Why does any function get thinner as $x$ is multiplied by a constant?












5












$begingroup$


Example:



$$cos(x)$$



Graph of y=cos(x)



$$cos(8x)$$



Graph of y=cos(8x)



"Thinner" might not be the correct term. But I just want to know why does changing $x$ to $8x$ make it look like that?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It's simple, what the first function used to do in an interval from $0$ to $1$ the new function does in the interval from $0$ to $1/8$. In particular if cos rounds once from $-pi/2$ to $pi/2$ the new function will do that from $-pi/16$ to $pi/16$. So it will round 8 times in the interval to $-pi/2$ to $pi/2$. Which makes it look thinner.
    $endgroup$
    – Yanko
    Feb 15 at 9:43










  • $begingroup$
    @Yanko Why are you answering in a comment?
    $endgroup$
    – Arthur
    Feb 15 at 9:46






  • 2




    $begingroup$
    indeed, this is like applying the function after a rescale in the x direction. I.e. consider the function $r: mathbb{R} to mathbb{R},x mapsto ax$ then your functions are just $fcirc r = f(r(x))$ and so just rescaled the whole grid. Also, please observe that if the constant is smaller than 1 it actually gets "fatter" and if it is negative it gets mirrored
    $endgroup$
    – Enkidu
    Feb 15 at 9:47










  • $begingroup$
    @Arthur It's too "non-formal" for me to post it as an answer. I don't mind if someone else turns this into an answer.
    $endgroup$
    – Yanko
    Feb 15 at 9:50










  • $begingroup$
    @Yanko There is no requirement here that answers are formal and I see nothing wrong with yours. And while you may not care about the points, getting an actual answer post upvoted and / or accepted will take this question off the unanswered queue and you will have done a little part in tidying up this place. Comments do not help in that regard. Answers do.
    $endgroup$
    – Arthur
    Feb 15 at 9:52


















5












$begingroup$


Example:



$$cos(x)$$



Graph of y=cos(x)



$$cos(8x)$$



Graph of y=cos(8x)



"Thinner" might not be the correct term. But I just want to know why does changing $x$ to $8x$ make it look like that?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It's simple, what the first function used to do in an interval from $0$ to $1$ the new function does in the interval from $0$ to $1/8$. In particular if cos rounds once from $-pi/2$ to $pi/2$ the new function will do that from $-pi/16$ to $pi/16$. So it will round 8 times in the interval to $-pi/2$ to $pi/2$. Which makes it look thinner.
    $endgroup$
    – Yanko
    Feb 15 at 9:43










  • $begingroup$
    @Yanko Why are you answering in a comment?
    $endgroup$
    – Arthur
    Feb 15 at 9:46






  • 2




    $begingroup$
    indeed, this is like applying the function after a rescale in the x direction. I.e. consider the function $r: mathbb{R} to mathbb{R},x mapsto ax$ then your functions are just $fcirc r = f(r(x))$ and so just rescaled the whole grid. Also, please observe that if the constant is smaller than 1 it actually gets "fatter" and if it is negative it gets mirrored
    $endgroup$
    – Enkidu
    Feb 15 at 9:47










  • $begingroup$
    @Arthur It's too "non-formal" for me to post it as an answer. I don't mind if someone else turns this into an answer.
    $endgroup$
    – Yanko
    Feb 15 at 9:50










  • $begingroup$
    @Yanko There is no requirement here that answers are formal and I see nothing wrong with yours. And while you may not care about the points, getting an actual answer post upvoted and / or accepted will take this question off the unanswered queue and you will have done a little part in tidying up this place. Comments do not help in that regard. Answers do.
    $endgroup$
    – Arthur
    Feb 15 at 9:52
















5












5








5





$begingroup$


Example:



$$cos(x)$$



Graph of y=cos(x)



$$cos(8x)$$



Graph of y=cos(8x)



"Thinner" might not be the correct term. But I just want to know why does changing $x$ to $8x$ make it look like that?










share|cite|improve this question











$endgroup$




Example:



$$cos(x)$$



Graph of y=cos(x)



$$cos(8x)$$



Graph of y=cos(8x)



"Thinner" might not be the correct term. But I just want to know why does changing $x$ to $8x$ make it look like that?







