Mixing
Can I describe a matrix multiplication as a transformation of one matrix by another?
$begingroup$
Let's say I have two matrices $A$ and $B$, and I want to transform my vector $x$ by the product of $A$ and $B$. I could either first transform it by $A$ and then $B$, or multiply $A$ and $B$ $(A*B=C)$ and then transform $x$ by $C$. Would it be correct to say that in the second case I transformed $A$ using $B$ and got $C$, then used $C$ to transform my vector?
Can matrix multiplication be described as a transformation of the first matrix by the second?
matrices linear-transformations
edited Jan 29 at 22:27
timtfj
2,503420
asked Oct 23 '18 at 11:14
Conny DagoConny Dago
1225
$endgroup$
add a comment |
$begingroup$
Let's say I have two matrices $A$ and $B$, and I want to transform my vector $x$ by the product of $A$ and $B$. I could either first transform it by $A$ and then $B$, or multiply $A$ and $B$ $(A*B=C)$ and then transform $x$ by $C$. Would it be correct to say that in the second case I transformed $A$ using $B$ and got $C$, then used $C$ to transform my vector?
Can matrix multiplication be described as a transformation of the first matrix by the second?
matrices linear-transformations
edited Jan 29 at 22:27
timtfj
2,503420
asked Oct 23 '18 at 11:14
Conny DagoConny Dago
1225
$endgroup$
$begingroup$
Related: jpmccarthymaths.com/2017/09/29/…
$endgroup$
– JP McCarthy
Oct 23 '18 at 12:50
add a comment |
$begingroup$
Let's say I have two matrices $A$ and $B$, and I want to transform my vector $x$ by the product of $A$ and $B$. I could either first transform it by $A$ and then $B$, or multiply $A$ and $B$ $(A*B=C)$ and then transform $x$ by $C$. Would it be correct to say that in the second case I transformed $A$ using $B$ and got $C$, then used $C$ to transform my vector?
Can matrix multiplication be described as a transformation of the first matrix by the second?
matrices linear-transformations
edited Jan 29 at 22:27
timtfj
2,503420
asked Oct 23 '18 at 11:14
Conny DagoConny Dago
1225
$endgroup$
Let's say I have two matrices $A$ and $B$, and I want to transform my vector $x$ by the product of $A$ and $B$. I could either first transform it by $A$ and then $B$, or multiply $A$ and $B$ $(A*B=C)$ and then transform $x$ by $C$. Would it be correct to say that in the second case I transformed $A$ using $B$ and got $C$, then used $C$ to transform my vector?
Can matrix multiplication be described as a transformation of the first matrix by the second?
matrices linear-transformations
matrices linear-transformations
edited Jan 29 at 22:27
timtfj
2,503420
asked Oct 23 '18 at 11:14
Conny DagoConny Dago
1225
edited Jan 29 at 22:27
timtfj
2,503420
asked Oct 23 '18 at 11:14
Conny DagoConny Dago
1225
edited Jan 29 at 22:27
timtfj
2,503420
edited Jan 29 at 22:27
timtfj
2,503420
edited Jan 29 at 22:27
timtfj
2,503420
2,503420
asked Oct 23 '18 at 11:14
Conny DagoConny Dago
1225
asked Oct 23 '18 at 11:14
Conny DagoConny Dago
1225
asked Oct 23 '18 at 11:14
Conny DagoConny Dago
1225
1225
$begingroup$
Related: jpmccarthymaths.com/2017/09/29/…
$endgroup$
– JP McCarthy
Oct 23 '18 at 12:50
add a comment |
$begingroup$
Related: jpmccarthymaths.com/2017/09/29/…
$endgroup$
– JP McCarthy
Oct 23 '18 at 12:50
$begingroup$
Related: jpmccarthymaths.com/2017/09/29/…
$endgroup$
– JP McCarthy
Oct 23 '18 at 12:50
$begingroup$
Related: jpmccarthymaths.com/2017/09/29/…
$endgroup$
– JP McCarthy
Oct 23 '18 at 12:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First of all, that's a great question.
Second, it's a little bit muddled, so I'll try to sort it out for you.
