Plot of $arctanleft(frac{x-1}{x+1}right) - arctanleft(xright)$












0












$begingroup$


So this gives a strange "step" plot, kind of like a signum plot. I came across a formula for the substraction of atans, however i would like to know whether there is a simpler way of achieving this plot.






trigonometry graphing-functions







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  • $begingroup$
    math.stackexchange.com/questions/1837410/…
    $endgroup$
    – lab bhattacharjee
    Jan 13 at 17:27
















0












$begingroup$


So this gives a strange "step" plot, kind of like a signum plot. I came across a formula for the substraction of atans, however i would like to know whether there is a simpler way of achieving this plot.






trigonometry graphing-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    math.stackexchange.com/questions/1837410/…
    $endgroup$
    – lab bhattacharjee
    Jan 13 at 17:27














0












0








0





$begingroup$


So this gives a strange "step" plot, kind of like a signum plot. I came across a formula for the substraction of atans, however i would like to know whether there is a simpler way of achieving this plot.






trigonometry graphing-functions







share|cite|improve this question











$endgroup$




So this gives a strange "step" plot, kind of like a signum plot. I came across a formula for the substraction of atans, however i would like to know whether there is a simpler way of achieving this plot.






trigonometry graphing-functions




trigonometry graphing-functions






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share|cite|improve this question








edited Jan 13 at 15:04









jameselmore

4,39432035




4,39432035










asked Jan 13 at 14:51









Jan WójcickiJan Wójcicki

1




1












  • $begingroup$
    math.stackexchange.com/questions/1837410/…
    $endgroup$
    – lab bhattacharjee
    Jan 13 at 17:27


















  • $begingroup$
    math.stackexchange.com/questions/1837410/…
    $endgroup$
    – lab bhattacharjee
    Jan 13 at 17:27
















$begingroup$
math.stackexchange.com/questions/1837410/…
$endgroup$
– lab bhattacharjee
Jan 13 at 17:27




$begingroup$
math.stackexchange.com/questions/1837410/…
$endgroup$
– lab bhattacharjee
Jan 13 at 17:27










3 Answers
3






active

oldest

votes


















1












$begingroup$

Note that $$arctan(x)-arctan(y)=arctan(frac{x-y}{1+xy})$$ if $$xy>-1$$
or $$pi+arctan(frac{x-y}{1+xy})$$ for $$x>0$$ and $$xy<-1$$
or$$pi-arctan(frac{x-y}{1+xy})$$ for $$x<0$$ and $$xy<-1$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    go to
    https://develop.open.wolframcloud.com/app/ -> "create a new notebook" -> write:



    Plot[ArcTan[(x-1)/(x+1)]-ArcTan[x],{x,a,b}].



    where $ a,b $ are the upper and lower bounds of the plot.



    the plot will be: $$
    left{
    begin{array}{c}y=frac{3}{4}pi for x<-1 \
    y= -frac{pi}{4} for x>-1
    end{array}
    right.
    $$






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      You might want to see the equivalent expression of the expression $arctan(x)+arctan(y)$. You can arrive at this expression using the known formula for $tan(A+B)$. In case you're unfamiliar with it, you might wanna recall $sin(A+B)$ and $cos(A+B)$ ( Click Here to read beep-boops' answer to Help me prove: sin(A+B) = sinA cosB + cosA sinB ) and the definition of $tan(varphi)$ as $dfrac{sin(varphi)}{cos(varphi)}$.



      First of all let us simplify and clean up things a bit. Let us make the following substitutions:



      $$begin{pmatrix} arctan(x) \ arctan(y)end{pmatrix}=begin{pmatrix} alpha \
      beta end{pmatrix} tag1$$



      Now, $$tan(alpha pm beta)=dfrac{tan(alpha)pmtan(beta)}{1mptan(alpha)tan(beta)} tag2$$ Replace $alpha$ and $beta$ with their equivalents in equation $(2)$ as in $(1)$ and taking the arctangent on both sides of equation $(2)$ to obtain: $$arctan(x)pmarctan(y)=arctan Biggl(dfrac{xpm y}{1mp xy}Biggr) +npitag3$$



