How to plot “chainsaw” functions in Maple?












1












$begingroup$


I want to plot piecewise function. For example,



begin{cases}
0, & 0le t<t_1 \
12e^{-(t-t_1)}, & t_1le t<t_2
end{cases}



The problem is that function is periodic (let $t_1$=4 and $t_2$=6, period $6$). For one period, it's pretty easy:



f := 12*exp(-t+4);

g := 0;

piece := piecewise(0 <= t and t < 4, g, 4 < t and t <= 6, f);

plot(piece, t = 0 .. 6);


enter image description here



But I don't know how to make it periodic. Of course, I can add some new functions in piecewise, but I'm sure that more adequate way to plot periodic fuctions exists. Any help appreciated!






maple







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Consider $frac6piarctan(tan(fracpi6x+fracpi2))+3$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 9 '18 at 10:44


















1












$begingroup$


I want to plot piecewise function. For example,



begin{cases}
0, & 0le t<t_1 \
12e^{-(t-t_1)}, & t_1le t<t_2
end{cases}



The problem is that function is periodic (let $t_1$=4 and $t_2$=6, period $6$). For one period, it's pretty easy:



f := 12*exp(-t+4);

g := 0;

piece := piecewise(0 <= t and t < 4, g, 4 < t and t <= 6, f);

plot(piece, t = 0 .. 6);


enter image description here



But I don't know how to make it periodic. Of course, I can add some new functions in piecewise, but I'm sure that more adequate way to plot periodic fuctions exists. Any help appreciated!






maple







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Consider $frac6piarctan(tan(fracpi6x+fracpi2))+3$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 9 '18 at 10:44
















1












1








1





$begingroup$


I want to plot piecewise function. For example,



begin{cases}
0, & 0le t<t_1 \
12e^{-(t-t_1)}, & t_1le t<t_2
end{cases}



The problem is that function is periodic (let $t_1$=4 and $t_2$=6, period $6$). For one period, it's pretty easy:



f := 12*exp(-t+4);

g := 0;

piece := piecewise(0 <= t and t < 4, g, 4 < t and t <= 6, f);

plot(piece, t = 0 .. 6);


enter image description here



But I don't know how to make it periodic. Of course, I can add some new functions in piecewise, but I'm sure that more adequate way to plot periodic fuctions exists. Any help appreciated!






maple







share|cite|improve this question











$endgroup$




I want to plot piecewise function. For example,



begin{cases}
0, & 0le t<t_1 \
12e^{-(t-t_1)}, & t_1le t<t_2
end{cases}



The problem is that function is periodic (let $t_1$=4 and $t_2$=6, period $6$). For one period, it's pretty easy:



f := 12*exp(-t+4);

g := 0;

piece := piecewise(0 <= t and t < 4, g, 4 < t and t <= 6, f);

plot(piece, t = 0 .. 6);


enter image description here



But I don't know how to make it periodic. Of course, I can add some new functions in piecewise, but I'm sure that more adequate way to plot periodic fuctions exists. Any help appreciated!






maple




maple






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 10:49









Jean-Claude Arbaut

14.7k63464




14.7k63464










asked Dec 9 '18 at 10:34









Kelly ShepphardKelly Shepphard

2298




2298








  • 1




    $begingroup$
    Consider $frac6piarctan(tan(fracpi6x+fracpi2))+3$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 9 '18 at 10:44
















  • 1




    $begingroup$
    Consider $frac6piarctan(tan(fracpi6x+fracpi2))+3$.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 9 '18 at 10:44










1




1




$begingroup$
Consider $frac6piarctan(tan(fracpi6x+fracpi2))+3$.
$endgroup$
– Jean-Claude Arbaut
Dec 9 '18 at 10:44






$begingroup$
Consider $frac6piarctan(tan(fracpi6x+fracpi2))+3$.
$endgroup$
– Jean-Claude Arbaut
Dec 9 '18 at 10:44












2 Answers
2






active

oldest

votes


















1












$begingroup$

You should not have to manually copy any part of the original expression into a new operator (manually, eg, cut & paste), to accomplish this. The unapply command is useful for that kind of thing.