functions trigonometry constants






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 15 at 11:49









Asaf Karagila

306k33438769




306k33438769










asked Feb 15 at 9:41









MrRobot9MrRobot9

1324




1324








  • 1




    $begingroup$
    It's simple, what the first function used to do in an interval from $0$ to $1$ the new function does in the interval from $0$ to $1/8$. In particular if cos rounds once from $-pi/2$ to $pi/2$ the new function will do that from $-pi/16$ to $pi/16$. So it will round 8 times in the interval to $-pi/2$ to $pi/2$. Which makes it look thinner.
    $endgroup$
    – Yanko
    Feb 15 at 9:43










  • $begingroup$
    @Yanko Why are you answering in a comment?
    $endgroup$
    – Arthur
    Feb 15 at 9:46






  • 2




    $begingroup$
    indeed, this is like applying the function after a rescale in the x direction. I.e. consider the function $r: mathbb{R} to mathbb{R},x mapsto ax$ then your functions are just $fcirc r = f(r(x))$ and so just rescaled the whole grid. Also, please observe that if the constant is smaller than 1 it actually gets "fatter" and if it is negative it gets mirrored
    $endgroup$
    – Enkidu
    Feb 15 at 9:47










  • $begingroup$
    @Arthur It's too "non-formal" for me to post it as an answer. I don't mind if someone else turns this into an answer.
    $endgroup$
    – Yanko
    Feb 15 at 9:50










  • $begingroup$
    @Yanko There is no requirement here that answers are formal and I see nothing wrong with yours. And while you may not care about the points, getting an actual answer post upvoted and / or accepted will take this question off the unanswered queue and you will have done a little part in tidying up this place. Comments do not help in that regard. Answers do.
    $endgroup$
    – Arthur
    Feb 15 at 9:52
















  • 1




    $begingroup$
    It's simple, what the first function used to do in an interval from $0$ to $1$ the new function does in the interval from $0$ to $1/8$. In particular if cos rounds once from $-pi/2$ to $pi/2$ the new function will do that from $-pi/16$ to $pi/16$. So it will round 8 times in the interval to $-pi/2$ to $pi/2$. Which makes it look thinner.
    $endgroup$
    – Yanko
    Feb 15 at 9:43










  • $begingroup$
    @Yanko Why are you answering in a comment?
    $endgroup$
    – Arthur
    Feb 15 at 9:46






  • 2




    $begingroup$
    indeed, this is like applying the function after a rescale in the x direction. I.e. consider the function $r: mathbb{R} to mathbb{R},x mapsto ax$ then your functions are just $fcirc r = f(r(x))$ and so just rescaled the whole grid. Also, please observe that if the constant is smaller than 1 it actually gets "fatter" and if it is negative it gets mirrored
    $endgroup$
    – Enkidu
    Feb 15 at 9:47










  • $begingroup$
    @Arthur It's too "non-formal" for me to post it as an answer. I don't mind if someone else turns this into an answer.
    $endgroup$
    – Yanko
    Feb 15 at 9:50










  • $begingroup$
    @Yanko There is no requirement here that answers are formal and I see nothing wrong with yours. And while you may not care about the points, getting an actual answer post upvoted and / or accepted will take this question off the unanswered queue and you will have done a little part in tidying up this place. Comments do not help in that regard. Answers do.
    $endgroup$
    – Arthur
    Feb 15 at 9:52










1




1




$begingroup$
It's simple, what the first function used to do in an interval from $0$ to $1$ the new function does in the interval from $0$ to $1/8$. In particular if cos rounds once from $-pi/2$ to $pi/2$ the new function will do that from $-pi/16$ to $pi/16$. So it will round 8 times in the interval to $-pi/2$ to $pi/2$. Which makes it look thinner.
$endgroup$
– Yanko
Feb 15 at 9:43




$begingroup$
It's simple, what the first function used to do in an interval from $0$ to $1$ the new function does in the interval from $0$ to $1/8$. In particular if cos rounds once from $-pi/2$ to $pi/2$ the new function will do that from $-pi/16$ to $pi/16$. So it will round 8 times in the interval to $-pi/2$ to $pi/2$. Which makes it look thinner.
$endgroup$
– Yanko
Feb 15 at 9:43












