Why does any function get thinner as $x$ is multiplied by a constant?
$begingroup$
Example:
$$cos(x)$$
$$cos(8x)$$
"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
$endgroup$
|
show 1 more comment
$begingroup$
Example:
$$cos(x)$$
$$cos(8x)$$
"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
$endgroup$
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.
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– Yanko
Feb 15 at 9:43
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@Yanko Why are you answering in a comment?
$endgroup$
– Arthur
Feb 15 at 9:46
2
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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
|
show 1 more comment
$begingroup$
Example:
$$cos(x)$$
$$cos(8x)$$
"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
$endgroup$
Example:
$$cos(x)$$
$$cos(8x)$$
"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
functions trigonometry constants
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
|
show 1 more comment
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
|
show 1 more comment
4 Answers
4
active
oldest
votes
$begingroup$
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.
$endgroup$
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Welcome to MSE!
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– YiFan
Feb 15 at 12:16
add a comment |
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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.
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add a comment |
$begingroup$
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.
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add a comment |
$begingroup$
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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add a comment |
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4 Answers
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4 Answers
4
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oldest
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$begingroup$
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.
$endgroup$
$begingroup$
Welcome to MSE!
$endgroup$
– YiFan
Feb 15 at 12:16
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
Welcome to MSE!
$endgroup$
– YiFan
Feb 15 at 12:16
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
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add a comment |
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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.
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add a comment |
$begingroup$
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.
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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.
answered Feb 15 at 9:54
community wiki
Arthur
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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.
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add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Feb 15 at 11:53
YiFanYiFan
4,7501727
4,7501727
add a comment |
add a comment |
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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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add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered Feb 15 at 11:01
Paras KhoslaParas Khosla
2,178222
2,178222
add a comment |
add a comment |
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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.
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– Yanko
Feb 15 at 9:43
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@Yanko Why are you answering in a comment?
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– Arthur
Feb 15 at 9:46
2
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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
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– Enkidu
Feb 15 at 9:47
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@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.
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– Yanko
Feb 15 at 9:50
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@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.
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– Arthur
Feb 15 at 9:52