Let's use "arrows with flags" ($mapsto$) to indicate where transformations take things. So when you transform $x$ by $A$ and then $B$, you have
$$
x mapsto Ax mapsto B(Ax)
$$
i.e., you multiply $x$ by $A$ (on the left) and then multiply that by $B$ (on the left). Because you can regard $x$ as an $n times 1$ matrix, this last item is really a product of three matrices ($A, B, $ and $x$), and matrix multiplication is associative, so we can rewrite and say that our composed operation is this:
$$
x mapsto Ax mapsto B(Ax) = (BA)x
$$
So the result of first transforming by $A$, and then transforming by $B$, is that you've transformed $x$ by the matrix $BA$. Notice how the order of the matrices is not what you expected in your question (namely, $AB$). But the associative law says that the one I've written is the one it has to be.
I'll just note that the transformation $x mapsto Sx$ (for any matrix $S$) is a linear transformation: if we write
$$
T(x) = Sx,
$$
then $T(x+y) = T(x) + T(y)$, and $T(cx) = c T(x)$ for any real number $c$. For that to make sense, we need to know a way to add vectors (so that $x+y$ makes sense) and a way to multiply them by real numbers, so that $cx$ makes sense. Hold that thought.
For later use, let's call the transformation defined by multiplication by the matrix $J$ by the name $T_J$, so that
$$
T_j : Bbb R^n to Bbb R^n : x mapsto Jx.
$$
So then my first diagram and the comment after it can be summarized by saying that
$$
T_B(T_A(x)) = T_{BA}(x)
$$
for any vector $x in Bbb R^n$.
You also asked another question, which was far more fun, in a way:
You've got an $n times n$ matrix, $A$, from which you can generate a transformation of vectors. We can think of the set of all such matrices, which I'll call $M_{nn}$, as a set, and $A$ is just one element of this set. Now "left multiplication by $B$" is an operation on this set: it takes a matrix $Q$ and turns it into $BQ$, which is another matrix:
begin{align}
M_{nn} &to M_{nn}\
Q &mapsto BQ
end{align}
Let's give this transformation a name, $H_B$, so
$$
H_B: M_{nn} to M_{nn} : Q mapsto BQ
$$
It turns out that $H_B$ is also a linear transformation. For that to make sense, we need to know how to add elements of $M_{nn}$, i.e., how to add matrices, and how to multiply a matrix by a real number. Fortunately, we know both, and can write $H_B(Q + R) = H_B(Q) + H_B(R)$, for instance, which we prove using the distributive law for matrix multiplication.
And now your later observation amounts to saying this:
$$
T_{H_B(A)} = T_B circ T_A
$$
or, in terms that involve an actual vector, $x in Bbb R^n$, it amounts to saying
$$
T_{H_B(A)}(x) = T_B ( T_A (x)).
$$
But the former version is really more interesting: it says that the transformation $H$, acting on the set of all linear-transformations-on-$Bbb R^n$, has a certain property that's related to composition-of-functions. So instead of discussing individual vectors, you've actually made a statement about a higher-level object, namely transformations.
This is typical of the kind of "abstraction" that happens a lot as you advance in mathematics. Kudos to you for thinking of it early!
answered Oct 23 '18 at 11:31
John HughesJohn Hughes
65.1k24293
$endgroup$
add a comment |
$begingroup$
Yes, there are certainly ways to see this. Gilbert Strang discusses various interpretations of matrix multiplication in this lecture Well, I'm pretty sure it's this lecture, but I haven't watched the videos recently.
I think you got a little confused with the or order of operations. It seems like you are computing $ABx,$ so you would multiply $x$ by $B$ first, not by $A$. If you want to consider $A$ as being transformed by $B$, think of $A$ as $<A_1,A_2,dots,A_n>$ where the $A_i$ are the columns of $A$. Then it's easy to see that the columns of $BA$ are just $<BA_1, BA_2, dots,BA_n>.$ That is, we just apply $B$ to the columns of $A$. Other possible interpretations are given in the lecture.
answered Oct 23 '18 at 12:03
saulspatzsaulspatz
17.1k31435
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
First of all, that's a great question.
Second, it's a little bit muddled, so I'll try to sort it out for you.
