Mixing
What exactly is a 'dummy variable'?
$begingroup$
I was watching an online course (The Calculus You Need - MIT OpenCourseWare), when (around 03:08), the lecturer (Gilbert Strang) says that he doesn't "care what that dummy variable is" (the variable x associated with the function y). He made the following change in the video:
$$frac{d}{dx}int_{0}^{x}y(x) = y(x)Rightarrow text{ change } Rightarrowfrac{d}{dx}int_{0}^{x}y(t)dt = y(x)$$
I don't why the notation of that variable doesn't cause any ambiguity here (I'm assuming Strang wanted to mean that), once x is clearly creating a dependency relation with the bound and the variable associated with the function y.
Ultimately... why it doesn't make a difference to call the variable associated with y by x (in this case)? What exactly is a 'dummy variable'?
real-analysis calculus integration riemann-integration
edited Jan 23 at 8:28
Lorenzo B.
1,8602520
asked Dec 21 '18 at 20:59
Nicholas YamasakiNicholas Yamasaki
234
$endgroup$
add a comment |
$begingroup$
I was watching an online course (The Calculus You Need - MIT OpenCourseWare), when (around 03:08), the lecturer (Gilbert Strang) says that he doesn't "care what that dummy variable is" (the variable x associated with the function y). He made the following change in the video:
$$frac{d}{dx}int_{0}^{x}y(x) = y(x)Rightarrow text{ change } Rightarrowfrac{d}{dx}int_{0}^{x}y(t)dt = y(x)$$
I don't why the notation of that variable doesn't cause any ambiguity here (I'm assuming Strang wanted to mean that), once x is clearly creating a dependency relation with the bound and the variable associated with the function y.
Ultimately... why it doesn't make a difference to call the variable associated with y by x (in this case)? What exactly is a 'dummy variable'?
real-analysis calculus integration riemann-integration
edited Jan 23 at 8:28
Lorenzo B.
1,8602520
asked Dec 21 '18 at 20:59
Nicholas YamasakiNicholas Yamasaki
234
$endgroup$
1
$begingroup$
Did Strang write $displaystyle dfrac{text d}{text dx}intlimits _{0}^{x}y(x)text dx$?
$endgroup$
– Git Gud
Dec 21 '18 at 21:06
$begingroup$
Oh, thanks, he didn't. But still, could you clarify me why you asked that?
$endgroup$
– Nicholas Yamasaki
Dec 21 '18 at 21:13
2
$begingroup$
Note that Strang immediately erased the expression $int_0^x y(x)$, replacing it with $int_0^x y(t) , dt$. Probably he did this because it is not correct to use the same letter for the dummy variable as for the number $x$ which has already been introduced. (The letter $x$ should not be used for two different things at the same time.) By the way, the "dummy variable" is just a piece of notation and we can eliminate it from the notation entirely if we like by writing $int_0^x y$ (for example). Spivak uses this notation (omitting the dummy variable) in chapter 3 of Calculus on Manifolds.
$endgroup$
– littleO
Dec 22 '18 at 2:24
$begingroup$
Thank you, littleO. But if you don't mind me asking one more question again... why is that it doesn't make sense to create a function where the bound and the $y$ variable are dependent? A function like $A(x) = int_{0}^{x}y(x)dx$ (let's say $y(x) = x^2$) and then you just put $x = 2$ and get $A(2) = 8$? I know this reasoning is a contradiction with the fundamental theorem of calculus (once you should get $A(2) = 8/3$), and by that, it seems that it's impossible to create any kind of relation with the bound and the $y$ variable... why is that?
$endgroup$
– Nicholas Yamasaki
Dec 30 '18 at 19:12
add a comment |
$begingroup$
I was watching an online course (The Calculus You Need - MIT OpenCourseWare), when (around 03:08), the lecturer (Gilbert Strang) says that he doesn't "care what that dummy variable is" (the variable x associated with the function y). He made the following change in the video:
$$frac{d}{dx}int_{0}^{x}y(x) = y(x)Rightarrow text{ change } Rightarrowfrac{d}{dx}int_{0}^{x}y(t)dt = y(x)$$
I don't why the notation of that variable doesn't cause any ambiguity here (I'm assuming Strang wanted to mean that), once x is clearly creating a dependency relation with the bound and the variable associated with the function y.
Ultimately... why it doesn't make a difference to call the variable associated with y by x (in this case)? What exactly is a 'dummy variable'?
real-analysis calculus integration riemann-integration
edited Jan 23 at 8:28
Lorenzo B.
