Mixing
Prove this property of softmax function?
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Consider the softmax function $f colon mathbb{R}^k rightarrow mathbb{R}^k $ defined by $f((x_1, ldots, x_n)) = (y_1, ldots, y_n)$ where $y_i = frac{e^{x_i}}{sum_j e^{x_j}}$.
Can you show that for every $x, Delta y,delta$, there exists $epsilon$ and $Delta x < delta$ such that $f(x + Delta x) = f(x) + epsilon cdot Delta y$? Also, is there a name for this property for general function? Assume there exists $x'$ such that $y + Delta y = f(x')$.
real-analysis linear-algebra functional-analysis
asked Jan 31 at 21:33
listenerlistener
205
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add a comment |
$begingroup$
Consider the softmax function $f colon mathbb{R}^k rightarrow mathbb{R}^k $ defined by $f((x_1, ldots, x_n)) = (y_1, ldots, y_n)$ where $y_i = frac{e^{x_i}}{sum_j e^{x_j}}$.
Can you show that for every $x, Delta y,delta$, there exists $epsilon$ and $Delta x < delta$ such that $f(x + Delta x) = f(x) + epsilon cdot Delta y$? Also, is there a name for this property for general function? Assume there exists $x'$ such that $y + Delta y = f(x')$.
real-analysis linear-algebra functional-analysis
asked Jan 31 at 21:33
listenerlistener
205
$endgroup$
add a comment |
$begingroup$
Consider the softmax function $f colon mathbb{R}^k rightarrow mathbb{R}^k $ defined by $f((x_1, ldots, x_n)) = (y_1, ldots, y_n)$ where $y_i = frac{e^{x_i}}{sum_j e^{x_j}}$.
Can you show that for every $x, Delta y,delta$, there exists $epsilon$ and $Delta x < delta$ such that $f(x + Delta x) = f(x) + epsilon cdot Delta y$? Also, is there a name for this property for general function? Assume there exists $x'$ such that $y + Delta y = f(x')$.
real-analysis linear-algebra functional-analysis
asked Jan 31 at 21:33
listenerlistener
205
$endgroup$
Consider the softmax function $f colon mathbb{R}^k rightarrow mathbb{R}^k $ defined by $f((x_1, ldots, x_n)) = (y_1, ldots, y_n)$ where $y_i = frac{e^{x_i}}{sum_j e^{x_j}}$.
Can you show that for every $x, Delta y,delta$, there exists $epsilon$ and $Delta x < delta$ such that $f(x + Delta x) = f(x) + epsilon cdot Delta y$? Also, is there a name for this property for general function? Assume there exists $x'$ such that $y + Delta y = f(x')$.
real-analysis linear-algebra functional-analysis
real-analysis linear-algebra functional-analysis
asked Jan 31 at 21:33
listenerlistener
205
asked Jan 31 at 21:33
listenerlistener
205
edited Feb 1 at 5:44
listener
asked Jan 31 at 21:33
listenerlistener
205
asked Jan 31 at 21:33
listenerlistener
205
asked Jan 31 at 21:33
listenerlistener
205
205
add a comment |
add a comment |
1 Answer
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I'm not so sure that you can make that statement. By the construction of the softmax, the sum of its entries is equal to $1$, in other words:
$$sum_{i=1}^k [f(x + Delta x)]_i = 1$$
and
$$sum_{i=1}^k [f(x)]_i = 1$$
However, from what you've written,
$$
f(x+Delta x) = f(x) + epsilon Delta y
$$
It would imply that
$$sum_{i=1}^k [f(x) + epsilon Delta y]_i > 1$$
which leads us to a contradiction if we sum over the entries of $f(x + Delta x)$:
$$
sum_{i=1}^k [f(x+Delta x)]_i = sum_{i=1}^k [f(x) + epsilon Delta y]_i > 1
$$
answered Feb 1 at 0:12
Sean LeeSean Lee
749214
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$begingroup$
updated the question. I meant for any valid vector $Delta y$.