Let's use "arrows with flags" ($mapsto$) to indicate where transformations take things. So when you transform $x$ by $A$ and then $B$, you have
$$
x mapsto Ax mapsto B(Ax)
$$
i.e., you multiply $x$ by $A$ (on the left) and then multiply that by $B$ (on the left). Because you can regard $x$ as an $n times 1$ matrix, this last item is really a product of three matrices ($A, B, $ and $x$), and matrix multiplication is associative, so we can rewrite and say that our composed operation is this:
$$
x mapsto Ax mapsto B(Ax) = (BA)x
$$
So the result of first transforming by $A$, and then transforming by $B$, is that you've transformed $x$ by the matrix $BA$. Notice how the order of the matrices is not what you expected in your question (namely, $AB$). But the associative law says that the one I've written is the one it has to be.
I'll just note that the transformation $x mapsto Sx$ (for any matrix $S$) is a linear transformation: if we write
$$
T(x) = Sx,
$$
then $T(x+y) = T(x) + T(y)$, and $T(cx) = c T(x)$ for any real number $c$. For that to make sense, we need to know a way to add vectors (so that $x+y$ makes sense) and a way to multiply them by real numbers, so that $cx$ makes sense. Hold that thought.
For later use, let's call the transformation defined by multiplication by the matrix $J$ by the name $T_J$, so that
$$
T_j : Bbb R^n to Bbb R^n : x mapsto Jx.
$$
So then my first diagram and the comment after it can be summarized by saying that
$$
T_B(T_A(x)) = T_{BA}(x)
$$
for any vector $x in Bbb R^n$.
You also asked another question, which was far more fun, in a way:
You've got an $n times n$ matrix, $A$, from which you can generate a transformation of vectors. We can think of the set of all such matrices, which I'll call $M_{nn}$, as a set, and $A$ is just one element of this set. Now "left multiplication by $B$" is an operation on this set: it takes a matrix $Q$ and turns it into $BQ$, which is another matrix:
begin{align}
M_{nn} &to M_{nn}\
Q &mapsto BQ
end{align}
Let's give this transformation a name, $H_B$, so
$$
H_B: M_{nn} to M_{nn} : Q mapsto BQ
$$
It turns out that $H_B$ is also a linear transformation. For that to make sense, we need to know how to add elements of $M_{nn}$, i.e., how to add matrices, and how to multiply a matrix by a real number. Fortunately, we know both, and can write $H_B(Q + R) = H_B(Q) + H_B(R)$, for instance, which we prove using the distributive law for matrix multiplication.
And now your later observation amounts to saying this:
$$
T_{H_B(A)} = T_B circ T_A
$$
or, in terms that involve an actual vector, $x in Bbb R^n$, it amounts to saying
$$
T_{H_B(A)}(x) = T_B ( T_A (x)).
$$
But the former version is really more interesting: it says that the transformation $H$, acting on the set of all linear-transformations-on-$Bbb R^n$, has a certain property that's related to composition-of-functions. So instead of discussing individual vectors, you've actually made a statement about a higher-level object, namely transformations.
This is typical of the kind of "abstraction" that happens a lot as you advance in mathematics. Kudos to you for thinking of it early!
answered Oct 23 '18 at 11:31
John HughesJohn Hughes
65.1k24293
$endgroup$
add a comment |
$begingroup$
First of all, that's a great question.
Second, it's a little bit muddled, so I'll try to sort it out for you.
Let's use "arrows with flags" ($mapsto$) to indicate where transformations take things. So when you transform $x$ by $A$ and then $B$, you have
$$
x mapsto Ax mapsto B(Ax)
$$
i.e., you multiply $x$ by $A$ (on the left) and then multiply that by $B$ (on the left). Because you can regard $x$ as an $n times 1$ matrix, this last item is really a product of three matrices ($A, B, $ and $x$), and matrix multiplication is associative, so we can rewrite and say that our composed operation is this:
$$
x mapsto Ax mapsto B(Ax) = (BA)x
$$
So the result of first transforming by $A$, and then transforming by $B$, is that you've transformed $x$ by the matrix $BA$. Notice how the order of the matrices is not what you expected in your question (namely, $AB$). But the associative law says that the one I've written is the one it has to be.
I'll just note that the transformation $x mapsto Sx$ (for any matrix $S$) is a linear transformation: if we write
$$
T(x) = Sx,
$$
then $T(x+y) = T(x) + T(y)$, and $T(cx) = c T(x)$ for any real number $c$. For that to make sense, we need to know a way to add vectors (so that $x+y$ makes sense) and a way to multiply them by real numbers, so that $cx$ makes sense. Hold that thought.