$begingroup$
@Yanko Why are you answering in a comment?
$endgroup$
– Arthur
Feb 15 at 9:46




$begingroup$
@Yanko Why are you answering in a comment?
$endgroup$
– Arthur
Feb 15 at 9:46




2




2




$begingroup$
indeed, this is like applying the function after a rescale in the x direction. I.e. consider the function $r: mathbb{R} to mathbb{R},x mapsto ax$ then your functions are just $fcirc r = f(r(x))$ and so just rescaled the whole grid. Also, please observe that if the constant is smaller than 1 it actually gets "fatter" and if it is negative it gets mirrored
$endgroup$
– Enkidu
Feb 15 at 9:47




$begingroup$
indeed, this is like applying the function after a rescale in the x direction. I.e. consider the function $r: mathbb{R} to mathbb{R},x mapsto ax$ then your functions are just $fcirc r = f(r(x))$ and so just rescaled the whole grid. Also, please observe that if the constant is smaller than 1 it actually gets "fatter" and if it is negative it gets mirrored
$endgroup$
– Enkidu
Feb 15 at 9:47












$begingroup$
@Arthur It's too "non-formal" for me to post it as an answer. I don't mind if someone else turns this into an answer.
$endgroup$
– Yanko
Feb 15 at 9:50




$begingroup$
@Arthur It's too "non-formal" for me to post it as an answer. I don't mind if someone else turns this into an answer.
$endgroup$
– Yanko
Feb 15 at 9:50












$begingroup$
@Yanko There is no requirement here that answers are formal and I see nothing wrong with yours. And while you may not care about the points, getting an actual answer post upvoted and / or accepted will take this question off the unanswered queue and you will have done a little part in tidying up this place. Comments do not help in that regard. Answers do.
$endgroup$
– Arthur
Feb 15 at 9:52






$begingroup$
@Yanko There is no requirement here that answers are formal and I see nothing wrong with yours. And while you may not care about the points, getting an actual answer post upvoted and / or accepted will take this question off the unanswered queue and you will have done a little part in tidying up this place. Comments do not help in that regard. Answers do.
$endgroup$
– Arthur
Feb 15 at 9:52












4 Answers
4






active

oldest

votes


















3












$begingroup$

Taking a stab at a non-mathematical answer (well, minimally mathematical I guess). It makes intuitive sense to me, but I can't quite explain it mathematically.



The first thing to notice is that this is unrelated to using goniometric functions. It applies to any function, even ones as simple as $y = x$. The only difference is that it's less clear at first sight that $y = 8.x$ is a squashed ("thinner") version of $y = x$.

Most people perceive the difference between the two graphs as a rotation instead of a horizontal squashing.



However, if you were to color-code the graph (e.g. red-green-blue-red-... for all integer values of $x$ (rounded down)), you will see that it is in fact squashed and not rotated.





Think of the x axis as measure of physical distance, let's say kilometers. From your starting point ($x=0$), the Eiffel tower is 10 kilometers ahead ($x=10$), and Big Ben is another 20 kilometers further ($x=30$). Try to visually imagine the monuments on the x axis.



                                                                 ^
A |o|
/- | |
-------------------------------------------------------------------------------------> (km)
0 1 2 3 4 5 6 7 8 9 10 . . . . . . . . . 20 . . . . . . . . . 30 . . . . . . . .


I apologize for the mediocre artwork.



Now I'm going to introduce a new unit, the Flatermeter, which happens to be exactly equal to 10km. What would our graph now look like if the X axis expresses distance in Flatermeters?



From your starting point ($x=0$), the Eiffel tower is 10 kilometers ahead, which is 1 Flatermeter ($x=1$), and Big Ben is another 20 kilometers further, which is another 2 Flatermeters ($x=3$). Which would look like this:



        ^
A |o|
/- | |
-------------------------------------------------------------------------------------> (Fm)
0 1 2 3 4 5 6 7 8 9 10 . . . . . . . . . 20 . . . . . . . . . 30 . . . . . . . .


Notice how everything bunched up together, and all the distances shrunk by a factor of 10. Also notice that you could replace $Fm$ by $10.km$ as they are equal values.