You started out by giving us this:



restart;

f := 12*exp(-t+4):

g := 0:

piece := piecewise(0 <= t and t < 4, g, 4 < t and t <= 6, f):



plot(piece, t = 0 .. 6, size=[200,200]);


enter image description here



Now let's construct an operator from that, which behaves like the supplied piecewise, with a period of our choice.



We'll use a re-usable constructor for this purpose.



makeperiodic := proc(expr, var, skip)

  local T, r;

  r := skip/2;

  unapply( 'eval'(expr, var=r+'frem'(T+r,skip)), [T], numeric);

end proc:


Here is the construction of the periodic operator, and quick check.



foo := makeperiodic( piece, t, 6 ):



foo(5);

                      4.414553294



foo(11);

                      4.414553294


This operator returns unevaluated when its argument is not numeric, by design.



foo(x);

                         foo(x)


Now for some plots,



# operator form

plot(foo, -12 .. 12, size=[600,200]);


enter image description here



# expression form (unevaluated function call)

plot(foo(x), x=-12 .. 12, size=[600,200]);


enter image description here



# shift two to the left

plot(foo(x+2), x=-12 .. 12, size=[600,200]);


enter image description here



And we could do a similar thing for some other choice of period,



bar := makeperiodic( piece, t, 5 ):

plot(bar, -10 .. 10, size=[500,200]);

plot(bar(t), t=-10 .. 10, size=[500,200]);

# shift 4 to the right

plot(bar(t-4), t=-10 .. 10, size=[500,200]);





share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for new detailed approach to solving this task!
    $endgroup$
    – Kelly Shepphard
    Dec 10 '18 at 9:52



















2












$begingroup$

You may try:



 f:=t->12*exp(-t+4)*Heaviside((t-4)*(6-t)):

 s:=t->t-floor(t):

 plot(f(6*s(t/6)),t=0..25);


It is easy to modify it with other values of $t_1$ and $t_2$.



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The function looks periodic, but it's not.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 9 '18 at 10:51






  • 1




    $begingroup$
    @Jean-ClaudeArbaut I see your point and I edited my answer. Do you like it now?
    $endgroup$
    – Robert Z
    Dec 9 '18 at 11:27






  • 1




    $begingroup$
    @KellyShepphard Yes we can generalize to more pieces. For each piece you use a Heaviside step function and then you add together all the pieces.
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:12








  • 1




    $begingroup$
    if you have two pieces $f1$ on $[a,b]$ and $f2$ on $[c,d]$ then f=f1*Heaviside((x-a)(b-x))+f2*Heaviside((x-c)(d-x)). Is it clear?
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:22








  • 1




    $begingroup$
    Yes, that is the period.
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:28











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2 Answers
2






active

oldest

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You should not have to manually copy any part of the original expression into a new operator (manually, eg, cut & paste), to accomplish this. The unapply command is useful for that kind of thing.



You started out by giving us this:



restart;

f := 12*exp(-t+4):

g := 0:

piece := piecewise(0 <= t and t < 4, g, 4 < t and t <= 6, f):



plot(piece, t = 0 .. 6, size=[200,200]);


enter image description here



Now let's construct an operator from that, which behaves like the supplied piecewise, with a period of our choice.



We'll use a re-usable constructor for this purpose.



makeperiodic := proc(expr, var, skip)

  local T, r;

  r := skip/2;

  unapply( 'eval'(expr, var=r+'frem'(T+r,skip)), [T], numeric);

end proc:


Here is the construction of the periodic operator, and quick check.



foo := makeperiodic( piece, t, 6 ):



foo(5);

                      4.414553294



foo(11);

                      4.414553294


This operator returns unevaluated when its argument is not numeric, by design.



foo(x);

                         foo(x)


Now for some plots,



# operator form

plot(foo, -12 .. 12, size=[600,200]);


enter image description here



# expression form (unevaluated function call)

plot(foo(x), x=-12 .. 12, size=[600,200]);


enter image description here



# shift two to the left

plot(foo(x+2), x=-12 .. 12, size=[600,200]);


enter image description here



And we could do a similar thing for some other choice of period,



bar := makeperiodic( piece, t, 5 ):

plot(bar, -10 .. 10, size=[500,200]);

plot(bar(t), t=-10 .. 10, size=[500,200]);

# shift 4 to the right

plot(bar(t-4), t=-10 .. 10, size=[500,200]);





share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for new detailed approach to solving this task!
    $endgroup$
    – Kelly Shepphard
    Dec 10 '18 at 9:52
















1












$begingroup$

You should not have to manually copy any part of the original expression into a new operator (manually, eg, cut & paste), to accomplish this. The unapply command is useful for that kind of thing.