For later use, let's call the transformation defined by multiplication by the matrix $J$ by the name $T_J$, so that
$$
T_j : Bbb R^n to Bbb R^n : x mapsto Jx.
$$
So then my first diagram and the comment after it can be summarized by saying that
$$
T_B(T_A(x)) = T_{BA}(x)
$$
for any vector $x in Bbb R^n$.
You also asked another question, which was far more fun, in a way:
You've got an $n times n$ matrix, $A$, from which you can generate a transformation of vectors. We can think of the set of all such matrices, which I'll call $M_{nn}$, as a set, and $A$ is just one element of this set. Now "left multiplication by $B$" is an operation on this set: it takes a matrix $Q$ and turns it into $BQ$, which is another matrix:
begin{align}
M_{nn} &to M_{nn}\
Q &mapsto BQ
end{align}
Let's give this transformation a name, $H_B$, so
$$
H_B: M_{nn} to M_{nn} : Q mapsto BQ
$$
It turns out that $H_B$ is also a linear transformation. For that to make sense, we need to know how to add elements of $M_{nn}$, i.e., how to add matrices, and how to multiply a matrix by a real number. Fortunately, we know both, and can write $H_B(Q + R) = H_B(Q) + H_B(R)$, for instance, which we prove using the distributive law for matrix multiplication.
And now your later observation amounts to saying this:
$$
T_{H_B(A)} = T_B circ T_A
$$
or, in terms that involve an actual vector, $x in Bbb R^n$, it amounts to saying
$$
T_{H_B(A)}(x) = T_B ( T_A (x)).
$$
But the former version is really more interesting: it says that the transformation $H$, acting on the set of all linear-transformations-on-$Bbb R^n$, has a certain property that's related to composition-of-functions. So instead of discussing individual vectors, you've actually made a statement about a higher-level object, namely transformations.
This is typical of the kind of "abstraction" that happens a lot as you advance in mathematics. Kudos to you for thinking of it early!
answered Oct 23 '18 at 11:31
John HughesJohn Hughes
65.1k24293
$endgroup$
add a comment |
$begingroup$
First of all, that's a great question.
Second, it's a little bit muddled, so I'll try to sort it out for you.
Let's use "arrows with flags" ($mapsto$) to indicate where transformations take things. So when you transform $x$ by $A$ and then $B$, you have
$$
x mapsto Ax mapsto B(Ax)
$$
i.e., you multiply $x$ by $A$ (on the left) and then multiply that by $B$ (on the left). Because you can regard $x$ as an $n times 1$ matrix, this last item is really a product of three matrices ($A, B, $ and $x$), and matrix multiplication is associative, so we can rewrite and say that our composed operation is this:
$$
x mapsto Ax mapsto B(Ax) = (BA)x
$$
So the result of first transforming by $A$, and then transforming by $B$, is that you've transformed $x$ by the matrix $BA$. Notice how the order of the matrices is not what you expected in your question (namely, $AB$). But the associative law says that the one I've written is the one it has to be.
I'll just note that the transformation $x mapsto Sx$ (for any matrix $S$) is a linear transformation: if we write
$$
T(x) = Sx,
$$
then $T(x+y) = T(x) + T(y)$, and $T(cx) = c T(x)$ for any real number $c$. For that to make sense, we need to know a way to add vectors (so that $x+y$ makes sense) and a way to multiply them by real numbers, so that $cx$ makes sense. Hold that thought.
For later use, let's call the transformation defined by multiplication by the matrix $J$ by the name $T_J$, so that
$$
T_j : Bbb R^n to Bbb R^n : x mapsto Jx.
$$
So then my first diagram and the comment after it can be summarized by saying that
$$
T_B(T_A(x)) = T_{BA}(x)
$$
for any vector $x in Bbb R^n$.
You also asked another question, which was far more fun, in a way:
You've got an $n times n$ matrix, $A$, from which you can generate a transformation of vectors. We can think of the set of all such matrices, which I'll call $M_{nn}$, as a set, and $A$ is just one element of this set. Now "left multiplication by $B$" is an operation on this set: it takes a matrix $Q$ and turns it into $BQ$, which is another matrix:
begin{align}
M_{nn} &to M_{nn}\
Q &mapsto BQ
end{align}
Let's give this transformation a name, $H_B$, so
$$
H_B: M_{nn} to M_{nn} : Q mapsto BQ
$$
It turns out that $H_B$ is also a linear transformation. For that to make sense, we need to know how to add elements of $M_{nn}$, i.e., how to add matrices, and how to multiply a matrix by a real number. Fortunately, we know both, and can write $H_B(Q + R) = H_B(Q) + H_B(R)$, for instance, which we prove using the distributive law for matrix multiplication.