The original $y = f(km)$ was quite wide. But the $y = f(Fm)$, which is the same as $y = f(10km)$ has bunched everything up much closer (which is what you're calling "thinner" in your question).



When you take a graph (e.g. $y = x$), and then artificially inflate the "step size" (= value of x) by a factor $k$ (e.g. $y = k.x$), then the graph will run through its shape $k$ times faster. Depending on how you visualize the graph, this has one of two (visual) consequences:




  • The markings on the x axis move further apart (by a factor of $k$) and the graph has the exact same shape, visually speaking.

  • The markings on the x axis stay the same and the graph itself horizontally shrinks (by a factor of $k$), visually speaking.


Your example deals with the latter scenario.






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    Welcome to MSE!
    $endgroup$
    – YiFan
    Feb 15 at 12:16



















7












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From the comment by Yanko above:



It's simple, what the first function used to do in an interval from $0$ to $1$ the new function does in the interval from $0$ to $1/8$. In particular if cos rounds once from $−π/2$ to $π/2$ the new function will do that from $−π/16$ to $π/16$. So it will round $8$ times in the interval to $−π/2$ to $π/2$. Which makes it look thinner.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    The plot of a graph is really just a set of points $S={(x,y)mid y=f(x)}$. Let's say you turned $x$ into $ax$ for a constant $a$. Then surely, $S$ will not in general remain the same. The new set will instead contain of $(x/a,y)$ for every $(x,y)$ that used to be in $S$, since now, $y=f(a(x/a))=f(x)$ which fufills the definition of a point being on the graph of a function. So the action of "making $f(x)$ become $f(ax)$" takes each point $(x,y)$ to $(x/a,y)$, hence "compressing" the $x$-axis.






    share|cite|improve this answer









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      0












      $begingroup$

      Multiplying the argument of a trigonometric function by a constant changes its period, which precisely is the distance between two consecutive local maxima or local minima which in either case must be equal.



      Consider the general sinusoidal wave $y=Asin(ax+b)+C$. The period of this wave or trigonometric function is given by $2pi/a$.



      In your case, define $A_1:y=cos x$ and $A_2:y=cos 8x$. By merely inspecting the expressions one can observe that these two waves must have a difference of periods (because the coefficients of $x$ is different in both the cases). Period of $A_1$ is $2pi$, however that of $A_2$ is $pi/4$. That is why you observe such a change in graphs of these functions.






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        4 Answers
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        active

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        4 Answers
        4






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        oldest

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        active

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        3












        $begingroup$

        Taking a stab at a non-mathematical answer (well, minimally mathematical I guess). It makes intuitive sense to me, but I can't quite explain it mathematically.



        The first thing to notice is that this is unrelated to using goniometric functions. It applies to any function, even ones as simple as $y = x$. The only difference is that it's less clear at first sight that $y = 8.x$ is a squashed ("thinner") version of $y = x$.

        Most people perceive the difference between the two graphs as a rotation instead of a horizontal squashing.



        However, if you were to color-code the graph (e.g. red-green-blue-red-... for all integer values of $x$ (rounded down)), you will see that it is in fact squashed and not rotated.





        Think of the x axis as measure of physical distance, let's say kilometers. From your starting point ($x=0$), the Eiffel tower is 10 kilometers ahead ($x=10$), and Big Ben is another 20 kilometers further ($x=30$). Try to visually imagine the monuments on the x axis.



                                                                         ^
        A |o|
        /- | |
        -------------------------------------------------------------------------------------> (km)
        0 1 2 3 4 5 6 7 8 9 10 . . . . . . . . . 20 . . . . . . . . . 30 . . . . . . . .


        I apologize for the mediocre artwork.



        Now I'm going to introduce a new unit, the Flatermeter, which happens to be exactly equal to 10km. What would our graph now look like if the X axis expresses distance in Flatermeters?



        From your starting point ($x=0$), the Eiffel tower is 10 kilometers ahead, which is 1 Flatermeter ($x=1$), and Big Ben is another 20 kilometers further, which is another 2 Flatermeters ($x=3$). Which would look like this:



                ^
        A |o|
        /- | |
        -------------------------------------------------------------------------------------> (Fm)
        0 1 2 3 4 5 6 7 8 9 10 . . . . . . . . . 20 . . . . . . . . . 30 . . . . . . . .


