Derivative of cumulative sum function
$begingroup$
A cumulative sum is a sequence of partial sums:
Applying a cumulative sum to ${a, b, c, d}$ gives ${a, a+b, a+b+c, a+b+c+d}$.
A more formal notation, the cumulative sum function takes an $N$-dimensional vector and produces another $N$-dimensional vector:
$C(a) : begin{bmatrix}a_1 \ vdots \a_Nend{bmatrix} xrightarrow{} begin{bmatrix}C_1 \vdots\C_Nend{bmatrix}$
For example:
$begin{bmatrix}1\2\4\5end{bmatrix} xrightarrow{} begin{bmatrix}1 \3\7\12\end{bmatrix}$
The per-element formula is:
$C_N = sum_{j=1}^N a_j$
Calculating the partial derivatives with respect to $a_i$ when $i = j$ and $i ne j$:
$i = j$:
$frac{partial( sum_{j=1}^N a_j)}{partial a_i} = sum_{j=1}^N 1$
and $i ne j$:
$frac{partial( sum_{j=1}^N a_j)}{partial a_i} = sum_{j=1}^N 0$
Does this look correct?
calculus differential
edited Jan 11 at 6:11
Lee
330111
asked Jan 11 at 6:04
user3564870user3564870
61
$endgroup$
add a comment |
$begingroup$
A cumulative sum is a sequence of partial sums:
Applying a cumulative sum to ${a, b, c, d}$ gives ${a, a+b, a+b+c, a+b+c+d}$.
A more formal notation, the cumulative sum function takes an $N$-dimensional vector and produces another $N$-dimensional vector:
$C(a) : begin{bmatrix}a_1 \ vdots \a_Nend{bmatrix} xrightarrow{} begin{bmatrix}C_1 \vdots\C_Nend{bmatrix}$
For example:
$begin{bmatrix}1\2\4\5end{bmatrix} xrightarrow{} begin{bmatrix}1 \3\7\12\end{bmatrix}$
The per-element formula is:
$C_N = sum_{j=1}^N a_j$
Calculating the partial derivatives with respect to $a_i$ when $i = j$ and $i ne j$:
$i = j$:
$frac{partial( sum_{j=1}^N a_j)}{partial a_i} = sum_{j=1}^N 1$
and $i ne j$:
$frac{partial( sum_{j=1}^N a_j)}{partial a_i} = sum_{j=1}^N 0$
Does this look correct?
calculus differential
edited Jan 11 at 6:11
Lee
330111
asked Jan 11 at 6:04
user3564870user3564870
61
$endgroup$
1
$begingroup$
Welcome the Mathematics Stack Exchange community! Great question! A quick tour of the site will help you get the most of your time here.
$endgroup$
– dantopa
Jan 11 at 6:08
add a comment |
$begingroup$
A cumulative sum is a sequence of partial sums:
Applying a cumulative sum to ${a, b, c, d}$ gives ${a, a+b, a+b+c, a+b+c+d}$.
A more formal notation, the cumulative sum function takes an $N$-dimensional vector and produces another $N$-dimensional vector:
$C(a) : begin{bmatrix}a_1 \ vdots \a_Nend{bmatrix} xrightarrow{} begin{bmatrix}C_1 \vdots\C_Nend{bmatrix}$
For example:
$begin{bmatrix}1\2\4\5end{bmatrix} xrightarrow{} begin{bmatrix}1 \3\7\12\end{bmatrix}$
The per-element formula is:
$C_N = sum_{j=1}^N a_j$
Calculating the partial derivatives with respect to $a_i$ when $i = j$ and $i ne j$:
$i = j$:
$frac{partial( sum_{j=1}^N a_j)}{partial a_i} = sum_{j=1}^N 1$
and $i ne j$:
$frac{partial( sum_{j=1}^N a_j)}{partial a_i} = sum_{j=1}^N 0$
Does this look correct?
calculus differential
edited Jan 11 at 6:11
Lee
330111
asked Jan 11 at 6:04
user3564870user3564870
61
$endgroup$
A cumulative sum is a sequence of partial sums:
Applying a cumulative sum to ${a, b, c, d}$ gives ${a, a+b, a+b+c, a+b+c+d}$.
