$
                      But what if I would like to mean is that, by varying a and b, y(t) also change accordingly?
                      $endgroup$
                      – Nyaruko
                      Feb 24 at 12:20












                      $begingroup$
                      Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
                      $endgroup$
                      – idriskameni
                      Feb 24 at 12:25




                      $begingroup$
                      Then, I would express it more or less as you did. $$y(t)=M(a,b,t)$$
                      $endgroup$
                      – idriskameni
                      Feb 24 at 12:25












                      $begingroup$
                      @Nyaruko If you change $a$ and $b$ the same way you change $t$ then $y$ becomes a function taking a 3-tuple as an argument.
                      $endgroup$
                      – Garmekain
                      Feb 24 at 16:04




                      $begingroup$
                      @Nyaruko If you change $a$ and $b$ the same way you change $t$ then $y$ becomes a function taking a 3-tuple as an argument.
                      $endgroup$
                      – Garmekain
                      Feb 24 at 16:04


















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