        Notice how everything bunched up together, and all the distances shrunk by a factor of 10. Also notice that you could replace $Fm$ by $10.km$ as they are equal values.



        The original $y = f(km)$ was quite wide. But the $y = f(Fm)$, which is the same as $y = f(10km)$ has bunched everything up much closer (which is what you're calling "thinner" in your question).



        When you take a graph (e.g. $y = x$), and then artificially inflate the "step size" (= value of x) by a factor $k$ (e.g. $y = k.x$), then the graph will run through its shape $k$ times faster. Depending on how you visualize the graph, this has one of two (visual) consequences:




        • The markings on the x axis move further apart (by a factor of $k$) and the graph has the exact same shape, visually speaking.

        • The markings on the x axis stay the same and the graph itself horizontally shrinks (by a factor of $k$), visually speaking.


        Your example deals with the latter scenario.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Welcome to MSE!
          $endgroup$
          – YiFan
          Feb 15 at 12:16
















        3












        $begingroup$

        Taking a stab at a non-mathematical answer (well, minimally mathematical I guess). It makes intuitive sense to me, but I can't quite explain it mathematically.



        The first thing to notice is that this is unrelated to using goniometric functions. It applies to any function, even ones as simple as $y = x$. The only difference is that it's less clear at first sight that $y = 8.x$ is a squashed ("thinner") version of $y = x$.

        Most people perceive the difference between the two graphs as a rotation instead of a horizontal squashing.



        However, if you were to color-code the graph (e.g. red-green-blue-red-... for all integer values of $x$ (rounded down)), you will see that it is in fact squashed and not rotated.





        Think of the x axis as measure of physical distance, let's say kilometers. From your starting point ($x=0$), the Eiffel tower is 10 kilometers ahead ($x=10$), and Big Ben is another 20 kilometers further ($x=30$). Try to visually imagine the monuments on the x axis.



                                                                         ^
        A |o|
        /- | |
        -------------------------------------------------------------------------------------> (km)
        0 1 2 3 4 5 6 7 8 9 10 . . . . . . . . . 20 . . . . . . . . . 30 . . . . . . . .


        I apologize for the mediocre artwork.



        Now I'm going to introduce a new unit, the Flatermeter, which happens to be exactly equal to 10km. What would our graph now look like if the X axis expresses distance in Flatermeters?



        From your starting point ($x=0$), the Eiffel tower is 10 kilometers ahead, which is 1 Flatermeter ($x=1$), and Big Ben is another 20 kilometers further, which is another 2 Flatermeters ($x=3$). Which would look like this:



                ^
        A |o|
        /- | |
        -------------------------------------------------------------------------------------> (Fm)
        0 1 2 3 4 5 6 7 8 9 10 . . . . . . . . . 20 . . . . . . . . . 30 . . . . . . . .


        Notice how everything bunched up together, and all the distances shrunk by a factor of 10. Also notice that you could replace $Fm$ by $10.km$ as they are equal values.



        The original $y = f(km)$ was quite wide. But the $y = f(Fm)$, which is the same as $y = f(10km)$ has bunched everything up much closer (which is what you're calling "thinner" in your question).



        When you take a graph (e.g. $y = x$), and then artificially inflate the "step size" (= value of x) by a factor $k$ (e.g. $y = k.x$), then the graph will run through its shape $k$ times faster. Depending on how you visualize the graph, this has one of two (visual) consequences:




        • The markings on the x axis move further apart (by a factor of $k$) and the graph has the exact same shape, visually speaking.

        • The markings on the x axis stay the same and the graph itself horizontally shrinks (by a factor of $k$), visually speaking.


        Your example deals with the latter scenario.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Welcome to MSE!
          $endgroup$
          – YiFan
          Feb 15 at 12:16














        3












        3








        3





        $begingroup$

        Taking a stab at a non-mathematical answer (well, minimally mathematical I guess). It makes intuitive sense to me, but I can't quite explain it mathematically.