1,8602520
asked Dec 21 '18 at 20:59
Nicholas YamasakiNicholas Yamasaki
234
$endgroup$
I was watching an online course (The Calculus You Need - MIT OpenCourseWare), when (around 03:08), the lecturer (Gilbert Strang) says that he doesn't "care what that dummy variable is" (the variable x associated with the function y). He made the following change in the video:
$$frac{d}{dx}int_{0}^{x}y(x) = y(x)Rightarrow text{ change } Rightarrowfrac{d}{dx}int_{0}^{x}y(t)dt = y(x)$$
I don't why the notation of that variable doesn't cause any ambiguity here (I'm assuming Strang wanted to mean that), once x is clearly creating a dependency relation with the bound and the variable associated with the function y.
Ultimately... why it doesn't make a difference to call the variable associated with y by x (in this case)? What exactly is a 'dummy variable'?
real-analysis calculus integration riemann-integration
real-analysis calculus integration riemann-integration
edited Jan 23 at 8:28
Lorenzo B.
1,8602520
asked Dec 21 '18 at 20:59
Nicholas YamasakiNicholas Yamasaki
234
edited Jan 23 at 8:28
Lorenzo B.
1,8602520
asked Dec 21 '18 at 20:59
Nicholas YamasakiNicholas Yamasaki
234
edited Jan 23 at 8:28
Lorenzo B.
1,8602520
edited Jan 23 at 8:28
Lorenzo B.
1,8602520
edited Jan 23 at 8:28
Lorenzo B.
1,8602520
1,8602520
asked Dec 21 '18 at 20:59
Nicholas YamasakiNicholas Yamasaki
234
asked Dec 21 '18 at 20:59
Nicholas YamasakiNicholas Yamasaki
234
asked Dec 21 '18 at 20:59
Nicholas YamasakiNicholas Yamasaki
234
234
1
$begingroup$
Did Strang write $displaystyle dfrac{text d}{text dx}intlimits _{0}^{x}y(x)text dx$?
$endgroup$
– Git Gud
Dec 21 '18 at 21:06
$begingroup$
Oh, thanks, he didn't. But still, could you clarify me why you asked that?
$endgroup$
– Nicholas Yamasaki
Dec 21 '18 at 21:13
2
$begingroup$
Note that Strang immediately erased the expression $int_0^x y(x)$, replacing it with $int_0^x y(t) , dt$. Probably he did this because it is not correct to use the same letter for the dummy variable as for the number $x$ which has already been introduced. (The letter $x$ should not be used for two different things at the same time.) By the way, the "dummy variable" is just a piece of notation and we can eliminate it from the notation entirely if we like by writing $int_0^x y$ (for example). Spivak uses this notation (omitting the dummy variable) in chapter 3 of Calculus on Manifolds.
$endgroup$
– littleO
Dec 22 '18 at 2:24
$begingroup$
Thank you, littleO. But if you don't mind me asking one more question again... why is that it doesn't make sense to create a function where the bound and the $y$ variable are dependent? A function like $A(x) = int_{0}^{x}y(x)dx$ (let's say $y(x) = x^2$) and then you just put $x = 2$ and get $A(2) = 8$? I know this reasoning is a contradiction with the fundamental theorem of calculus (once you should get $A(2) = 8/3$), and by that, it seems that it's impossible to create any kind of relation with the bound and the $y$ variable... why is that?
$endgroup$
– Nicholas Yamasaki
Dec 30 '18 at 19:12
add a comment |
1
$begingroup$
Did Strang write $displaystyle dfrac{text d}{text dx}intlimits _{0}^{x}y(x)text dx$?
$endgroup$
– Git Gud
Dec 21 '18 at 21:06
$begingroup$
Oh, thanks, he didn't. But still, could you clarify me why you asked that?
$endgroup$
– Nicholas Yamasaki
Dec 21 '18 at 21:13
2
$begingroup$
Note that Strang immediately erased the expression $int_0^x y(x)$, replacing it with $int_0^x y(t) , dt$. Probably he did this because it is not correct to use the same letter for the dummy variable as for the number $x$ which has already been introduced. (The letter $x$ should not be used for two different things at the same time.) By the way, the "dummy variable" is just a piece of notation and we can eliminate it from the notation entirely if we like by writing $int_0^x y$ (for example). Spivak uses this notation (omitting the dummy variable) in chapter 3 of Calculus on Manifolds.
$endgroup$
– littleO
Dec 22 '18 at 2:24
$begingroup$
Thank you, littleO. But if you don't mind me asking one more question again... why is that it doesn't make sense to create a function where the bound and the $y$ variable are dependent? A function like $A(x) = int_{0}^{x}y(x)dx$ (let's say $y(x) = x^2$) and then you just put $x = 2$ and get $A(2) = 8$? I know this reasoning is a contradiction with the fundamental theorem of calculus (once you should get $A(2) = 8/3$), and by that, it seems that it's impossible to create any kind of relation with the bound and the $y$ variable... why is that?
$endgroup$
– Nicholas Yamasaki
Dec 30 '18 at 19:12
1
1
$begingroup$
Did Strang write $displaystyle dfrac{text d}{text dx}intlimits _{0}^{x}y(x)text dx$?