      With the function the OP has stated. We get $f(x)=arctan Biggl(dfrac{frac{x-1}{x+1}-x}{1+xfrac{x-1}{x+1}} Biggr)$ which simplifies to $f(x)=-arctan(1)$ and gives you the plot defined in a piecewise fashion as: $$f(x)=begin{cases}
      frac{3pi}{4} & x le -1 \
      -frac{pi}{4} & x ge -1 \
      end{cases}
      $$






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        You might want to add $npi$ on the right-hand side of equation $(3)$, especially since it turns out $f(x)$ is $-arctan(1) + pi$ rather than $-arctan(1)$ on part of its domain.
        $endgroup$
        – David K
        Jan 13 at 18:23











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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      Note that $$arctan(x)-arctan(y)=arctan(frac{x-y}{1+xy})$$ if $$xy>-1$$
      or $$pi+arctan(frac{x-y}{1+xy})$$ for $$x>0$$ and $$xy<-1$$
      or$$pi-arctan(frac{x-y}{1+xy})$$ for $$x<0$$ and $$xy<-1$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Note that $$arctan(x)-arctan(y)=arctan(frac{x-y}{1+xy})$$ if $$xy>-1$$
        or $$pi+arctan(frac{x-y}{1+xy})$$ for $$x>0$$ and $$xy<-1$$
        or$$pi-arctan(frac{x-y}{1+xy})$$ for $$x<0$$ and $$xy<-1$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Note that $$arctan(x)-arctan(y)=arctan(frac{x-y}{1+xy})$$ if $$xy>-1$$
          or $$pi+arctan(frac{x-y}{1+xy})$$ for $$x>0$$ and $$xy<-1$$
          or$$pi-arctan(frac{x-y}{1+xy})$$ for $$x<0$$ and $$xy<-1$$






          share|cite|improve this answer









          $endgroup$



          Note that $$arctan(x)-arctan(y)=arctan(frac{x-y}{1+xy})$$ if $$xy>-1$$
          or $$pi+arctan(frac{x-y}{1+xy})$$ for $$x>0$$ and $$xy<-1$$
          or$$pi-arctan(frac{x-y}{1+xy})$$ for $$x<0$$ and $$xy<-1$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 13 at 15:02









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          74.6k42865




          74.6k42865























              0












              $begingroup$

              go to
              https://develop.open.wolframcloud.com/app/ -> "create a new notebook" -> write:



              Plot[ArcTan[(x-1)/(x+1)]-ArcTan[x],{x,a,b}].



              where $ a,b $ are the upper and lower bounds of the plot.



              the plot will be: $$
              left{
              begin{array}{c}y=frac{3}{4}pi for x<-1 \
              y= -frac{pi}{4} for x>-1
              end{array}
              right.
              $$






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                go to
                https://develop.open.wolframcloud.com/app/ -> "create a new notebook" -> write:



                Plot[ArcTan[(x-1)/(x+1)]-ArcTan[x],{x,a,b}].



                where $ a,b $ are the upper and lower bounds of the plot.



                the plot will be: $$
                left{
                begin{array}{c}y=frac{3}{4}pi for x<-1 \
                y= -frac{pi}{4} for x>-1
                end{array}
                right.
                $$






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  go to
                  https://develop.open.wolframcloud.com/app/ -> "create a new notebook" -> write:



                  Plot[ArcTan[(x-1)/(x+1)]-ArcTan[x],{x,a,b}].



                  where $ a,b $ are the upper and lower bounds of the plot.



                  the plot will be: $$
                  left{
                  begin{array}{c}y=frac{3}{4}pi for x<-1 \
                  y= -frac{pi}{4} for x>-1
                  end{array}
                  right.
                  $$






                  share|cite|improve this answer











                  $endgroup$



                  go to
                  https://develop.open.wolframcloud.com/app/ -> "create a new notebook" -> write:



                  Plot[ArcTan[(x-1)/(x+1)]-ArcTan[x],{x,a,b}].



                  where $ a,b $ are the upper and lower bounds of the plot.



                  the plot will be: $$
                  left{
                  begin{array}{c}y=frac{3}{4}pi for x<-1 \
                  y= -frac{pi}{4} for x>-1
                  end{array}
                  right.
                  $$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 13 at 15:24

























                  answered Jan 13 at 15:14









                  Patrick DanziPatrick Danzi

                  439113




                  439113























                      0












                      $begingroup$

                      You might want to see the equivalent expression of the expression $arctan(x)+arctan(y)$. You can arrive at this expression using the known formula for $tan(A+B)$. In case you're unfamiliar with it, you might wanna recall $sin(A+B)$ and $cos(A+B)$ ( Click Here to read beep-boops' answer to Help me prove: sin(A+B) = sinA cosB + cosA sinB ) and the definition of $tan(varphi)$ as $dfrac{sin(varphi)}{cos(varphi)}$.