        The first thing to notice is that this is unrelated to using goniometric functions. It applies to any function, even ones as simple as $y = x$. The only difference is that it's less clear at first sight that $y = 8.x$ is a squashed ("thinner") version of $y = x$.

        Most people perceive the difference between the two graphs as a rotation instead of a horizontal squashing.



        However, if you were to color-code the graph (e.g. red-green-blue-red-... for all integer values of $x$ (rounded down)), you will see that it is in fact squashed and not rotated.





        Think of the x axis as measure of physical distance, let's say kilometers. From your starting point ($x=0$), the Eiffel tower is 10 kilometers ahead ($x=10$), and Big Ben is another 20 kilometers further ($x=30$). Try to visually imagine the monuments on the x axis.



                                                                         ^
        A |o|
        /- | |
        -------------------------------------------------------------------------------------> (km)
        0 1 2 3 4 5 6 7 8 9 10 . . . . . . . . . 20 . . . . . . . . . 30 . . . . . . . .


        I apologize for the mediocre artwork.



        Now I'm going to introduce a new unit, the Flatermeter, which happens to be exactly equal to 10km. What would our graph now look like if the X axis expresses distance in Flatermeters?



        From your starting point ($x=0$), the Eiffel tower is 10 kilometers ahead, which is 1 Flatermeter ($x=1$), and Big Ben is another 20 kilometers further, which is another 2 Flatermeters ($x=3$). Which would look like this:



                ^
        A |o|
        /- | |
        -------------------------------------------------------------------------------------> (Fm)
        0 1 2 3 4 5 6 7 8 9 10 . . . . . . . . . 20 . . . . . . . . . 30 . . . . . . . .


        Notice how everything bunched up together, and all the distances shrunk by a factor of 10. Also notice that you could replace $Fm$ by $10.km$ as they are equal values.



        The original $y = f(km)$ was quite wide. But the $y = f(Fm)$, which is the same as $y = f(10km)$ has bunched everything up much closer (which is what you're calling "thinner" in your question).



        When you take a graph (e.g. $y = x$), and then artificially inflate the "step size" (= value of x) by a factor $k$ (e.g. $y = k.x$), then the graph will run through its shape $k$ times faster. Depending on how you visualize the graph, this has one of two (visual) consequences:




        • The markings on the x axis move further apart (by a factor of $k$) and the graph has the exact same shape, visually speaking.

        • The markings on the x axis stay the same and the graph itself horizontally shrinks (by a factor of $k$), visually speaking.


        Your example deals with the latter scenario.






        share|cite|improve this answer











        $endgroup$



        Taking a stab at a non-mathematical answer (well, minimally mathematical I guess). It makes intuitive sense to me, but I can't quite explain it mathematically.



        The first thing to notice is that this is unrelated to using goniometric functions. It applies to any function, even ones as simple as $y = x$. The only difference is that it's less clear at first sight that $y = 8.x$ is a squashed ("thinner") version of $y = x$.

        Most people perceive the difference between the two graphs as a rotation instead of a horizontal squashing.



        However, if you were to color-code the graph (e.g. red-green-blue-red-... for all integer values of $x$ (rounded down)), you will see that it is in fact squashed and not rotated.





        Think of the x axis as measure of physical distance, let's say kilometers. From your starting point ($x=0$), the Eiffel tower is 10 kilometers ahead ($x=10$), and Big Ben is another 20 kilometers further ($x=30$). Try to visually imagine the monuments on the x axis.



                                                                         ^
        A |o|
        /- | |
        -------------------------------------------------------------------------------------> (km)
        0 1 2 3 4 5 6 7 8 9 10 . . . . . . . . . 20 . . . . . . . . . 30 . . . . . . . .


        I apologize for the mediocre artwork.



        Now I'm going to introduce a new unit, the Flatermeter, which happens to be exactly equal to 10km. What would our graph now look like if the X axis expresses distance in Flatermeters?



        From your starting point ($x=0$), the Eiffel tower is 10 kilometers ahead, which is 1 Flatermeter ($x=1$), and Big Ben is another 20 kilometers further, which is another 2 Flatermeters ($x=3$). Which would look like this:



                ^
        A |o|
        /- | |
        -------------------------------------------------------------------------------------> (Fm)
        0 1 2 3 4 5 6 7 8 9 10 . . . . . . . . . 20 . . . . . . . . . 30 . . . . . . . .