$endgroup$
– Git Gud
Dec 21 '18 at 21:06
$begingroup$
Did Strang write $displaystyle dfrac{text d}{text dx}intlimits _{0}^{x}y(x)text dx$?
$endgroup$
– Git Gud
Dec 21 '18 at 21:06
$begingroup$
Oh, thanks, he didn't. But still, could you clarify me why you asked that?
$endgroup$
– Nicholas Yamasaki
Dec 21 '18 at 21:13
$begingroup$
Oh, thanks, he didn't. But still, could you clarify me why you asked that?
$endgroup$
– Nicholas Yamasaki
Dec 21 '18 at 21:13
2
2
$begingroup$
Note that Strang immediately erased the expression $int_0^x y(x)$, replacing it with $int_0^x y(t) , dt$. Probably he did this because it is not correct to use the same letter for the dummy variable as for the number $x$ which has already been introduced. (The letter $x$ should not be used for two different things at the same time.) By the way, the "dummy variable" is just a piece of notation and we can eliminate it from the notation entirely if we like by writing $int_0^x y$ (for example). Spivak uses this notation (omitting the dummy variable) in chapter 3 of Calculus on Manifolds.
$endgroup$
– littleO
Dec 22 '18 at 2:24
$begingroup$
Note that Strang immediately erased the expression $int_0^x y(x)$, replacing it with $int_0^x y(t) , dt$. Probably he did this because it is not correct to use the same letter for the dummy variable as for the number $x$ which has already been introduced. (The letter $x$ should not be used for two different things at the same time.) By the way, the "dummy variable" is just a piece of notation and we can eliminate it from the notation entirely if we like by writing $int_0^x y$ (for example). Spivak uses this notation (omitting the dummy variable) in chapter 3 of Calculus on Manifolds.
$endgroup$
– littleO
Dec 22 '18 at 2:24
$begingroup$
Thank you, littleO. But if you don't mind me asking one more question again... why is that it doesn't make sense to create a function where the bound and the $y$ variable are dependent? A function like $A(x) = int_{0}^{x}y(x)dx$ (let's say $y(x) = x^2$) and then you just put $x = 2$ and get $A(2) = 8$? I know this reasoning is a contradiction with the fundamental theorem of calculus (once you should get $A(2) = 8/3$), and by that, it seems that it's impossible to create any kind of relation with the bound and the $y$ variable... why is that?
$endgroup$
– Nicholas Yamasaki
Dec 30 '18 at 19:12
$begingroup$
Thank you, littleO. But if you don't mind me asking one more question again... why is that it doesn't make sense to create a function where the bound and the $y$ variable are dependent? A function like $A(x) = int_{0}^{x}y(x)dx$ (let's say $y(x) = x^2$) and then you just put $x = 2$ and get $A(2) = 8$? I know this reasoning is a contradiction with the fundamental theorem of calculus (once you should get $A(2) = 8/3$), and by that, it seems that it's impossible to create any kind of relation with the bound and the $y$ variable... why is that?
$endgroup$
– Nicholas Yamasaki
Dec 30 '18 at 19:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have asked two questions, one about whether the name of a variable matters, the other about dummy variables. They're different but related.
Strictly speaking, it does not matter what you name your variables. If you have a function $f$ from, say, $mathbb{R}$ to itself then the elements in both the domain and the codomain are real numbers. You may often see this described as "$y = f(x)$". That usually does no damage, and sometimes helps keep the meaning of numbers straight - those called "$x$" are in the domain while those called "$y$" are in the codomain. But that's not at all required and is sometimes confusing. You should talk about the function $sin$, not the function $sin(x)$.
Now for dummy variables. The expression
$$
sum_{n=1}^3 n^2
$$
means just
$$
1^2 + 2^2 +3^2 .
$$
There is no "$n$" in it. You could write the same thing as
$$
sum_{@=1}^3 @^2 .
$$
The $n$ and the $@$ are dummy variables - they are not there. The same is true for the "variable of integration":
$$
int_1^2 sin(x) dx = int_1^2 sin(@) d@ = int_1^2 sin .
$$
The last of these is unambiguous when you work with $sin$ as a real valued function of a real variable.
That said, there are times when a (dummy) variable of integration is helpful. With it you can distinguish between
$$
int_1^2 sin(xt^2) dx text{ and } int_1^2 sin(xt^2) dt .
$$
The first is a number that depends on the value of $t$, the second a number that depends on the value of $x$.
