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Solve $fracdda_kbig[sum_k=1^p-a_kcdot y(n-k)big]$
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$fracdda_khaty(n) = fracdda_kbig[sum_k=1^p-a_kcdot y(n-k)big]$ is a formula that is part of this paper (p. 6, Formula (6)). It says the solution is:
$fracdda_khaty(n) = -y(n-k)$.
BUT isn't the solution $fracdda_khaty(n) = sum_k=1^p-y(n-k)$ ?
derivatives summation
asked Jul 22 at 18:15
Alon
104
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up vote
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$fracdda_khaty(n) = fracdda_kbig[sum_k=1^p-a_kcdot y(n-k)big]$ is a formula that is part of this paper (p. 6, Formula (6)). It says the solution is:
$fracdda_khaty(n) = -y(n-k)$.
BUT isn't the solution $fracdda_khaty(n) = sum_k=1^p-y(n-k)$ ?
derivatives summation
asked Jul 22 at 18:15
Alon
104
What happens when you differentiate $-a_k+1cdot y(n-k)$ with respect to $a_k$?
– mrtaurho
Jul 22 at 18:19
If the ak's are independent from on to the other, when you derive by one of the ak, all the other terms are 0?
– Ismasou
Jul 22 at 18:20
@mrtaurho if I understand correctly, this would yield $-y(n-k)$
– Alon
Jul 23 at 10:21
@Ismasou well, that's what I've figured, too.. But how does the paper derive its answer? I mean, the topic is Linear Prediction and this is a thing in science since many years... I don't think it could be wrong...
– Alon
Jul 23 at 10:22
@Alon Yes, exactly. And since this is the given solution I would say it is right. Hence the differential is outside of the sum it should only effect one element - The one to which the $fracdda_k$ refers to.
– mrtaurho
Jul 23 at 10:23
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$fracdda_khaty(n) = fracdda_kbig[sum_k=1^p-a_kcdot y(n-k)big]$ is a formula that is part of this paper (p. 6, Formula (6)). It says the solution is:
$fracdda_khaty(n) = -y(n-k)$.
BUT isn't the solution $fracdda_khaty(n) = sum_k=1^p-y(n-k)$ ?
derivatives summation
asked Jul 22 at 18:15
Alon
104
$fracdda_khaty(n) = fracdda_kbig[sum_k=1^p-a_kcdot y(n-k)big]$ is a formula that is part of this paper (p. 6, Formula (6)). It says the solution is:
$fracdda_khaty(n) = -y(n-k)$.
BUT isn't the solution $fracdda_khaty(n) = sum_k=1^p-y(n-k)$ ?
derivatives summation
asked Jul 22 at 18:15
Alon
104
asked Jul 22 at 18:15
Alon
104
asked Jul 22 at 18:15
Alon
104
asked Jul 22 at 18:15
Alon
104
104
What happens when you differentiate $-a_k+1cdot y(n-k)$ with respect to $a_k$?
– mrtaurho
Jul 22 at 18:19
If the ak's are independent from on to the other, when you derive by one of the ak, all the other terms are 0?
– Ismasou
Jul 22 at 18:20
@mrtaurho if I understand correctly, this would yield $-y(n-k)$
– Alon
Jul 23 at 10:21
@Ismasou well, that's what I've figured, too.. But how does the paper derive its answer? I mean, the topic is Linear Prediction and this is a thing in science since many years... I don't think it could be wrong...
– Alon
Jul 23 at 10:22
@Alon Yes, exactly. And since this is the given solution I would say it is right. Hence the differential is outside of the sum it should only effect one element - The one to which the $fracdda_k$ refers to.
– mrtaurho
Jul 23 at 10:23
add a comment |Â
What happens when you differentiate $-a_k+1cdot y(n-k)$ with respect to $a_k$?
– mrtaurho
Jul 22 at 18:19
If the ak's are independent from on to the other, when you derive by one of the ak, all the other terms are 0?
– Ismasou
Jul 22 at 18:20
@mrtaurho if I understand correctly, this would yield $-y(n-k)$
– Alon
Jul 23 at 10:21
@Ismasou well, that's what I've figured, too.. But how does the paper derive its answer? I mean, the topic is Linear Prediction and this is a thing in science since many years... I don't think it could be wrong...
– Alon
Jul 23 at 10:22
@Alon Yes, exactly. And since this is the given solution I would say it is right. Hence the differential is outside of the sum it should only effect one element - The one to which the $fracdda_k$ refers to.
– mrtaurho
Jul 23 at 10:23
What happens when you differentiate $-a_k+1cdot y(n-k)$ with respect to $a_k$?