And now your later observation amounts to saying this:
$$
T_{H_B(A)} = T_B circ T_A
$$
or, in terms that involve an actual vector, $x in Bbb R^n$, it amounts to saying
$$
T_{H_B(A)}(x) = T_B ( T_A (x)).
$$
But the former version is really more interesting: it says that the transformation $H$, acting on the set of all linear-transformations-on-$Bbb R^n$, has a certain property that's related to composition-of-functions. So instead of discussing individual vectors, you've actually made a statement about a higher-level object, namely transformations.
This is typical of the kind of "abstraction" that happens a lot as you advance in mathematics. Kudos to you for thinking of it early!
answered Oct 23 '18 at 11:31
John HughesJohn Hughes
65.1k24293
$endgroup$
First of all, that's a great question.
Second, it's a little bit muddled, so I'll try to sort it out for you.
Let's use "arrows with flags" ($mapsto$) to indicate where transformations take things. So when you transform $x$ by $A$ and then $B$, you have
$$
x mapsto Ax mapsto B(Ax)
$$
i.e., you multiply $x$ by $A$ (on the left) and then multiply that by $B$ (on the left). Because you can regard $x$ as an $n times 1$ matrix, this last item is really a product of three matrices ($A, B, $ and $x$), and matrix multiplication is associative, so we can rewrite and say that our composed operation is this:
$$
x mapsto Ax mapsto B(Ax) = (BA)x
$$
So the result of first transforming by $A$, and then transforming by $B$, is that you've transformed $x$ by the matrix $BA$. Notice how the order of the matrices is not what you expected in your question (namely, $AB$). But the associative law says that the one I've written is the one it has to be.
I'll just note that the transformation $x mapsto Sx$ (for any matrix $S$) is a linear transformation: if we write
$$
T(x) = Sx,
$$
then $T(x+y) = T(x) + T(y)$, and $T(cx) = c T(x)$ for any real number $c$. For that to make sense, we need to know a way to add vectors (so that $x+y$ makes sense) and a way to multiply them by real numbers, so that $cx$ makes sense. Hold that thought.
For later use, let's call the transformation defined by multiplication by the matrix $J$ by the name $T_J$, so that
$$
T_j : Bbb R^n to Bbb R^n : x mapsto Jx.
$$
So then my first diagram and the comment after it can be summarized by saying that
$$
T_B(T_A(x)) = T_{BA}(x)
$$
for any vector $x in Bbb R^n$.
You also asked another question, which was far more fun, in a way:
You've got an $n times n$ matrix, $A$, from which you can generate a transformation of vectors. We can think of the set of all such matrices, which I'll call $M_{nn}$, as a set, and $A$ is just one element of this set. Now "left multiplication by $B$" is an operation on this set: it takes a matrix $Q$ and turns it into $BQ$, which is another matrix:
begin{align}
M_{nn} &to M_{nn}\
Q &mapsto BQ
end{align}
Let's give this transformation a name, $H_B$, so
$$
H_B: M_{nn} to M_{nn} : Q mapsto BQ
$$
It turns out that $H_B$ is also a linear transformation. For that to make sense, we need to know how to add elements of $M_{nn}$, i.e., how to add matrices, and how to multiply a matrix by a real number. Fortunately, we know both, and can write $H_B(Q + R) = H_B(Q) + H_B(R)$, for instance, which we prove using the distributive law for matrix multiplication.
And now your later observation amounts to saying this:
$$
T_{H_B(A)} = T_B circ T_A
$$
or, in terms that involve an actual vector, $x in Bbb R^n$, it amounts to saying
$$
T_{H_B(A)}(x) = T_B ( T_A (x)).
$$
But the former version is really more interesting: it says that the transformation $H$, acting on the set of all linear-transformations-on-$Bbb R^n$, has a certain property that's related to composition-of-functions. So instead of discussing individual vectors, you've actually made a statement about a higher-level object, namely transformations.