The dummy variable $x$ and the $dx$ are quite important when you think about applications of integrals in geometry and physics. If you imagine the integral of $sin$ as calculating the area under the sine curve then the expression
$$
sin(x) dx
$$
is the (infinitesimal) area of a rectangle with height $sin(x)$ and (infinitesimal) base $dx$. Each of those quantities has units length and their product has units area. The integral sign is Leibniz's elongated "S", for "sum".
answered Dec 21 '18 at 21:19
Ethan BolkerEthan Bolker
45k553120
$endgroup$
$begingroup$
Hm.. got it. Just one more thing.. what is the difference between these two expressions? $$int_{0}^{x} y(x) dx$$ and $$int_{0}^{x} y(t) dt$$
$endgroup$
– Nicholas Yamasaki
Dec 21 '18 at 23:13
3
$begingroup$
They are essentially the same, but writing the first one would be very bad expository taste. I hope never to see it anywhere.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 0:40
1
$begingroup$
Yes. But the notation is so weird and (in my view) unacceptable that you shouldn't spend any time or energy trying to parse it correctly.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 1:12
1
$begingroup$
In both cases an antiderivative is the function $f$ that returns the a third of the cube of its input (you can call that $x^3/3$ if you like) and both times $A(2) = f(2)-f(0) = 2^3/3 - 0^3/3 = 8/3$. When you substitute $2$ for $x$ you don't substitute for the dummy variable of integration.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 2:00
1
$begingroup$
Your confusion was caused by the author's careless reuse of "$x$". They shouldn't have done that.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 12:26
|
show 3 more comments
$begingroup$
To add to what was already said, I find it helpful to think in programming terms. What is a dummy variable? It's a local variable with very limited scope - a sum, integral, or other similar construct - and it had better not appear anywhere outside that. If it does, as in that initial expression $frac{d}{dx}int_0^x y(x)$, we'll get errors and unpredictable results. For example, here? In order to get something that parses, we have to take that integral with respect to some other variable - which means that inside the integral, $x$ and $y(x)$ are constants. We then get $int_0^x y(x),dt=xy(x)$, so its derivative with respect to $x$ is $y(x)+xy'(x)$, and the solution of the equation is for $y(x)$ to be a constant function.
The equation after the change is clearly what was intended all along, and it doesn't mean the same thing as the equation before the change. It doesn't matter what we call the dummy variable - as long as we're careful not to conflict it with a variable name on the outside and cause out-of-scope errors.
As for leaving the $d?$ off the integral? I'm not a fan of that; it's a delimiter, which clearly marks what we're integrating, and what variable we're integrating with respect to. As far as I'm concerned, it's part of the integral sign. The only way I'd ever leave it off is if the dummy variable is left out entirely. $int_a^b f(x),dx$ or $int_a^b f$ are OK and unambiguous; $int_a^b f(x)$ is malformed.
answered Dec 21 '18 at 22:43
jmerryjmerry
13.7k1629
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have asked two questions, one about whether the name of a variable matters, the other about dummy variables. They're different but related.
Strictly speaking, it does not matter what you name your variables. If you have a function $f$ from, say, $mathbb{R}$ to itself then the elements in both the domain and the codomain are real numbers. You may often see this described as "$y = f(x)$". That usually does no damage, and sometimes helps keep the meaning of numbers straight - those called "$x$" are in the domain while those called "$y$" are in the codomain. But that's not at all required and is sometimes confusing. You should talk about the function $sin$, not the function $sin(x)$.
Now for dummy variables. The expression
$$
sum_{n=1}^3 n^2
$$
means just
$$
1^2 + 2^2 +3^2 .
$$
There is no "$n$" in it. You could write the same thing as
$$
sum_{@=1}^3 @^2 .
$$
The $n$ and the $@$ are dummy variables - they are not there. The same is true for the "variable of integration":
$$
int_1^2 sin(x) dx = int_1^2 sin(@) d@ = int_1^2 sin .
$$
The last of these is unambiguous when you work with $sin$ as a real valued function of a real variable.
That said, there are times when a (dummy) variable of integration is helpful. With it you can distinguish between
$$
int_1^2 sin(xt^2) dx text{ and } int_1^2 sin(xt^2) dt .
$$
The first is a number that depends on the value of $t$, the second a number that depends on the value of $x$.
The dummy variable $x$ and the $dx$ are quite important when you think about applications of integrals in geometry and physics. If you imagine the integral of $sin$ as calculating the area under the sine curve then the expression
$$
sin(x) dx
$$
is the (infinitesimal) area of a rectangle with height $sin(x)$ and (infinitesimal) base $dx$. Each of those quantities has units length and their product has units area. The integral sign is Leibniz's elongated "S", for "sum".