– mrtaurho
Jul 22 at 18:19
What happens when you differentiate $-a_k+1cdot y(n-k)$ with respect to $a_k$?
– mrtaurho
Jul 22 at 18:19
If the ak's are independent from on to the other, when you derive by one of the ak, all the other terms are 0?
– Ismasou
Jul 22 at 18:20
If the ak's are independent from on to the other, when you derive by one of the ak, all the other terms are 0?
– Ismasou
Jul 22 at 18:20
@mrtaurho if I understand correctly, this would yield $-y(n-k)$
– Alon
Jul 23 at 10:21
@mrtaurho if I understand correctly, this would yield $-y(n-k)$
– Alon
Jul 23 at 10:21
@Ismasou well, that's what I've figured, too.. But how does the paper derive its answer? I mean, the topic is Linear Prediction and this is a thing in science since many years... I don't think it could be wrong...
– Alon
Jul 23 at 10:22
@Ismasou well, that's what I've figured, too.. But how does the paper derive its answer? I mean, the topic is Linear Prediction and this is a thing in science since many years... I don't think it could be wrong...
– Alon
Jul 23 at 10:22
@Alon Yes, exactly. And since this is the given solution I would say it is right. Hence the differential is outside of the sum it should only effect one element - The one to which the $fracdda_k$ refers to.
– mrtaurho
Jul 23 at 10:23
@Alon Yes, exactly. And since this is the given solution I would say it is right. Hence the differential is outside of the sum it should only effect one element - The one to which the $fracdda_k$ refers to.
– mrtaurho
Jul 23 at 10:23
add a comment |Â
1 Answer
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I guess there is a little propblem concerning the differential operator. My guess is that the $a_k$ there refers to one of the indices, lets call it $k'$ with the restriction $1 leq k' leq p$, and so the sum becomes
$$fracdda_k'haty(n)~=~fracdda_k'left[sum_k=1^p -a_kcdot y(n-k)right]~=~fracdda_k' left[ -a_1cdot y(n-1)-a_2cdot y(n-2)-~cdots - a_k'cdot y(n-k')-cdots -a_p-1cdot y(n-p-1)-a_pcdot y(n-p)right]$$
By differentiation of each of the sum elements with respect to $a_k'$ leads to
$$0 + 0 + cdots -y(n-k') + cdots + 0 + 0 $$
To put this all together yields to
$$fracdda_k'haty(n)~=~-y(n-k')$$
The problem is that the indices of the sum and one single variable are denoted with the same letter.
Otherwise, if the differential would be part of every single sum element, your first guess would be right.
answered Jul 23 at 10:40
mrtaurho
740219
1
Good answer.. for everyone seeking the same advice: There is an excellent paper by Makhoul "Linear Prediction: A Tutorial Review" that does the steps a little more granular for beginners to understand.
– Alon
Jul 28 at 18:40
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I guess there is a little propblem concerning the differential operator. My guess is that the $a_k$ there refers to one of the indices, lets call it $k'$ with the restriction $1 leq k' leq p$, and so the sum becomes
$$fracdda_k'haty(n)~=~fracdda_k'left[sum_k=1^p -a_kcdot y(n-k)right]~=~fracdda_k' left[ -a_1cdot y(n-1)-a_2cdot y(n-2)-~cdots - a_k'cdot y(n-k')-cdots -a_p-1cdot y(n-p-1)-a_pcdot y(n-p)right]$$
By differentiation of each of the sum elements with respect to $a_k'$ leads to
$$0 + 0 + cdots -y(n-k') + cdots + 0 + 0 $$
To put this all together yields to
$$fracdda_k'haty(n)~=~-y(n-k')$$
The problem is that the indices of the sum and one single variable are denoted with the same letter.
Otherwise, if the differential would be part of every single sum element, your first guess would be right.
answered Jul 23 at 10:40
mrtaurho
740219
1
Good answer.. for everyone seeking the same advice: There is an excellent paper by Makhoul "Linear Prediction: A Tutorial Review" that does the steps a little more granular for beginners to understand.
– Alon
Jul 28 at 18:40
add a comment |Â
up vote
0
down vote
accepted
I guess there is a little propblem concerning the differential operator. My guess is that the $a_k$ there refers to one of the indices, lets call it $k'$ with the restriction $1 leq k' leq p$, and so the sum becomes
$$fracdda_k'haty(n)~=~fracdda_k'left[sum_k=1^p -a_kcdot y(n-k)right]~=~fracdda_k' left[ -a_1cdot y(n-1)-a_2cdot y(n-2)-~cdots - a_k'cdot y(n-k')-cdots -a_p-1cdot y(n-p-1)-a_pcdot y(n-p)right]$$
By differentiation of each of the sum elements with respect to $a_k'$ leads to
$$0 + 0 + cdots -y(n-k') + cdots + 0 + 0 $$
To put this all together yields to
$$fracdda_k'haty(n)~=~-y(n-k')$$
The problem is that the indices of the sum and one single variable are denoted with the same letter.