This is typical of the kind of "abstraction" that happens a lot as you advance in mathematics. Kudos to you for thinking of it early!
answered Oct 23 '18 at 11:31
John HughesJohn Hughes
65.1k24293
edited Oct 23 '18 at 11:51
answered Oct 23 '18 at 11:31
John HughesJohn Hughes
65.1k24293
answered Oct 23 '18 at 11:31
John HughesJohn Hughes
65.1k24293
answered Oct 23 '18 at 11:31
John HughesJohn Hughes
65.1k24293
65.1k24293
add a comment |
add a comment |
$begingroup$
Yes, there are certainly ways to see this. Gilbert Strang discusses various interpretations of matrix multiplication in this lecture Well, I'm pretty sure it's this lecture, but I haven't watched the videos recently.
I think you got a little confused with the or order of operations. It seems like you are computing $ABx,$ so you would multiply $x$ by $B$ first, not by $A$. If you want to consider $A$ as being transformed by $B$, think of $A$ as $<A_1,A_2,dots,A_n>$ where the $A_i$ are the columns of $A$. Then it's easy to see that the columns of $BA$ are just $<BA_1, BA_2, dots,BA_n>.$ That is, we just apply $B$ to the columns of $A$. Other possible interpretations are given in the lecture.
answered Oct 23 '18 at 12:03
saulspatzsaulspatz
17.1k31435
$endgroup$
add a comment |
$begingroup$
Yes, there are certainly ways to see this. Gilbert Strang discusses various interpretations of matrix multiplication in this lecture Well, I'm pretty sure it's this lecture, but I haven't watched the videos recently.
I think you got a little confused with the or order of operations. It seems like you are computing $ABx,$ so you would multiply $x$ by $B$ first, not by $A$. If you want to consider $A$ as being transformed by $B$, think of $A$ as $<A_1,A_2,dots,A_n>$ where the $A_i$ are the columns of $A$. Then it's easy to see that the columns of $BA$ are just $<BA_1, BA_2, dots,BA_n>.$ That is, we just apply $B$ to the columns of $A$. Other possible interpretations are given in the lecture.
answered Oct 23 '18 at 12:03
saulspatzsaulspatz
17.1k31435
$endgroup$
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Yes, there are certainly ways to see this. Gilbert Strang discusses various interpretations of matrix multiplication in this lecture Well, I'm pretty sure it's this lecture, but I haven't watched the videos recently.
I think you got a little confused with the or order of operations. It seems like you are computing $ABx,$ so you would multiply $x$ by $B$ first, not by $A$. If you want to consider $A$ as being transformed by $B$, think of $A$ as $<A_1,A_2,dots,A_n>$ where the $A_i$ are the columns of $A$. Then it's easy to see that the columns of $BA$ are just $<BA_1, BA_2, dots,BA_n>.$ That is, we just apply $B$ to the columns of $A$. Other possible interpretations are given in the lecture.
answered Oct 23 '18 at 12:03
saulspatzsaulspatz
17.1k31435
$endgroup$
Yes, there are certainly ways to see this. Gilbert Strang discusses various interpretations of matrix multiplication in this lecture Well, I'm pretty sure it's this lecture, but I haven't watched the videos recently.
I think you got a little confused with the or order of operations. It seems like you are computing $ABx,$ so you would multiply $x$ by $B$ first, not by $A$. If you want to consider $A$ as being transformed by $B$, think of $A$ as $<A_1,A_2,dots,A_n>$ where the $A_i$ are the columns of $A$. Then it's easy to see that the columns of $BA$ are just $<BA_1, BA_2, dots,BA_n>.$ That is, we just apply $B$ to the columns of $A$. Other possible interpretations are given in the lecture.
answered Oct 23 '18 at 12:03
saulspatzsaulspatz
17.1k31435
answered Oct 23 '18 at 12:03
saulspatzsaulspatz
17.1k31435
answered Oct 23 '18 at 12:03
saulspatzsaulspatz
17.1k31435
answered Oct 23 '18 at 12:03
saulspatzsaulspatz
17.1k31435
17.1k31435
add a comment |
add a comment |
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Related: jpmccarthymaths.com/2017/09/29/…
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– JP McCarthy
Oct 23 '18 at 12:50