You started out by giving us this:



restart;

f := 12*exp(-t+4):

g := 0:

piece := piecewise(0 <= t and t < 4, g, 4 < t and t <= 6, f):



plot(piece, t = 0 .. 6, size=[200,200]);


enter image description here



Now let's construct an operator from that, which behaves like the supplied piecewise, with a period of our choice.



We'll use a re-usable constructor for this purpose.



makeperiodic := proc(expr, var, skip)

  local T, r;

  r := skip/2;

  unapply( 'eval'(expr, var=r+'frem'(T+r,skip)), [T], numeric);

end proc:


Here is the construction of the periodic operator, and quick check.



foo := makeperiodic( piece, t, 6 ):



foo(5);

                      4.414553294



foo(11);

                      4.414553294


This operator returns unevaluated when its argument is not numeric, by design.



foo(x);

                         foo(x)


Now for some plots,



# operator form

plot(foo, -12 .. 12, size=[600,200]);


enter image description here



# expression form (unevaluated function call)

plot(foo(x), x=-12 .. 12, size=[600,200]);


enter image description here



# shift two to the left

plot(foo(x+2), x=-12 .. 12, size=[600,200]);


enter image description here



And we could do a similar thing for some other choice of period,



bar := makeperiodic( piece, t, 5 ):

plot(bar, -10 .. 10, size=[500,200]);

plot(bar(t), t=-10 .. 10, size=[500,200]);

# shift 4 to the right

plot(bar(t-4), t=-10 .. 10, size=[500,200]);





share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much for new detailed approach to solving this task!
    $endgroup$
    – Kelly Shepphard
    Dec 10 '18 at 9:52














1












1








1





$begingroup$

You should not have to manually copy any part of the original expression into a new operator (manually, eg, cut & paste), to accomplish this. The unapply command is useful for that kind of thing.



You started out by giving us this:



restart;

f := 12*exp(-t+4):

g := 0:

piece := piecewise(0 <= t and t < 4, g, 4 < t and t <= 6, f):



plot(piece, t = 0 .. 6, size=[200,200]);


enter image description here



Now let's construct an operator from that, which behaves like the supplied piecewise, with a period of our choice.



We'll use a re-usable constructor for this purpose.



makeperiodic := proc(expr, var, skip)

  local T, r;

  r := skip/2;

  unapply( 'eval'(expr, var=r+'frem'(T+r,skip)), [T], numeric);

end proc:


Here is the construction of the periodic operator, and quick check.



foo := makeperiodic( piece, t, 6 ):



foo(5);

                      4.414553294



foo(11);

                      4.414553294


This operator returns unevaluated when its argument is not numeric, by design.



foo(x);

                         foo(x)


Now for some plots,



# operator form

plot(foo, -12 .. 12, size=[600,200]);


enter image description here



# expression form (unevaluated function call)

plot(foo(x), x=-12 .. 12, size=[600,200]);


enter image description here



# shift two to the left

plot(foo(x+2), x=-12 .. 12, size=[600,200]);


enter image description here



And we could do a similar thing for some other choice of period,



bar := makeperiodic( piece, t, 5 ):

plot(bar, -10 .. 10, size=[500,200]);

plot(bar(t), t=-10 .. 10, size=[500,200]);

# shift 4 to the right

plot(bar(t-4), t=-10 .. 10, size=[500,200]);





share|cite|improve this answer









$endgroup$



You should not have to manually copy any part of the original expression into a new operator (manually, eg, cut & paste), to accomplish this. The unapply command is useful for that kind of thing.



You started out by giving us this:



restart;

f := 12*exp(-t+4):

g := 0:

piece := piecewise(0 <= t and t < 4, g, 4 < t and t <= 6, f):



plot(piece, t = 0 .. 6, size=[200,200]);


enter image description here



Now let's construct an operator from that, which behaves like the supplied piecewise, with a period of our choice.