Otherwise, if the differential would be part of every single sum element, your first guess would be right.
answered Jul 23 at 10:40
mrtaurho
740219
1
Good answer.. for everyone seeking the same advice: There is an excellent paper by Makhoul "Linear Prediction: A Tutorial Review" that does the steps a little more granular for beginners to understand.
– Alon
Jul 28 at 18:40
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I guess there is a little propblem concerning the differential operator. My guess is that the $a_k$ there refers to one of the indices, lets call it $k'$ with the restriction $1 leq k' leq p$, and so the sum becomes
$$fracdda_k'haty(n)~=~fracdda_k'left[sum_k=1^p -a_kcdot y(n-k)right]~=~fracdda_k' left[ -a_1cdot y(n-1)-a_2cdot y(n-2)-~cdots - a_k'cdot y(n-k')-cdots -a_p-1cdot y(n-p-1)-a_pcdot y(n-p)right]$$
By differentiation of each of the sum elements with respect to $a_k'$ leads to
$$0 + 0 + cdots -y(n-k') + cdots + 0 + 0 $$
To put this all together yields to
$$fracdda_k'haty(n)~=~-y(n-k')$$
The problem is that the indices of the sum and one single variable are denoted with the same letter.
Otherwise, if the differential would be part of every single sum element, your first guess would be right.
answered Jul 23 at 10:40
mrtaurho
740219
I guess there is a little propblem concerning the differential operator. My guess is that the $a_k$ there refers to one of the indices, lets call it $k'$ with the restriction $1 leq k' leq p$, and so the sum becomes
$$fracdda_k'haty(n)~=~fracdda_k'left[sum_k=1^p -a_kcdot y(n-k)right]~=~fracdda_k' left[ -a_1cdot y(n-1)-a_2cdot y(n-2)-~cdots - a_k'cdot y(n-k')-cdots -a_p-1cdot y(n-p-1)-a_pcdot y(n-p)right]$$
By differentiation of each of the sum elements with respect to $a_k'$ leads to
$$0 + 0 + cdots -y(n-k') + cdots + 0 + 0 $$
To put this all together yields to
$$fracdda_k'haty(n)~=~-y(n-k')$$
The problem is that the indices of the sum and one single variable are denoted with the same letter.
Otherwise, if the differential would be part of every single sum element, your first guess would be right.
answered Jul 23 at 10:40
mrtaurho
740219
answered Jul 23 at 10:40
mrtaurho
740219
answered Jul 23 at 10:40
mrtaurho
740219
answered Jul 23 at 10:40
mrtaurho
740219
740219
1
Good answer.. for everyone seeking the same advice: There is an excellent paper by Makhoul "Linear Prediction: A Tutorial Review" that does the steps a little more granular for beginners to understand.
– Alon
Jul 28 at 18:40
add a comment |Â
1
Good answer.. for everyone seeking the same advice: There is an excellent paper by Makhoul "Linear Prediction: A Tutorial Review" that does the steps a little more granular for beginners to understand.
– Alon
Jul 28 at 18:40
1
1
Good answer.. for everyone seeking the same advice: There is an excellent paper by Makhoul "Linear Prediction: A Tutorial Review" that does the steps a little more granular for beginners to understand.
– Alon
Jul 28 at 18:40
Good answer.. for everyone seeking the same advice: There is an excellent paper by Makhoul "Linear Prediction: A Tutorial Review" that does the steps a little more granular for beginners to understand.
– Alon
Jul 28 at 18:40
add a comment |Â
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What happens when you differentiate $-a_k+1cdot y(n-k)$ with respect to $a_k$?
– mrtaurho
Jul 22 at 18:19
If the ak's are independent from on to the other, when you derive by one of the ak, all the other terms are 0?
– Ismasou
Jul 22 at 18:20
@mrtaurho if I understand correctly, this would yield $-y(n-k)$
– Alon
Jul 23 at 10:21
@Ismasou well, that's what I've figured, too.. But how does the paper derive its answer? I mean, the topic is Linear Prediction and this is a thing in science since many years... I don't think it could be wrong...
– Alon
Jul 23 at 10:22
@Alon Yes, exactly. And since this is the given solution I would say it is right. Hence the differential is outside of the sum it should only effect one element - The one to which the $fracdda_k$ refers to.
– mrtaurho
Jul 23 at 10:23