A more formal notation, the cumulative sum function takes an $N$-dimensional vector and produces another $N$-dimensional vector:
$C(a) : begin{bmatrix}a_1 \ vdots \a_Nend{bmatrix} xrightarrow{} begin{bmatrix}C_1 \vdots\C_Nend{bmatrix}$
For example:
$begin{bmatrix}1\2\4\5end{bmatrix} xrightarrow{} begin{bmatrix}1 \3\7\12\end{bmatrix}$
The per-element formula is:
$C_N = sum_{j=1}^N a_j$
Calculating the partial derivatives with respect to $a_i$ when $i = j$ and $i ne j$:
$i = j$:
$frac{partial( sum_{j=1}^N a_j)}{partial a_i} = sum_{j=1}^N 1$
and $i ne j$:
$frac{partial( sum_{j=1}^N a_j)}{partial a_i} = sum_{j=1}^N 0$
Does this look correct?
calculus differential
calculus differential
edited Jan 11 at 6:11
Lee
330111
asked Jan 11 at 6:04
user3564870user3564870
61
edited Jan 11 at 6:11
Lee
330111
asked Jan 11 at 6:04
user3564870user3564870
61
edited Jan 11 at 6:11
Lee
330111
edited Jan 11 at 6:11
Lee
330111
edited Jan 11 at 6:11
Lee
330111
330111
asked Jan 11 at 6:04
user3564870user3564870
61
asked Jan 11 at 6:04
user3564870user3564870
61
asked Jan 11 at 6:04
user3564870user3564870
61
61
1
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Welcome the Mathematics Stack Exchange community! Great question! A quick tour of the site will help you get the most of your time here.
$endgroup$
– dantopa
Jan 11 at 6:08
add a comment |
1
$begingroup$
Welcome the Mathematics Stack Exchange community! Great question! A quick tour of the site will help you get the most of your time here.
$endgroup$
– dantopa
Jan 11 at 6:08
1
1
$begingroup$
Welcome the Mathematics Stack Exchange community! Great question! A quick tour of the site will help you get the most of your time here.
$endgroup$
– dantopa
Jan 11 at 6:08
$begingroup$
Welcome the Mathematics Stack Exchange community! Great question! A quick tour of the site will help you get the most of your time here.
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– dantopa
Jan 11 at 6:08
add a comment |
1 Answer
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$frac{partial C_N}{partial a_i}=frac{partial C_N}{partial a_j}=1$ if $C_N=a_1+cdots+a_N$ and $i,jin{1,2,cdots N}$ for both $i=j$ and $ineq j$.
$frac{partial C_N}{partial a_i}=0$ if $inotin{1,2,cdots,N}$
answered Jan 11 at 6:21
LeeLee
330111
$endgroup$
$begingroup$
Thanks Lee. I'm not so sure of it being 1 in all cases. Say $g_i$ = $a_j$, then the derivative of $g_i$ with respect to $a_j$ is 1 only if i = j, because that's the only time $g_i$ has $a_j$ in it. By the way, I'm modeling this after a similar derivation of the softmax function found at: eli.thegreenplace.net/2016/…
$endgroup$
– user3564870
Jan 11 at 15:58
add a comment |
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$begingroup$
$frac{partial C_N}{partial a_i}=frac{partial C_N}{partial a_j}=1$ if $C_N=a_1+cdots+a_N$ and $i,jin{1,2,cdots N}$ for both $i=j$ and $ineq j$.