        Notice how everything bunched up together, and all the distances shrunk by a factor of 10. Also notice that you could replace $Fm$ by $10.km$ as they are equal values.



        The original $y = f(km)$ was quite wide. But the $y = f(Fm)$, which is the same as $y = f(10km)$ has bunched everything up much closer (which is what you're calling "thinner" in your question).



        When you take a graph (e.g. $y = x$), and then artificially inflate the "step size" (= value of x) by a factor $k$ (e.g. $y = k.x$), then the graph will run through its shape $k$ times faster. Depending on how you visualize the graph, this has one of two (visual) consequences:




        • The markings on the x axis move further apart (by a factor of $k$) and the graph has the exact same shape, visually speaking.

        • The markings on the x axis stay the same and the graph itself horizontally shrinks (by a factor of $k$), visually speaking.


        Your example deals with the latter scenario.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Feb 15 at 12:43

























        answered Feb 15 at 12:13









        FlaterFlater

        1463




        1463












        • $begingroup$
          Welcome to MSE!
          $endgroup$
          – YiFan
          Feb 15 at 12:16


















        • $begingroup$
          Welcome to MSE!
          $endgroup$
          – YiFan
          Feb 15 at 12:16
















        $begingroup$
        Welcome to MSE!
        $endgroup$
        – YiFan
        Feb 15 at 12:16




        $begingroup$
        Welcome to MSE!
        $endgroup$
        – YiFan
        Feb 15 at 12:16











        7












        $begingroup$

        From the comment by Yanko above:



        It's simple, what the first function used to do in an interval from $0$ to $1$ the new function does in the interval from $0$ to $1/8$. In particular if cos rounds once from $−π/2$ to $π/2$ the new function will do that from $−π/16$ to $π/16$. So it will round $8$ times in the interval to $−π/2$ to $π/2$. Which makes it look thinner.






        share|cite|improve this answer











        $endgroup$


















          7












          $begingroup$

          From the comment by Yanko above:



          It's simple, what the first function used to do in an interval from $0$ to $1$ the new function does in the interval from $0$ to $1/8$. In particular if cos rounds once from $−π/2$ to $π/2$ the new function will do that from $−π/16$ to $π/16$. So it will round $8$ times in the interval to $−π/2$ to $π/2$. Which makes it look thinner.






          share|cite|improve this answer











          $endgroup$
















            7












            7








            7





            $begingroup$

            From the comment by Yanko above:



            It's simple, what the first function used to do in an interval from $0$ to $1$ the new function does in the interval from $0$ to $1/8$. In particular if cos rounds once from $−π/2$ to $π/2$ the new function will do that from $−π/16$ to $π/16$. So it will round $8$ times in the interval to $−π/2$ to $π/2$. Which makes it look thinner.






            share|cite|improve this answer











            $endgroup$



            From the comment by Yanko above:



            It's simple, what the first function used to do in an interval from $0$ to $1$ the new function does in the interval from $0$ to $1/8$. In particular if cos rounds once from $−π/2$ to $π/2$ the new function will do that from $−π/16$ to $π/16$. So it will round $8$ times in the interval to $−π/2$ to $π/2$. Which makes it look thinner.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            answered Feb 15 at 9:54


























            community wiki





            Arthur
























                1












                $begingroup$

                The plot of a graph is really just a set of points $S={(x,y)mid y=f(x)}$. Let's say you turned $x$ into $ax$ for a constant $a$. Then surely, $S$ will not in general remain the same. The new set will instead contain of $(x/a,y)$ for every $(x,y)$ that used to be in $S$, since now, $y=f(a(x/a))=f(x)$ which fufills the definition of a point being on the graph of a function. So the action of "making $f(x)$ become $f(ax)$" takes each point $(x,y)$ to $(x/a,y)$, hence "compressing" the $x$-axis.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  The plot of a graph is really just a set of points $S={(x,y)mid y=f(x)}$. Let's say you turned $x$ into $ax$ for a constant $a$. Then surely, $S$ will not in general remain the same. The new set will instead contain of $(x/a,y)$ for every $(x,y)$ that used to be in $S$, since now, $y=f(a(x/a))=f(x)$ which fufills the definition of a point being on the graph of a function. So the action of "making $f(x)$ become $f(ax)$" takes each point $(x,y)$ to $(x/a,y)$, hence "compressing" the $x$-axis.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    The plot of a graph is really just a set of points $S={(x,y)mid y=f(x)}$. Let's say you turned $x$ into $ax$ for a constant $a$. Then surely, $S$ will not in general remain the same. The new set will instead contain of $(x/a,y)$ for every $(x,y)$ that used to be in $S$, since now, $y=f(a(x/a))=f(x)$ which fufills the definition of a point being on the graph of a function. So the action of "making $f(x)$ become $f(ax)$" takes each point $(x,y)$ to $(x/a,y)$, hence "compressing" the $x$-axis.