answered Dec 21 '18 at 21:19
Ethan BolkerEthan Bolker
45k553120
$endgroup$
$begingroup$
Hm.. got it. Just one more thing.. what is the difference between these two expressions? $$int_{0}^{x} y(x) dx$$ and $$int_{0}^{x} y(t) dt$$
$endgroup$
– Nicholas Yamasaki
Dec 21 '18 at 23:13
3
$begingroup$
They are essentially the same, but writing the first one would be very bad expository taste. I hope never to see it anywhere.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 0:40
1
$begingroup$
Yes. But the notation is so weird and (in my view) unacceptable that you shouldn't spend any time or energy trying to parse it correctly.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 1:12
1
$begingroup$
In both cases an antiderivative is the function $f$ that returns the a third of the cube of its input (you can call that $x^3/3$ if you like) and both times $A(2) = f(2)-f(0) = 2^3/3 - 0^3/3 = 8/3$. When you substitute $2$ for $x$ you don't substitute for the dummy variable of integration.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 2:00
1
$begingroup$
Your confusion was caused by the author's careless reuse of "$x$". They shouldn't have done that.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 12:26
|
show 3 more comments
$begingroup$
You have asked two questions, one about whether the name of a variable matters, the other about dummy variables. They're different but related.
Strictly speaking, it does not matter what you name your variables. If you have a function $f$ from, say, $mathbb{R}$ to itself then the elements in both the domain and the codomain are real numbers. You may often see this described as "$y = f(x)$". That usually does no damage, and sometimes helps keep the meaning of numbers straight - those called "$x$" are in the domain while those called "$y$" are in the codomain. But that's not at all required and is sometimes confusing. You should talk about the function $sin$, not the function $sin(x)$.
Now for dummy variables. The expression
$$
sum_{n=1}^3 n^2
$$
means just
$$
1^2 + 2^2 +3^2 .
$$
There is no "$n$" in it. You could write the same thing as
$$
sum_{@=1}^3 @^2 .
$$
The $n$ and the $@$ are dummy variables - they are not there. The same is true for the "variable of integration":
$$
int_1^2 sin(x) dx = int_1^2 sin(@) d@ = int_1^2 sin .
$$
The last of these is unambiguous when you work with $sin$ as a real valued function of a real variable.
That said, there are times when a (dummy) variable of integration is helpful. With it you can distinguish between
$$
int_1^2 sin(xt^2) dx text{ and } int_1^2 sin(xt^2) dt .
$$
The first is a number that depends on the value of $t$, the second a number that depends on the value of $x$.
The dummy variable $x$ and the $dx$ are quite important when you think about applications of integrals in geometry and physics. If you imagine the integral of $sin$ as calculating the area under the sine curve then the expression
$$
sin(x) dx
$$
is the (infinitesimal) area of a rectangle with height $sin(x)$ and (infinitesimal) base $dx$. Each of those quantities has units length and their product has units area. The integral sign is Leibniz's elongated "S", for "sum".
answered Dec 21 '18 at 21:19
Ethan BolkerEthan Bolker
45k553120
$endgroup$
$begingroup$
Hm.. got it. Just one more thing.. what is the difference between these two expressions? $$int_{0}^{x} y(x) dx$$ and $$int_{0}^{x} y(t) dt$$
$endgroup$
– Nicholas Yamasaki
Dec 21 '18 at 23:13
3
$begingroup$
They are essentially the same, but writing the first one would be very bad expository taste. I hope never to see it anywhere.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 0:40
1
$begingroup$
Yes. But the notation is so weird and (in my view) unacceptable that you shouldn't spend any time or energy trying to parse it correctly.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 1:12
1
$begingroup$
In both cases an antiderivative is the function $f$ that returns the a third of the cube of its input (you can call that $x^3/3$ if you like) and both times $A(2) = f(2)-f(0) = 2^3/3 - 0^3/3 = 8/3$. When you substitute $2$ for $x$ you don't substitute for the dummy variable of integration.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 2:00
1
$begingroup$
Your confusion was caused by the author's careless reuse of "$x$". They shouldn't have done that.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 12:26
|
show 3 more comments
$begingroup$
You have asked two questions, one about whether the name of a variable matters, the other about dummy variables. They're different but related.
Strictly speaking, it does not matter what you name your variables. If you have a function $f$ from, say, $mathbb{R}$ to itself then the elements in both the domain and the codomain are real numbers. You may often see this described as "$y = f(x)$". That usually does no damage, and sometimes helps keep the meaning of numbers straight - those called "$x$" are in the domain while those called "$y$" are in the codomain. But that's not at all required and is sometimes confusing. You should talk about the function $sin$, not the function $sin(x)$.
Now for dummy variables. The expression
$$
sum_{n=1}^3 n^2
$$
means just
$$
1^2 + 2^2 +3^2 .
$$
There is no "$n$" in it. You could write the same thing as
$$
sum_{@=1}^3 @^2 .
$$
The $n$ and the $@$ are dummy variables - they are not there. The same is true for the "variable of integration":
$$
int_1^2 sin(x) dx = int_1^2 sin(@) d@ = int_1^2 sin .
$$
The last of these is unambiguous when you work with $sin$ as a real valued function of a real variable.