                      First of all let us simplify and clean up things a bit. Let us make the following substitutions:



                      $$begin{pmatrix} arctan(x) \ arctan(y)end{pmatrix}=begin{pmatrix} alpha \
                      beta end{pmatrix} tag1$$



                      Now, $$tan(alpha pm beta)=dfrac{tan(alpha)pmtan(beta)}{1mptan(alpha)tan(beta)} tag2$$ Replace $alpha$ and $beta$ with their equivalents in equation $(2)$ as in $(1)$ and taking the arctangent on both sides of equation $(2)$ to obtain: $$arctan(x)pmarctan(y)=arctan Biggl(dfrac{xpm y}{1mp xy}Biggr) +npitag3$$



                      With the function the OP has stated. We get $f(x)=arctan Biggl(dfrac{frac{x-1}{x+1}-x}{1+xfrac{x-1}{x+1}} Biggr)$ which simplifies to $f(x)=-arctan(1)$ and gives you the plot defined in a piecewise fashion as: $$f(x)=begin{cases}
                      frac{3pi}{4} & x le -1 \
                      -frac{pi}{4} & x ge -1 \
                      end{cases}
                      $$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        You might want to add $npi$ on the right-hand side of equation $(3)$, especially since it turns out $f(x)$ is $-arctan(1) + pi$ rather than $-arctan(1)$ on part of its domain.
                        $endgroup$
                        – David K
                        Jan 13 at 18:23
















                      0












                      $begingroup$

                      You might want to see the equivalent expression of the expression $arctan(x)+arctan(y)$. You can arrive at this expression using the known formula for $tan(A+B)$. In case you're unfamiliar with it, you might wanna recall $sin(A+B)$ and $cos(A+B)$ ( Click Here to read beep-boops' answer to Help me prove: sin(A+B) = sinA cosB + cosA sinB ) and the definition of $tan(varphi)$ as $dfrac{sin(varphi)}{cos(varphi)}$.



                      First of all let us simplify and clean up things a bit. Let us make the following substitutions:



                      $$begin{pmatrix} arctan(x) \ arctan(y)end{pmatrix}=begin{pmatrix} alpha \
                      beta end{pmatrix} tag1$$



                      Now, $$tan(alpha pm beta)=dfrac{tan(alpha)pmtan(beta)}{1mptan(alpha)tan(beta)} tag2$$ Replace $alpha$ and $beta$ with their equivalents in equation $(2)$ as in $(1)$ and taking the arctangent on both sides of equation $(2)$ to obtain: $$arctan(x)pmarctan(y)=arctan Biggl(dfrac{xpm y}{1mp xy}Biggr) +npitag3$$



                      With the function the OP has stated. We get $f(x)=arctan Biggl(dfrac{frac{x-1}{x+1}-x}{1+xfrac{x-1}{x+1}} Biggr)$ which simplifies to $f(x)=-arctan(1)$ and gives you the plot defined in a piecewise fashion as: $$f(x)=begin{cases}
                      frac{3pi}{4} & x le -1 \
                      -frac{pi}{4} & x ge -1 \
                      end{cases}
                      $$






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        You might want to add $npi$ on the right-hand side of equation $(3)$, especially since it turns out $f(x)$ is $-arctan(1) + pi$ rather than $-arctan(1)$ on part of its domain.
                        $endgroup$
                        – David K
                        Jan 13 at 18:23














                      0












                      0








                      0





                      $begingroup$

                      You might want to see the equivalent expression of the expression $arctan(x)+arctan(y)$. You can arrive at this expression using the known formula for $tan(A+B)$. In case you're unfamiliar with it, you might wanna recall $sin(A+B)$ and $cos(A+B)$ ( Click Here to read beep-boops' answer to Help me prove: sin(A+B) = sinA cosB + cosA sinB ) and the definition of $tan(varphi)$ as $dfrac{sin(varphi)}{cos(varphi)}$.