                    share|cite|improve this answer









                    $endgroup$



                    The plot of a graph is really just a set of points $S={(x,y)mid y=f(x)}$. Let's say you turned $x$ into $ax$ for a constant $a$. Then surely, $S$ will not in general remain the same. The new set will instead contain of $(x/a,y)$ for every $(x,y)$ that used to be in $S$, since now, $y=f(a(x/a))=f(x)$ which fufills the definition of a point being on the graph of a function. So the action of "making $f(x)$ become $f(ax)$" takes each point $(x,y)$ to $(x/a,y)$, hence "compressing" the $x$-axis.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Feb 15 at 11:53









                    YiFanYiFan

                    4,7501727




                    4,7501727























                        0












                        $begingroup$

                        Multiplying the argument of a trigonometric function by a constant changes its period, which precisely is the distance between two consecutive local maxima or local minima which in either case must be equal.



                        Consider the general sinusoidal wave $y=Asin(ax+b)+C$. The period of this wave or trigonometric function is given by $2pi/a$.



                        In your case, define $A_1:y=cos x$ and $A_2:y=cos 8x$. By merely inspecting the expressions one can observe that these two waves must have a difference of periods (because the coefficients of $x$ is different in both the cases). Period of $A_1$ is $2pi$, however that of $A_2$ is $pi/4$. That is why you observe such a change in graphs of these functions.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Multiplying the argument of a trigonometric function by a constant changes its period, which precisely is the distance between two consecutive local maxima or local minima which in either case must be equal.



                          Consider the general sinusoidal wave $y=Asin(ax+b)+C$. The period of this wave or trigonometric function is given by $2pi/a$.



                          In your case, define $A_1:y=cos x$ and $A_2:y=cos 8x$. By merely inspecting the expressions one can observe that these two waves must have a difference of periods (because the coefficients of $x$ is different in both the cases). Period of $A_1$ is $2pi$, however that of $A_2$ is $pi/4$. That is why you observe such a change in graphs of these functions.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Multiplying the argument of a trigonometric function by a constant changes its period, which precisely is the distance between two consecutive local maxima or local minima which in either case must be equal.



                            Consider the general sinusoidal wave $y=Asin(ax+b)+C$. The period of this wave or trigonometric function is given by $2pi/a$.



                            In your case, define $A_1:y=cos x$ and $A_2:y=cos 8x$. By merely inspecting the expressions one can observe that these two waves must have a difference of periods (because the coefficients of $x$ is different in both the cases). Period of $A_1$ is $2pi$, however that of $A_2$ is $pi/4$. That is why you observe such a change in graphs of these functions.






                            share|cite|improve this answer









                            $endgroup$



                            Multiplying the argument of a trigonometric function by a constant changes its period, which precisely is the distance between two consecutive local maxima or local minima which in either case must be equal.



                            Consider the general sinusoidal wave $y=Asin(ax+b)+C$. The period of this wave or trigonometric function is given by $2pi/a$.



                            In your case, define $A_1:y=cos x$ and $A_2:y=cos 8x$. By merely inspecting the expressions one can observe that these two waves must have a difference of periods (because the coefficients of $x$ is different in both the cases). Period of $A_1$ is $2pi$, however that of $A_2$ is $pi/4$. That is why you observe such a change in graphs of these functions.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Feb 15 at 11:01









                            Paras KhoslaParas Khosla

                            2,178222




                            2,178222






























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