That said, there are times when a (dummy) variable of integration is helpful. With it you can distinguish between
$$
int_1^2 sin(xt^2) dx text{ and } int_1^2 sin(xt^2) dt .
$$
The first is a number that depends on the value of $t$, the second a number that depends on the value of $x$.
The dummy variable $x$ and the $dx$ are quite important when you think about applications of integrals in geometry and physics. If you imagine the integral of $sin$ as calculating the area under the sine curve then the expression
$$
sin(x) dx
$$
is the (infinitesimal) area of a rectangle with height $sin(x)$ and (infinitesimal) base $dx$. Each of those quantities has units length and their product has units area. The integral sign is Leibniz's elongated "S", for "sum".
answered Dec 21 '18 at 21:19
Ethan BolkerEthan Bolker
45k553120
$endgroup$
You have asked two questions, one about whether the name of a variable matters, the other about dummy variables. They're different but related.
Strictly speaking, it does not matter what you name your variables. If you have a function $f$ from, say, $mathbb{R}$ to itself then the elements in both the domain and the codomain are real numbers. You may often see this described as "$y = f(x)$". That usually does no damage, and sometimes helps keep the meaning of numbers straight - those called "$x$" are in the domain while those called "$y$" are in the codomain. But that's not at all required and is sometimes confusing. You should talk about the function $sin$, not the function $sin(x)$.
Now for dummy variables. The expression
$$
sum_{n=1}^3 n^2
$$
means just
$$
1^2 + 2^2 +3^2 .
$$
There is no "$n$" in it. You could write the same thing as
$$
sum_{@=1}^3 @^2 .
$$
The $n$ and the $@$ are dummy variables - they are not there. The same is true for the "variable of integration":
$$
int_1^2 sin(x) dx = int_1^2 sin(@) d@ = int_1^2 sin .
$$
The last of these is unambiguous when you work with $sin$ as a real valued function of a real variable.
That said, there are times when a (dummy) variable of integration is helpful. With it you can distinguish between
$$
int_1^2 sin(xt^2) dx text{ and } int_1^2 sin(xt^2) dt .
$$
The first is a number that depends on the value of $t$, the second a number that depends on the value of $x$.
The dummy variable $x$ and the $dx$ are quite important when you think about applications of integrals in geometry and physics. If you imagine the integral of $sin$ as calculating the area under the sine curve then the expression
$$
sin(x) dx
$$
is the (infinitesimal) area of a rectangle with height $sin(x)$ and (infinitesimal) base $dx$. Each of those quantities has units length and their product has units area. The integral sign is Leibniz's elongated "S", for "sum".
answered Dec 21 '18 at 21:19
Ethan BolkerEthan Bolker
45k553120
edited Dec 22 '18 at 2:15
answered Dec 21 '18 at 21:19
Ethan BolkerEthan Bolker
45k553120
answered Dec 21 '18 at 21:19
Ethan BolkerEthan Bolker
45k553120
answered Dec 21 '18 at 21:19
Ethan BolkerEthan Bolker
45k553120
45k553120
$begingroup$
Hm.. got it. Just one more thing.. what is the difference between these two expressions? $$int_{0}^{x} y(x) dx$$ and $$int_{0}^{x} y(t) dt$$
$endgroup$
– Nicholas Yamasaki
Dec 21 '18 at 23:13
3
$begingroup$
They are essentially the same, but writing the first one would be very bad expository taste. I hope never to see it anywhere.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 0:40
1
$begingroup$
Yes. But the notation is so weird and (in my view) unacceptable that you shouldn't spend any time or energy trying to parse it correctly.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 1:12
1
$begingroup$
In both cases an antiderivative is the function $f$ that returns the a third of the cube of its input (you can call that $x^3/3$ if you like) and both times $A(2) = f(2)-f(0) = 2^3/3 - 0^3/3 = 8/3$. When you substitute $2$ for $x$ you don't substitute for the dummy variable of integration.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 2:00
1
$begingroup$
Your confusion was caused by the author's careless reuse of "$x$". They shouldn't have done that.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 12:26
|
show 3 more comments
$begingroup$
Hm.. got it. Just one more thing.. what is the difference between these two expressions? $$int_{0}^{x} y(x) dx$$ and $$int_{0}^{x} y(t) dt$$
$endgroup$
– Nicholas Yamasaki
Dec 21 '18 at 23:13
3
$begingroup$
They are essentially the same, but writing the first one would be very bad expository taste. I hope never to see it anywhere.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 0:40
1
$begingroup$
Yes. But the notation is so weird and (in my view) unacceptable that you shouldn't spend any time or energy trying to parse it correctly.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 1:12
1
$begingroup$
In both cases an antiderivative is the function $f$ that returns the a third of the cube of its input (you can call that $x^3/3$ if you like) and both times $A(2) = f(2)-f(0) = 2^3/3 - 0^3/3 = 8/3$. When you substitute $2$ for $x$ you don't substitute for the dummy variable of integration.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 2:00
