Interchange of differentiation and summation
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I came across an example about interchange of differentiation and summation. Can anyone show me how to prove the equation in the picture? Thank you!
self-learning
asked Dec 12 '18 at 3:43
user7989204user7989204
1
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migrated from stats.stackexchange.com Dec 12 '18 at 12:48
This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.
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$begingroup$
I came across an example about interchange of differentiation and summation. Can anyone show me how to prove the equation in the picture? Thank you!
self-learning
asked Dec 12 '18 at 3:43
user7989204user7989204
1
$endgroup$
migrated from stats.stackexchange.com Dec 12 '18 at 12:48
This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.
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Just differentiation: $-frac{partial}{partialtheta}(1-theta)^x=x(1-theta)^{x-1}$
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– StubbornAtom
Dec 12 '18 at 6:05
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There is no prob!em with finite sums---convergence is not an issue
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– kjetil b halvorsen
Dec 12 '18 at 7:20
add a comment |
$begingroup$
I came across an example about interchange of differentiation and summation. Can anyone show me how to prove the equation in the picture? Thank you!
self-learning
asked Dec 12 '18 at 3:43
user7989204user7989204
1
$endgroup$
I came across an example about interchange of differentiation and summation. Can anyone show me how to prove the equation in the picture? Thank you!
self-learning
self-learning
asked Dec 12 '18 at 3:43
user7989204user7989204
1
asked Dec 12 '18 at 3:43
user7989204user7989204
1
asked Dec 12 '18 at 3:43
user7989204user7989204
1
asked Dec 12 '18 at 3:43
user7989204user7989204
1
asked Dec 12 '18 at 3:43
user7989204user7989204
1
1
migrated from stats.stackexchange.com Dec 12 '18 at 12:48
This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.
migrated from stats.stackexchange.com Dec 12 '18 at 12:48
This question came from our site for people interested in statistics, machine learning, data analysis, data mining, and data visualization.
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Just differentiation: $-frac{partial}{partialtheta}(1-theta)^x=x(1-theta)^{x-1}$
$endgroup$
– StubbornAtom
Dec 12 '18 at 6:05
$begingroup$
There is no prob!em with finite sums---convergence is not an issue
$endgroup$
– kjetil b halvorsen
Dec 12 '18 at 7:20
add a comment |
$begingroup$
Just differentiation: $-frac{partial}{partialtheta}(1-theta)^x=x(1-theta)^{x-1}$
$endgroup$
– StubbornAtom
Dec 12 '18 at 6:05
$begingroup$
There is no prob!em with finite sums---convergence is not an issue
$endgroup$
– kjetil b halvorsen
Dec 12 '18 at 7:20
$begingroup$
Just differentiation: $-frac{partial}{partialtheta}(1-theta)^x=x(1-theta)^{x-1}$
$endgroup$
– StubbornAtom
Dec 12 '18 at 6:05
$begingroup$
Just differentiation: $-frac{partial}{partialtheta}(1-theta)^x=x(1-theta)^{x-1}$
$endgroup$
– StubbornAtom
Dec 12 '18 at 6:05
$begingroup$
There is no prob!em with finite sums---convergence is not an issue
$endgroup$
– kjetil b halvorsen
Dec 12 '18 at 7:20
$begingroup$
There is no prob!em with finite sums---convergence is not an issue
$endgroup$
– kjetil b halvorsen
Dec 12 '18 at 7:20
add a comment |
1 Answer
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You do not interchange differentiation and summation. Obviuosly $sum_{x=0}^n theta x (1 - theta)^{x-1} = theta sum_{x=0}^n x (1 - theta)^{x-1}$. In the second step you use that $x (1 - theta)^{x-1} = -frac{partial}{partial theta} (1 - theta)^x$ for each summand.
answered Dec 12 '18 at 15:38
Paul FrostPaul Frost
11.1k3934
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$begingroup$
You do not interchange differentiation and summation. Obviuosly $sum_{x=0}^n theta x (1 - theta)^{x-1} = theta sum_{x=0}^n x (1 - theta)^{x-1}$. In the second step you use that $x (1 - theta)^{x-1} = -frac{partial}{partial theta} (1 - theta)^x$ for each summand.
answered Dec 12 '18 at 15:38
Paul FrostPaul Frost
11.1k3934
$endgroup$
add a comment |
$begingroup$
You do not interchange differentiation and summation. Obviuosly $sum_{x=0}^n theta x (1 - theta)^{x-1} = theta sum_{x=0}^n x (1 - theta)^{x-1}$. In the second step you use that $x (1 - theta)^{x-1} = -frac{partial}{partial theta} (1 - theta)^x$ for each summand.
answered Dec 12 '18 at 15:38
Paul FrostPaul Frost
11.1k3934
$endgroup$
add a comment |
$begingroup$
You do not interchange differentiation and summation. Obviuosly $sum_{x=0}^n theta x (1 - theta)^{x-1} = theta sum_{x=0}^n x (1 - theta)^{x-1}$. In the second step you use that $x (1 - theta)^{x-1} = -frac{partial}{partial theta} (1 - theta)^x$ for each summand.
answered Dec 12 '18 at 15:38
Paul FrostPaul Frost
11.1k3934
$endgroup$
You do not interchange differentiation and summation. Obviuosly $sum_{x=0}^n theta x (1 - theta)^{x-1} = theta sum_{x=0}^n x (1 - theta)^{x-1}$. In the second step you use that $x (1 - theta)^{x-1} = -frac{partial}{partial theta} (1 - theta)^x$ for each summand.
answered Dec 12 '18 at 15:38
Paul FrostPaul Frost
11.1k3934
answered Dec 12 '18 at 15:38
Paul FrostPaul Frost
11.1k3934
answered Dec 12 '18 at 15:38
Paul FrostPaul Frost
11.1k3934
answered Dec 12 '18 at 15:38
Paul FrostPaul Frost
11.1k3934
11.1k3934
add a comment |
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$begingroup$
Just differentiation: $-frac{partial}{partialtheta}(1-theta)^x=x(1-theta)^{x-1}$
$endgroup$
– StubbornAtom
Dec 12 '18 at 6:05
$begingroup$
There is no prob!em with finite sums---convergence is not an issue
$endgroup$
– kjetil b halvorsen
Dec 12 '18 at 7:20