$endgroup$
– listener
Feb 1 at 5:39
add a comment |
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1 Answer
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active
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votes
1 Answer
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active
oldest
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active
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$begingroup$
I'm not so sure that you can make that statement. By the construction of the softmax, the sum of its entries is equal to $1$, in other words:
$$sum_{i=1}^k [f(x + Delta x)]_i = 1$$
and
$$sum_{i=1}^k [f(x)]_i = 1$$
However, from what you've written,
$$
f(x+Delta x) = f(x) + epsilon Delta y
$$
It would imply that
$$sum_{i=1}^k [f(x) + epsilon Delta y]_i > 1$$
which leads us to a contradiction if we sum over the entries of $f(x + Delta x)$:
$$
sum_{i=1}^k [f(x+Delta x)]_i = sum_{i=1}^k [f(x) + epsilon Delta y]_i > 1
$$
answered Feb 1 at 0:12
Sean LeeSean Lee
749214
$endgroup$
$begingroup$
updated the question. I meant for any valid vector $Delta y$.
$endgroup$
– listener
Feb 1 at 5:39
add a comment |
$begingroup$
I'm not so sure that you can make that statement. By the construction of the softmax, the sum of its entries is equal to $1$, in other words:
$$sum_{i=1}^k [f(x + Delta x)]_i = 1$$
and
$$sum_{i=1}^k [f(x)]_i = 1$$
However, from what you've written,
$$
f(x+Delta x) = f(x) + epsilon Delta y
$$
It would imply that
$$sum_{i=1}^k [f(x) + epsilon Delta y]_i > 1$$
which leads us to a contradiction if we sum over the entries of $f(x + Delta x)$:
$$
sum_{i=1}^k [f(x+Delta x)]_i = sum_{i=1}^k [f(x) + epsilon Delta y]_i > 1
$$
answered Feb 1 at 0:12
Sean LeeSean Lee
749214
$endgroup$
$begingroup$
updated the question. I meant for any valid vector $Delta y$.
$endgroup$
– listener
Feb 1 at 5:39
add a comment |
$begingroup$
I'm not so sure that you can make that statement. By the construction of the softmax, the sum of its entries is equal to $1$, in other words:
$$sum_{i=1}^k [f(x + Delta x)]_i = 1$$
and
$$sum_{i=1}^k [f(x)]_i = 1$$
However, from what you've written,
$$
f(x+Delta x) = f(x) + epsilon Delta y
$$
It would imply that
$$sum_{i=1}^k [f(x) + epsilon Delta y]_i > 1$$
which leads us to a contradiction if we sum over the entries of $f(x + Delta x)$:
$$
sum_{i=1}^k [f(x+Delta x)]_i = sum_{i=1}^k [f(x) + epsilon Delta y]_i > 1
$$
answered Feb 1 at 0:12
Sean LeeSean Lee
749214
$endgroup$
I'm not so sure that you can make that statement. By the construction of the softmax, the sum of its entries is equal to $1$, in other words:
$$sum_{i=1}^k [f(x + Delta x)]_i = 1$$
and
$$sum_{i=1}^k [f(x)]_i = 1$$
However, from what you've written,
$$
f(x+Delta x) = f(x) + epsilon Delta y
$$
It would imply that
$$sum_{i=1}^k [f(x) + epsilon Delta y]_i > 1$$
which leads us to a contradiction if we sum over the entries of $f(x + Delta x)$:
$$
sum_{i=1}^k [f(x+Delta x)]_i = sum_{i=1}^k [f(x) + epsilon Delta y]_i > 1
$$
answered Feb 1 at 0:12
Sean LeeSean Lee
749214
answered Feb 1 at 0:12
Sean LeeSean Lee
749214
answered Feb 1 at 0:12
Sean LeeSean Lee
749214
answered Feb 1 at 0:12
Sean LeeSean Lee
749214
749214
$begingroup$
updated the question. I meant for any valid vector $Delta y$.
$endgroup$
– listener
Feb 1 at 5:39
add a comment |
$begingroup$
updated the question. I meant for any valid vector $Delta y$.
$endgroup$
– listener
Feb 1 at 5:39
$begingroup$
updated the question. I meant for any valid vector $Delta y$.
$endgroup$
– listener
Feb 1 at 5:39
$begingroup$
updated the question. I meant for any valid vector $Delta y$.
$endgroup$
– listener
Feb 1 at 5:39
add a comment |
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