1
$begingroup$
Your confusion was caused by the author's careless reuse of "$x$". They shouldn't have done that.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 12:26
$begingroup$
Hm.. got it. Just one more thing.. what is the difference between these two expressions? $$int_{0}^{x} y(x) dx$$ and $$int_{0}^{x} y(t) dt$$
$endgroup$
– Nicholas Yamasaki
Dec 21 '18 at 23:13
$begingroup$
Hm.. got it. Just one more thing.. what is the difference between these two expressions? $$int_{0}^{x} y(x) dx$$ and $$int_{0}^{x} y(t) dt$$
$endgroup$
– Nicholas Yamasaki
Dec 21 '18 at 23:13
3
3
$begingroup$
They are essentially the same, but writing the first one would be very bad expository taste. I hope never to see it anywhere.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 0:40
$begingroup$
They are essentially the same, but writing the first one would be very bad expository taste. I hope never to see it anywhere.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 0:40
1
1
$begingroup$
Yes. But the notation is so weird and (in my view) unacceptable that you shouldn't spend any time or energy trying to parse it correctly.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 1:12
$begingroup$
Yes. But the notation is so weird and (in my view) unacceptable that you shouldn't spend any time or energy trying to parse it correctly.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 1:12
1
1
$begingroup$
In both cases an antiderivative is the function $f$ that returns the a third of the cube of its input (you can call that $x^3/3$ if you like) and both times $A(2) = f(2)-f(0) = 2^3/3 - 0^3/3 = 8/3$. When you substitute $2$ for $x$ you don't substitute for the dummy variable of integration.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 2:00
$begingroup$
In both cases an antiderivative is the function $f$ that returns the a third of the cube of its input (you can call that $x^3/3$ if you like) and both times $A(2) = f(2)-f(0) = 2^3/3 - 0^3/3 = 8/3$. When you substitute $2$ for $x$ you don't substitute for the dummy variable of integration.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 2:00
1
1
$begingroup$
Your confusion was caused by the author's careless reuse of "$x$". They shouldn't have done that.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 12:26
$begingroup$
Your confusion was caused by the author's careless reuse of "$x$". They shouldn't have done that.
$endgroup$
– Ethan Bolker
Dec 22 '18 at 12:26
|
show 3 more comments
$begingroup$
To add to what was already said, I find it helpful to think in programming terms. What is a dummy variable? It's a local variable with very limited scope - a sum, integral, or other similar construct - and it had better not appear anywhere outside that. If it does, as in that initial expression $frac{d}{dx}int_0^x y(x)$, we'll get errors and unpredictable results. For example, here? In order to get something that parses, we have to take that integral with respect to some other variable - which means that inside the integral, $x$ and $y(x)$ are constants. We then get $int_0^x y(x),dt=xy(x)$, so its derivative with respect to $x$ is $y(x)+xy'(x)$, and the solution of the equation is for $y(x)$ to be a constant function.
The equation after the change is clearly what was intended all along, and it doesn't mean the same thing as the equation before the change. It doesn't matter what we call the dummy variable - as long as we're careful not to conflict it with a variable name on the outside and cause out-of-scope errors.
As for leaving the $d?$ off the integral? I'm not a fan of that; it's a delimiter, which clearly marks what we're integrating, and what variable we're integrating with respect to. As far as I'm concerned, it's part of the integral sign. The only way I'd ever leave it off is if the dummy variable is left out entirely. $int_a^b f(x),dx$ or $int_a^b f$ are OK and unambiguous; $int_a^b f(x)$ is malformed.
answered Dec 21 '18 at 22:43
jmerryjmerry
13.7k1629
$endgroup$
add a comment |
$begingroup$
To add to what was already said, I find it helpful to think in programming terms. What is a dummy variable? It's a local variable with very limited scope - a sum, integral, or other similar construct - and it had better not appear anywhere outside that. If it does, as in that initial expression $frac{d}{dx}int_0^x y(x)$, we'll get errors and unpredictable results. For example, here? In order to get something that parses, we have to take that integral with respect to some other variable - which means that inside the integral, $x$ and $y(x)$ are constants. We then get $int_0^x y(x),dt=xy(x)$, so its derivative with respect to $x$ is $y(x)+xy'(x)$, and the solution of the equation is for $y(x)$ to be a constant function.
The equation after the change is clearly what was intended all along, and it doesn't mean the same thing as the equation before the change. It doesn't matter what we call the dummy variable - as long as we're careful not to conflict it with a variable name on the outside and cause out-of-scope errors.