We'll use a re-usable constructor for this purpose.



makeperiodic := proc(expr, var, skip)

  local T, r;

  r := skip/2;

  unapply( 'eval'(expr, var=r+'frem'(T+r,skip)), [T], numeric);

end proc:


Here is the construction of the periodic operator, and quick check.



foo := makeperiodic( piece, t, 6 ):



foo(5);

                      4.414553294



foo(11);

                      4.414553294


This operator returns unevaluated when its argument is not numeric, by design.



foo(x);

                         foo(x)


Now for some plots,



# operator form

plot(foo, -12 .. 12, size=[600,200]);


enter image description here



# expression form (unevaluated function call)

plot(foo(x), x=-12 .. 12, size=[600,200]);


enter image description here



# shift two to the left

plot(foo(x+2), x=-12 .. 12, size=[600,200]);


enter image description here



And we could do a similar thing for some other choice of period,



bar := makeperiodic( piece, t, 5 ):

plot(bar, -10 .. 10, size=[500,200]);

plot(bar(t), t=-10 .. 10, size=[500,200]);

# shift 4 to the right

plot(bar(t-4), t=-10 .. 10, size=[500,200]);






share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 3:37









aceracer

3,615199




3,615199












  • $begingroup$
    Thank you very much for new detailed approach to solving this task!
    $endgroup$
    – Kelly Shepphard
    Dec 10 '18 at 9:52


















  • $begingroup$
    Thank you very much for new detailed approach to solving this task!
    $endgroup$
    – Kelly Shepphard
    Dec 10 '18 at 9:52
















$begingroup$
Thank you very much for new detailed approach to solving this task!
$endgroup$
– Kelly Shepphard
Dec 10 '18 at 9:52




$begingroup$
Thank you very much for new detailed approach to solving this task!
$endgroup$
– Kelly Shepphard
Dec 10 '18 at 9:52











2












$begingroup$

You may try:



 f:=t->12*exp(-t+4)*Heaviside((t-4)*(6-t)):

 s:=t->t-floor(t):

 plot(f(6*s(t/6)),t=0..25);


It is easy to modify it with other values of $t_1$ and $t_2$.



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The function looks periodic, but it's not.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 9 '18 at 10:51






  • 1




    $begingroup$
    @Jean-ClaudeArbaut I see your point and I edited my answer. Do you like it now?
    $endgroup$
    – Robert Z
    Dec 9 '18 at 11:27






  • 1




    $begingroup$
    @KellyShepphard Yes we can generalize to more pieces. For each piece you use a Heaviside step function and then you add together all the pieces.
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:12








  • 1




    $begingroup$
    if you have two pieces $f1$ on $[a,b]$ and $f2$ on $[c,d]$ then f=f1*Heaviside((x-a)(b-x))+f2*Heaviside((x-c)(d-x)). Is it clear?
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:22








  • 1




    $begingroup$
    Yes, that is the period.
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:28
















2












$begingroup$

You may try:



 f:=t->12*exp(-t+4)*Heaviside((t-4)*(6-t)):

 s:=t->t-floor(t):

 plot(f(6*s(t/6)),t=0..25);


It is easy to modify it with other values of $t_1$ and $t_2$.



enter image description here






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The function looks periodic, but it's not.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 9 '18 at 10:51






  • 1




    $begingroup$
    @Jean-ClaudeArbaut I see your point and I edited my answer. Do you like it now?
    $endgroup$
    – Robert Z
    Dec 9 '18 at 11:27






  • 1




    $begingroup$
    @KellyShepphard Yes we can generalize to more pieces. For each piece you use a Heaviside step function and then you add together all the pieces.
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:12








  • 1




    $begingroup$
    if you have two pieces $f1$ on $[a,b]$ and $f2$ on $[c,d]$ then f=f1*Heaviside((x-a)(b-x))+f2*Heaviside((x-c)(d-x)). Is it clear?
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:22








  • 1




    $begingroup$
    Yes, that is the period.
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:28














2












2








2





$begingroup$

You may try:



 f:=t->12*exp(-t+4)*Heaviside((t-4)*(6-t)):

 s:=t->t-floor(t):

 plot(f(6*s(t/6)),t=0..25);