                      First of all let us simplify and clean up things a bit. Let us make the following substitutions:



                      $$begin{pmatrix} arctan(x) \ arctan(y)end{pmatrix}=begin{pmatrix} alpha \
                      beta end{pmatrix} tag1$$



                      Now, $$tan(alpha pm beta)=dfrac{tan(alpha)pmtan(beta)}{1mptan(alpha)tan(beta)} tag2$$ Replace $alpha$ and $beta$ with their equivalents in equation $(2)$ as in $(1)$ and taking the arctangent on both sides of equation $(2)$ to obtain: $$arctan(x)pmarctan(y)=arctan Biggl(dfrac{xpm y}{1mp xy}Biggr) +npitag3$$



                      With the function the OP has stated. We get $f(x)=arctan Biggl(dfrac{frac{x-1}{x+1}-x}{1+xfrac{x-1}{x+1}} Biggr)$ which simplifies to $f(x)=-arctan(1)$ and gives you the plot defined in a piecewise fashion as: $$f(x)=begin{cases}
                      frac{3pi}{4} & x le -1 \
                      -frac{pi}{4} & x ge -1 \
                      end{cases}
                      $$






                      share|cite|improve this answer











                      $endgroup$



                      You might want to see the equivalent expression of the expression $arctan(x)+arctan(y)$. You can arrive at this expression using the known formula for $tan(A+B)$. In case you're unfamiliar with it, you might wanna recall $sin(A+B)$ and $cos(A+B)$ ( Click Here to read beep-boops' answer to Help me prove: sin(A+B) = sinA cosB + cosA sinB ) and the definition of $tan(varphi)$ as $dfrac{sin(varphi)}{cos(varphi)}$.



                      First of all let us simplify and clean up things a bit. Let us make the following substitutions:



                      $$begin{pmatrix} arctan(x) \ arctan(y)end{pmatrix}=begin{pmatrix} alpha \
                      beta end{pmatrix} tag1$$



                      Now, $$tan(alpha pm beta)=dfrac{tan(alpha)pmtan(beta)}{1mptan(alpha)tan(beta)} tag2$$ Replace $alpha$ and $beta$ with their equivalents in equation $(2)$ as in $(1)$ and taking the arctangent on both sides of equation $(2)$ to obtain: $$arctan(x)pmarctan(y)=arctan Biggl(dfrac{xpm y}{1mp xy}Biggr) +npitag3$$



                      With the function the OP has stated. We get $f(x)=arctan Biggl(dfrac{frac{x-1}{x+1}-x}{1+xfrac{x-1}{x+1}} Biggr)$ which simplifies to $f(x)=-arctan(1)$ and gives you the plot defined in a piecewise fashion as: $$f(x)=begin{cases}
                      frac{3pi}{4} & x le -1 \
                      -frac{pi}{4} & x ge -1 \
                      end{cases}
                      $$







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 13 at 18:26

























                      answered Jan 13 at 17:35









                      Paras KhoslaParas Khosla

                      13711




                      13711












                      • $begingroup$
                        You might want to add $npi$ on the right-hand side of equation $(3)$, especially since it turns out $f(x)$ is $-arctan(1) + pi$ rather than $-arctan(1)$ on part of its domain.
                        $endgroup$
                        – David K
                        Jan 13 at 18:23


















                      • $begingroup$
                        You might want to add $npi$ on the right-hand side of equation $(3)$, especially since it turns out $f(x)$ is $-arctan(1) + pi$ rather than $-arctan(1)$ on part of its domain.
                        $endgroup$
                        – David K
                        Jan 13 at 18:23
















                      $begingroup$
                      You might want to add $npi$ on the right-hand side of equation $(3)$, especially since it turns out $f(x)$ is $-arctan(1) + pi$ rather than $-arctan(1)$ on part of its domain.
                      $endgroup$
                      – David K
                      Jan 13 at 18:23




                      $begingroup$
                      You might want to add $npi$ on the right-hand side of equation $(3)$, especially since it turns out $f(x)$ is $-arctan(1) + pi$ rather than $-arctan(1)$ on part of its domain.
                      $endgroup$
                      – David K
                      Jan 13 at 18:23


















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