As for leaving the $d?$ off the integral? I'm not a fan of that; it's a delimiter, which clearly marks what we're integrating, and what variable we're integrating with respect to. As far as I'm concerned, it's part of the integral sign. The only way I'd ever leave it off is if the dummy variable is left out entirely. $int_a^b f(x),dx$ or $int_a^b f$ are OK and unambiguous; $int_a^b f(x)$ is malformed.
answered Dec 21 '18 at 22:43
jmerryjmerry
13.7k1629
$endgroup$
add a comment |
$begingroup$
To add to what was already said, I find it helpful to think in programming terms. What is a dummy variable? It's a local variable with very limited scope - a sum, integral, or other similar construct - and it had better not appear anywhere outside that. If it does, as in that initial expression $frac{d}{dx}int_0^x y(x)$, we'll get errors and unpredictable results. For example, here? In order to get something that parses, we have to take that integral with respect to some other variable - which means that inside the integral, $x$ and $y(x)$ are constants. We then get $int_0^x y(x),dt=xy(x)$, so its derivative with respect to $x$ is $y(x)+xy'(x)$, and the solution of the equation is for $y(x)$ to be a constant function.
The equation after the change is clearly what was intended all along, and it doesn't mean the same thing as the equation before the change. It doesn't matter what we call the dummy variable - as long as we're careful not to conflict it with a variable name on the outside and cause out-of-scope errors.
As for leaving the $d?$ off the integral? I'm not a fan of that; it's a delimiter, which clearly marks what we're integrating, and what variable we're integrating with respect to. As far as I'm concerned, it's part of the integral sign. The only way I'd ever leave it off is if the dummy variable is left out entirely. $int_a^b f(x),dx$ or $int_a^b f$ are OK and unambiguous; $int_a^b f(x)$ is malformed.
answered Dec 21 '18 at 22:43
jmerryjmerry
13.7k1629
$endgroup$
To add to what was already said, I find it helpful to think in programming terms. What is a dummy variable? It's a local variable with very limited scope - a sum, integral, or other similar construct - and it had better not appear anywhere outside that. If it does, as in that initial expression $frac{d}{dx}int_0^x y(x)$, we'll get errors and unpredictable results. For example, here? In order to get something that parses, we have to take that integral with respect to some other variable - which means that inside the integral, $x$ and $y(x)$ are constants. We then get $int_0^x y(x),dt=xy(x)$, so its derivative with respect to $x$ is $y(x)+xy'(x)$, and the solution of the equation is for $y(x)$ to be a constant function.
The equation after the change is clearly what was intended all along, and it doesn't mean the same thing as the equation before the change. It doesn't matter what we call the dummy variable - as long as we're careful not to conflict it with a variable name on the outside and cause out-of-scope errors.
As for leaving the $d?$ off the integral? I'm not a fan of that; it's a delimiter, which clearly marks what we're integrating, and what variable we're integrating with respect to. As far as I'm concerned, it's part of the integral sign. The only way I'd ever leave it off is if the dummy variable is left out entirely. $int_a^b f(x),dx$ or $int_a^b f$ are OK and unambiguous; $int_a^b f(x)$ is malformed.
answered Dec 21 '18 at 22:43
jmerryjmerry
13.7k1629
answered Dec 21 '18 at 22:43
jmerryjmerry
13.7k1629
answered Dec 21 '18 at 22:43
jmerryjmerry
13.7k1629
answered Dec 21 '18 at 22:43
jmerryjmerry
13.7k1629
13.7k1629
add a comment |
add a comment |
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1
$begingroup$
Did Strang write $displaystyle dfrac{text d}{text dx}intlimits _{0}^{x}y(x)text dx$?
$endgroup$
– Git Gud
Dec 21 '18 at 21:06
$begingroup$
Oh, thanks, he didn't. But still, could you clarify me why you asked that?
$endgroup$
– Nicholas Yamasaki
Dec 21 '18 at 21:13
2
$begingroup$
Note that Strang immediately erased the expression $int_0^x y(x)$, replacing it with $int_0^x y(t) , dt$. Probably he did this because it is not correct to use the same letter for the dummy variable as for the number $x$ which has already been introduced. (The letter $x$ should not be used for two different things at the same time.) By the way, the "dummy variable" is just a piece of notation and we can eliminate it from the notation entirely if we like by writing $int_0^x y$ (for example). Spivak uses this notation (omitting the dummy variable) in chapter 3 of Calculus on Manifolds.
$endgroup$
– littleO
Dec 22 '18 at 2:24
$begingroup$
Thank you, littleO. But if you don't mind me asking one more question again... why is that it doesn't make sense to create a function where the bound and the $y$ variable are dependent? A function like $A(x) = int_{0}^{x}y(x)dx$ (let's say $y(x) = x^2$) and then you just put $x = 2$ and get $A(2) = 8$? I know this reasoning is a contradiction with the fundamental theorem of calculus (once you should get $A(2) = 8/3$), and by that, it seems that it's impossible to create any kind of relation with the bound and the $y$ variable... why is that?
$endgroup$
– Nicholas Yamasaki
Dec 30 '18 at 19:12