Find a wave function with the given zeros












1












$begingroup$


Let $bin(0,pi)$, and let $kinmathbb{Z}$.



Find a function $f:mathbb{R}to[0,1]$ with:




  1. The set of zeros $f_0={x:x=2kpipm b}$

  2. The set of ones $f_1={x:x=2kpi}$

  3. The set of negative ones $f_{-1}={x:x=(2k+1)pi}$


In other words, what I'm looking for is some kind of wave function with the zeros adjusted to be closer to one peak or the other, without changing the amplitude. Here's a visual example:



enter image description here



In this case, $b$ would be somewhere around $2$ or so.



Can you help me figure out what this function $f$ is?






calculus functions trigonometry roots







share|cite|improve this question









$endgroup$












  • $begingroup$
    Look at $cos(x+rsin(x))$ and adjust the $r$. However, if the absolute value of $r$ exceeds $1$, one loses some of your criteria.
    $endgroup$
    – Torsten Schoeneberg
    Dec 16 '18 at 6:30










  • $begingroup$
    @TorstenSchoeneberg, that's definitely helpful, but it also doesn't give me the full range of control I'm looking for.
    $endgroup$
    – jippyjoe4
    Dec 16 '18 at 6:32
















1












$begingroup$


Let $bin(0,pi)$, and let $kinmathbb{Z}$.



Find a function $f:mathbb{R}to[0,1]$ with:




  1. The set of zeros $f_0={x:x=2kpipm b}$

  2. The set of ones $f_1={x:x=2kpi}$

  3. The set of negative ones $f_{-1}={x:x=(2k+1)pi}$


In other words, what I'm looking for is some kind of wave function with the zeros adjusted to be closer to one peak or the other, without changing the amplitude. Here's a visual example:



enter image description here



In this case, $b$ would be somewhere around $2$ or so.



Can you help me figure out what this function $f$ is?






calculus functions trigonometry roots







share|cite|improve this question









$endgroup$












  • $begingroup$
    Look at $cos(x+rsin(x))$ and adjust the $r$. However, if the absolute value of $r$ exceeds $1$, one loses some of your criteria.
    $endgroup$
    – Torsten Schoeneberg
    Dec 16 '18 at 6:30










  • $begingroup$
    @TorstenSchoeneberg, that's definitely helpful, but it also doesn't give me the full range of control I'm looking for.
    $endgroup$
    – jippyjoe4
    Dec 16 '18 at 6:32














1












1








1





$begingroup$


Let $bin(0,pi)$, and let $kinmathbb{Z}$.



Find a function $f:mathbb{R}to[0,1]$ with:




  1. The set of zeros $f_0={x:x=2kpipm b}$

  2. The set of ones $f_1={x:x=2kpi}$

  3. The set of negative ones $f_{-1}={x:x=(2k+1)pi}$


In other words, what I'm looking for is some kind of wave function with the zeros adjusted to be closer to one peak or the other, without changing the amplitude. Here's a visual example:



enter image description here



In this case, $b$ would be somewhere around $2$ or so.



Can you help me figure out what this function $f$ is?






calculus functions trigonometry roots







share|cite|improve this question









$endgroup$




Let $bin(0,pi)$, and let $kinmathbb{Z}$.



Find a function $f:mathbb{R}to[0,1]$ with:




  1. The set of zeros $f_0={x:x=2kpipm b}$

  2. The set of ones $f_1={x:x=2kpi}$

  3. The set of negative ones $f_{-1}={x:x=(2k+1)pi}$


In other words, what I'm looking for is some kind of wave function with the zeros adjusted to be closer to one peak or the other, without changing the amplitude. Here's a visual example:



enter image description here



In this case, $b$ would be somewhere around $2$ or so.



Can you help me figure out what this function $f$ is?






calculus functions trigonometry roots




calculus functions trigonometry roots






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 16 '18 at 6:12









jippyjoe4jippyjoe4

49518




49518












  • $begingroup$
    Look at $cos(x+rsin(x))$ and adjust the $r$. However, if the absolute value of $r$ exceeds $1$, one loses some of your criteria.
    $endgroup$
    – Torsten Schoeneberg
    Dec 16 '18 at 6:30










  • $begingroup$
    @TorstenSchoeneberg, that's definitely helpful, but it also doesn't give me the full range of control I'm looking for.
    $endgroup$
    – jippyjoe4
    Dec 16 '18 at 6:32


















  • $begingroup$
    Look at $cos(x+rsin(x))$ and adjust the $r$. However, if the absolute value of $r$ exceeds $1$, one loses some of your criteria.
    $endgroup$
    – Torsten Schoeneberg
    Dec 16 '18 at 6:30