$frac{partial C_N}{partial a_i}=0$ if $inotin{1,2,cdots,N}$
answered Jan 11 at 6:21
LeeLee
330111
$endgroup$
$begingroup$
Thanks Lee. I'm not so sure of it being 1 in all cases. Say $g_i$ = $a_j$, then the derivative of $g_i$ with respect to $a_j$ is 1 only if i = j, because that's the only time $g_i$ has $a_j$ in it. By the way, I'm modeling this after a similar derivation of the softmax function found at: eli.thegreenplace.net/2016/…
$endgroup$
– user3564870
Jan 11 at 15:58
add a comment |
$begingroup$
$frac{partial C_N}{partial a_i}=frac{partial C_N}{partial a_j}=1$ if $C_N=a_1+cdots+a_N$ and $i,jin{1,2,cdots N}$ for both $i=j$ and $ineq j$.
$frac{partial C_N}{partial a_i}=0$ if $inotin{1,2,cdots,N}$
answered Jan 11 at 6:21
LeeLee
330111
$endgroup$
$begingroup$
Thanks Lee. I'm not so sure of it being 1 in all cases. Say $g_i$ = $a_j$, then the derivative of $g_i$ with respect to $a_j$ is 1 only if i = j, because that's the only time $g_i$ has $a_j$ in it. By the way, I'm modeling this after a similar derivation of the softmax function found at: eli.thegreenplace.net/2016/…
$endgroup$
– user3564870
Jan 11 at 15:58
add a comment |
$begingroup$
$frac{partial C_N}{partial a_i}=frac{partial C_N}{partial a_j}=1$ if $C_N=a_1+cdots+a_N$ and $i,jin{1,2,cdots N}$ for both $i=j$ and $ineq j$.
$frac{partial C_N}{partial a_i}=0$ if $inotin{1,2,cdots,N}$
answered Jan 11 at 6:21
LeeLee
330111
$endgroup$
$frac{partial C_N}{partial a_i}=frac{partial C_N}{partial a_j}=1$ if $C_N=a_1+cdots+a_N$ and $i,jin{1,2,cdots N}$ for both $i=j$ and $ineq j$.
$frac{partial C_N}{partial a_i}=0$ if $inotin{1,2,cdots,N}$
answered Jan 11 at 6:21
LeeLee
330111
answered Jan 11 at 6:21
LeeLee
330111
answered Jan 11 at 6:21
LeeLee
330111
answered Jan 11 at 6:21
LeeLee
330111
330111
$begingroup$
Thanks Lee. I'm not so sure of it being 1 in all cases. Say $g_i$ = $a_j$, then the derivative of $g_i$ with respect to $a_j$ is 1 only if i = j, because that's the only time $g_i$ has $a_j$ in it. By the way, I'm modeling this after a similar derivation of the softmax function found at: eli.thegreenplace.net/2016/…
$endgroup$
– user3564870
Jan 11 at 15:58
add a comment |
$begingroup$
Thanks Lee. I'm not so sure of it being 1 in all cases. Say $g_i$ = $a_j$, then the derivative of $g_i$ with respect to $a_j$ is 1 only if i = j, because that's the only time $g_i$ has $a_j$ in it. By the way, I'm modeling this after a similar derivation of the softmax function found at: eli.thegreenplace.net/2016/…
$endgroup$
– user3564870
Jan 11 at 15:58
$begingroup$
Thanks Lee. I'm not so sure of it being 1 in all cases. Say $g_i$ = $a_j$, then the derivative of $g_i$ with respect to $a_j$ is 1 only if i = j, because that's the only time $g_i$ has $a_j$ in it. By the way, I'm modeling this after a similar derivation of the softmax function found at: eli.thegreenplace.net/2016/…
$endgroup$
– user3564870
Jan 11 at 15:58
$begingroup$
Thanks Lee. I'm not so sure of it being 1 in all cases. Say $g_i$ = $a_j$, then the derivative of $g_i$ with respect to $a_j$ is 1 only if i = j, because that's the only time $g_i$ has $a_j$ in it. By the way, I'm modeling this after a similar derivation of the softmax function found at: eli.thegreenplace.net/2016/…
$endgroup$
– user3564870
Jan 11 at 15:58
add a comment |
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Welcome the Mathematics Stack Exchange community! Great question! A quick tour of the site will help you get the most of your time here.
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– dantopa
Jan 11 at 6:08