It is easy to modify it with other values of $t_1$ and $t_2$.



enter image description here






share|cite|improve this answer











$endgroup$



You may try:



 f:=t->12*exp(-t+4)*Heaviside((t-4)*(6-t)):

 s:=t->t-floor(t):

 plot(f(6*s(t/6)),t=0..25);


It is easy to modify it with other values of $t_1$ and $t_2$.



enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 9 '18 at 11:26

























answered Dec 9 '18 at 10:37









Robert ZRobert Z

95.1k1063134




95.1k1063134












  • $begingroup$
    The function looks periodic, but it's not.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 9 '18 at 10:51






  • 1




    $begingroup$
    @Jean-ClaudeArbaut I see your point and I edited my answer. Do you like it now?
    $endgroup$
    – Robert Z
    Dec 9 '18 at 11:27






  • 1




    $begingroup$
    @KellyShepphard Yes we can generalize to more pieces. For each piece you use a Heaviside step function and then you add together all the pieces.
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:12








  • 1




    $begingroup$
    if you have two pieces $f1$ on $[a,b]$ and $f2$ on $[c,d]$ then f=f1*Heaviside((x-a)(b-x))+f2*Heaviside((x-c)(d-x)). Is it clear?
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:22








  • 1




    $begingroup$
    Yes, that is the period.
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:28


















  • $begingroup$
    The function looks periodic, but it's not.
    $endgroup$
    – Jean-Claude Arbaut
    Dec 9 '18 at 10:51






  • 1




    $begingroup$
    @Jean-ClaudeArbaut I see your point and I edited my answer. Do you like it now?
    $endgroup$
    – Robert Z
    Dec 9 '18 at 11:27






  • 1




    $begingroup$
    @KellyShepphard Yes we can generalize to more pieces. For each piece you use a Heaviside step function and then you add together all the pieces.
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:12








  • 1




    $begingroup$
    if you have two pieces $f1$ on $[a,b]$ and $f2$ on $[c,d]$ then f=f1*Heaviside((x-a)(b-x))+f2*Heaviside((x-c)(d-x)). Is it clear?
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:22








  • 1




    $begingroup$
    Yes, that is the period.
    $endgroup$
    – Robert Z
    Dec 9 '18 at 12:28
















$begingroup$
The function looks periodic, but it's not.
$endgroup$
– Jean-Claude Arbaut
Dec 9 '18 at 10:51




$begingroup$
The function looks periodic, but it's not.
$endgroup$
– Jean-Claude Arbaut
Dec 9 '18 at 10:51




1




1




$begingroup$
@Jean-ClaudeArbaut I see your point and I edited my answer. Do you like it now?
$endgroup$
– Robert Z
Dec 9 '18 at 11:27




$begingroup$
@Jean-ClaudeArbaut I see your point and I edited my answer. Do you like it now?
$endgroup$
– Robert Z
Dec 9 '18 at 11:27




1




1




$begingroup$
@KellyShepphard Yes we can generalize to more pieces. For each piece you use a Heaviside step function and then you add together all the pieces.
$endgroup$
– Robert Z
Dec 9 '18 at 12:12






$begingroup$
@KellyShepphard Yes we can generalize to more pieces. For each piece you use a Heaviside step function and then you add together all the pieces.
$endgroup$
– Robert Z
Dec 9 '18 at 12:12






1




1




$begingroup$
if you have two pieces $f1$ on $[a,b]$ and $f2$ on $[c,d]$ then f=f1*Heaviside((x-a)(b-x))+f2*Heaviside((x-c)(d-x)). Is it clear?
$endgroup$
– Robert Z
Dec 9 '18 at 12:22






$begingroup$
if you have two pieces $f1$ on $[a,b]$ and $f2$ on $[c,d]$ then f=f1*Heaviside((x-a)(b-x))+f2*Heaviside((x-c)(d-x)). Is it clear?
$endgroup$
– Robert Z
Dec 9 '18 at 12:22






1




1




$begingroup$
Yes, that is the period.
$endgroup$
– Robert Z
Dec 9 '18 at 12:28




$begingroup$
Yes, that is the period.
$endgroup$
– Robert Z
Dec 9 '18 at 12:28


















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