  • $begingroup$
    @TorstenSchoeneberg, that's definitely helpful, but it also doesn't give me the full range of control I'm looking for.
    $endgroup$
    – jippyjoe4
    Dec 16 '18 at 6:32
















$begingroup$
Look at $cos(x+rsin(x))$ and adjust the $r$. However, if the absolute value of $r$ exceeds $1$, one loses some of your criteria.
$endgroup$
– Torsten Schoeneberg
Dec 16 '18 at 6:30




$begingroup$
Look at $cos(x+rsin(x))$ and adjust the $r$. However, if the absolute value of $r$ exceeds $1$, one loses some of your criteria.
$endgroup$
– Torsten Schoeneberg
Dec 16 '18 at 6:30












$begingroup$
@TorstenSchoeneberg, that's definitely helpful, but it also doesn't give me the full range of control I'm looking for.
$endgroup$
– jippyjoe4
Dec 16 '18 at 6:32




$begingroup$
@TorstenSchoeneberg, that's definitely helpful, but it also doesn't give me the full range of control I'm looking for.
$endgroup$
– jippyjoe4
Dec 16 '18 at 6:32










2 Answers
2






active

oldest

votes


















1












$begingroup$

On the interval $[-pi, pi]$ you get such a wave by putting
$$f(x) = cos(pi cdot (|x|/pi)^{log_{b/pi}0.5}).$$



Now copy the period $[-pi,pi]$ by a piecewise definition, which one could probably hide by using the floor/remainder functions or something like that.



Note however that for $b<pi/2$, this function is not differentiable at $0$. To remedy that, one might use one of the infinitely differentiable bump functions inside there, instead of my crude powers of an absolute value, to go smoothly from $0$ to $1$ and crossing the value $1/2$ exactly at $b$.



Update: With the mollifier functions



$$h_r(x) = e^r cdot expleft(dfrac{-r}{1-(frac{x}{pi}-1)^2}right)$$



the functions



$$f_r(x) = cos(picdot h_r(|x|)$$



do the job smoothly, where you just have to compute the right $rin (0, infty)$ to match your $b$. Actually, since you want $h_r(b) = frac12$, you must set



$$r := dfrac{-log(2)cdot(frac{b^2}{pi^2}-frac{2b}{pi})}{1+frac{b^2}{pi^2}-frac{2b}{pi}}$$






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    Following Torsten's idea of using $cos(x+rsin x)$, you can get around the limitation $|r|leq1$ by iterating the same procedure again in the argument of sine, as in
    $$
    cos(x+rsin(x+sin x)),
    qquad
    cos(x+rsin(x+sin(x+sin x))),
    $$
    et cetera. Here is an image of $cos(x+sin(x+sin(x+sin(x+sin(x+sin x)))))$.



    enter image description here






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      On the interval $[-pi, pi]$ you get such a wave by putting
      $$f(x) = cos(pi cdot (|x|/pi)^{log_{b/pi}0.5}).$$



      Now copy the period $[-pi,pi]$ by a piecewise definition, which one could probably hide by using the floor/remainder functions or something like that.



      Note however that for $b<pi/2$, this function is not differentiable at $0$. To remedy that, one might use one of the infinitely differentiable bump functions inside there, instead of my crude powers of an absolute value, to go smoothly from $0$ to $1$ and crossing the value $1/2$ exactly at $b$.



      Update: With the mollifier functions



      $$h_r(x) = e^r cdot expleft(dfrac{-r}{1-(frac{x}{pi}-1)^2}right)$$



      the functions



      $$f_r(x) = cos(picdot h_r(|x|)$$



      do the job smoothly, where you just have to compute the right $rin (0, infty)$ to match your $b$. Actually, since you want $h_r(b) = frac12$, you must set



      $$r := dfrac{-log(2)cdot(frac{b^2}{pi^2}-frac{2b}{pi})}{1+frac{b^2}{pi^2}-frac{2b}{pi}}$$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        On the interval $[-pi, pi]$ you get such a wave by putting
        $$f(x) = cos(pi cdot (|x|/pi)^{log_{b/pi}0.5}).$$



        Now copy the period $[-pi,pi]$ by a piecewise definition, which one could probably hide by using the floor/remainder functions or something like that.



        Note however that for $b<pi/2$, this function is not differentiable at $0$. To remedy that, one might use one of the infinitely differentiable bump functions inside there, instead of my crude powers of an absolute value, to go smoothly from $0$ to $1$ and crossing the value $1/2$ exactly at $b$.



        Update: With the mollifier functions



        $$h_r(x) = e^r cdot expleft(dfrac{-r}{1-(frac{x}{pi}-1)^2}right)$$



        the functions



        $$f_r(x) = cos(picdot h_r(|x|)$$



        do the job smoothly, where you just have to compute the right $rin (0, infty)$ to match your $b$. Actually, since you want $h_r(b) = frac12$, you must set



        $$r := dfrac{-log(2)cdot(frac{b^2}{pi^2}-frac{2b}{pi})}{1+frac{b^2}{pi^2}-frac{2b}{pi}}$$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          On the interval $[-pi, pi]$ you get such a wave by putting
          $$f(x) = cos(pi cdot (|x|/pi)^{log_{b/pi}0.5}).$$



          Now copy the period $[-pi,pi]$ by a piecewise definition, which one could probably hide by using the floor/remainder functions or something like that.



          Note however that for $b<pi/2$, this function is not differentiable at $0$. To remedy that, one might use one of the infinitely differentiable bump functions inside there, instead of my crude powers of an absolute value, to go smoothly from $0$ to $1$ and crossing the value $1/2$ exactly at $b$.



          Update: With the mollifier functions



          $$h_r(x) = e^r cdot expleft(dfrac{-r}{1-(frac{x}{pi}-1)^2}right)$$



          the functions



          $$f_r(x) = cos(picdot h_r(|x|)$$



          do the job smoothly, where you just have to compute the right $rin (0, infty)$ to match your $b$. Actually, since you want $h_r(b) = frac12$, you must set



          $$r := dfrac{-log(2)cdot(frac{b^2}{pi^2}-frac{2b}{pi})}{1+frac{b^2}{pi^2}-frac{2b}{pi}}$$






          share|cite|improve this answer











          $endgroup$



          On the interval $[-pi, pi]$ you get such a wave by putting
          $$f(x) = cos(pi cdot (|x|/pi)^{log_{b/pi}0.5}).$$



          Now copy the period $[-pi,pi]$ by a piecewise definition, which one could probably hide by using the floor/remainder functions or something like that.



          Note however that for $b<pi/2$, this function is not differentiable at $0$. To remedy that, one might use one of the infinitely differentiable bump functions inside there, instead of my crude powers of an absolute value, to go smoothly from $0$ to $1$ and crossing the value $1/2$ exactly at $b$.



          Update: With the mollifier functions



          $$h_r(x) = e^r cdot expleft(dfrac{-r}{1-(frac{x}{pi}-1)^2}right)$$



          the functions



          $$f_r(x) = cos(picdot h_r(|x|)$$



          do the job smoothly, where you just have to compute the right $rin (0, infty)$ to match your $b$. Actually, since you want $h_r(b) = frac12$, you must set



          $$r := dfrac{-log(2)cdot(frac{b^2}{pi^2}-frac{2b}{pi})}{1+frac{b^2}{pi^2}-frac{2b}{pi}}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 18:09

























          answered Dec 16 '18 at 23:10









          Torsten SchoenebergTorsten Schoeneberg

          4,3212833




          4,3212833























              0












              $begingroup$

              Following Torsten's idea of using $cos(x+rsin x)$, you can get around the limitation $|r|leq1$ by iterating the same procedure again in the argument of sine, as in
              $$
              cos(x+rsin(x+sin x)),
              qquad
              cos(x+rsin(x+sin(x+sin x))),
              $$
              et cetera. Here is an image of $cos(x+sin(x+sin(x+sin(x+sin(x+sin x)))))$.



              enter image description here






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Following Torsten's idea of using $cos(x+rsin x)$, you can get around the limitation $|r|leq1$ by iterating the same procedure again in the argument of sine, as in
                $$
                cos(x+rsin(x+sin x)),
                qquad
                cos(x+rsin(x+sin(x+sin x))),
                $$
                et cetera. Here is an image of $cos(x+sin(x+sin(x+sin(x+sin(x+sin x)))))$.



                enter image description here






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Following Torsten's idea of using $cos(x+rsin x)$, you can get around the limitation $|r|leq1$ by iterating the same procedure again in the argument of sine, as in
                  $$
                  cos(x+rsin(x+sin x)),
                  qquad
                  cos(x+rsin(x+sin(x+sin x))),
                  $$
                  et cetera. Here is an image of $cos(x+sin(x+sin(x+sin(x+sin(x+sin x)))))$.



                  enter image description here






                  share|cite|improve this answer









                  $endgroup$



                  Following Torsten's idea of using $cos(x+rsin x)$, you can get around the limitation $|r|leq1$ by iterating the same procedure again in the argument of sine, as in
                  $$
                  cos(x+rsin(x+sin x)),
                  qquad
                  cos(x+rsin(x+sin(x+sin x))),
                  $$
                  et cetera. Here is an image of $cos(x+sin(x+sin(x+sin(x+sin(x+sin x)))))$.



                  enter image description here







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 17 '18 at 18:22









                  timurtimur

                  12.2k2144




                  12.2k2144






























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