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Solving the quadratic system $v_i = p_i + v_i sum_{j neq i} p_j v_{-j}$

Solving a system of nonlinear (quadratic) equationsSolving system of three quadratic equationsSolving Quadratic system of equationsSolving a quadratic systemSolving non linear system of equationSolving a non linear systemSolving a non-linear system of equations with multiple variablesSolution of a system of quadratic equationsSolving a system of semilinear equations: Will this method work?Solving a system of ternary quadratic equation

1

$begingroup$

Consider the system of equations for $v_i$
$$
v_i = p_i + v_i sum_{j neq i} p_j v_{-j} tag{1}
$$
where $i = pm 1, pm2 dots pm m$ and $sum_{i = -m}^{m} p_i = 1$. Apparently, the solution is given by
$$
v_i = frac{1}{2p_{-i}}left(-x + sqrt{x^2 + 4 p_i p_{-i}}right), tag{2.a}
$$
where $x$ is a solution of
$$
0 = 1 + (m - 1) x - sum_{i = 1}^{m}sqrt{x^2 + 4 p_i p_{-i}}. tag{2.b}
$$
I can't understand where Eq. (2) comes from. The source is Eq. (1) and the following paragraph in Ref. [1], but it's in Russian.

[1] E. B. Dynkin, M. B. Malyutov, “Random walk on groups with a finite number of generators”, Dokl. Akad. Nauk SSSR, 137:5 (1961), 1042–1045

systems-of-equations nonlinear-system

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asked Jun 30 '18 at 19:52

Mr. GMr. G

635521

$endgroup$

add a comment | 

1

$begingroup$

Consider the system of equations for $v_i$
$$
v_i = p_i + v_i sum_{j neq i} p_j v_{-j} tag{1}
$$
where $i = pm 1, pm2 dots pm m$ and $sum_{i = -m}^{m} p_i = 1$. Apparently, the solution is given by
$$
v_i = frac{1}{2p_{-i}}left(-x + sqrt{x^2 + 4 p_i p_{-i}}right), tag{2.a}
$$
where $x$ is a solution of
$$
0 = 1 + (m - 1) x - sum_{i = 1}^{m}sqrt{x^2 + 4 p_i p_{-i}}. tag{2.b}
$$
I can't understand where Eq. (2) comes from. The source is Eq. (1) and the following paragraph in Ref. [1], but it's in Russian.

[1] E. B. Dynkin, M. B. Malyutov, “Random walk on groups with a finite number of generators”, Dokl. Akad. Nauk SSSR, 137:5 (1961), 1042–1045

systems-of-equations nonlinear-system

share|cite|improve this question

asked Jun 30 '18 at 19:52

Mr. GMr. G

635521

$endgroup$

add a comment | 

$begingroup$

Consider the system of equations for $v_i$
$$
v_i = p_i + v_i sum_{j neq i} p_j v_{-j} tag{1}
$$
where $i = pm 1, pm2 dots pm m$ and $sum_{i = -m}^{m} p_i = 1$. Apparently, the solution is given by
$$
v_i = frac{1}{2p_{-i}}left(-x + sqrt{x^2 + 4 p_i p_{-i}}right), tag{2.a}
$$
where $x$ is a solution of
$$
0 = 1 + (m - 1) x - sum_{i = 1}^{m}sqrt{x^2 + 4 p_i p_{-i}}. tag{2.b}
$$
I can't understand where Eq. (2) comes from. The source is Eq. (1) and the following paragraph in Ref. [1], but it's in Russian.

[1] E. B. Dynkin, M. B. Malyutov, “Random walk on groups with a finite number of generators”, Dokl. Akad. Nauk SSSR, 137:5 (1961), 1042–1045

systems-of-equations nonlinear-system

share|cite|improve this question

asked Jun 30 '18 at 19:52

Mr. GMr. G

635521

$endgroup$

share|cite|improve this question

asked Jun 30 '18 at 19:52

Mr. GMr. G

635521

share|cite|improve this question

asked Jun 30 '18 at 19:52

Mr. GMr. G

635521

asked Jun 30 '18 at 19:52

Mr. GMr. G

635521

asked Jun 30 '18 at 19:52

Mr. GMr. G

635521

1 Answer
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0

$begingroup$

This was a while ago, but I guess I'll post the answer for future internet travelers. Observe that

$$
sum_{j neq i} p_{j} v_{-j} = left(sum_{j} p_{j} v_{-j}right) - p_{i} v_{-i} tag{3}
$$

Note that the quantity in brackets is independent of $i$. Therefore, let $y = sum_{j} p_{j} v_{-j}$. Substituting $(3)$ into $(1)$ along with the definition of $y$ gives

$$
v_i + p_i v_{i} v_{-i} = p_i + v_i y tag{4}
$$

Interchanging $i$ for $-i$ gives

$$
v_{-i} + p_{-i} v_{i} v_{-i} = p_{-i} + v_{-i} y tag{4}
$$

Now observe that $(4)$ and $(5)$ form a quadratic system for $v_i$ and $v_{-i}$. Solving and taking the positive root gives

$$
v_{i} = frac{1}{2 p_{-i}}left[-(1-y) + sqrt{(1-y)^2 + 4 p_i p_{-i}} right]
$$

Now, let $x = 1 - y$. We have

begin{align}
x &= 1 - sum_{j} p_{j} v_{-j} \
&= 1 - sum_{j = -m}^{m} frac{1}{2}left[-x + sqrt{x^2 + 4 p_j p_{-j}} right] \
&= 1 - sum_{j = 1}^{m}left[-x + sqrt{x^2 + 4 p_j p_{-j}} right] \
&= 1 + m x - sum_{j = 1}^{m}sqrt{x^2 + 4 p_j p_{-j}} \
0 &= 1 + (m - 1)x - sum_{j = 1}^{m}sqrt{x^2 + 4 p_j p_{-j}}
end{align}

as required.

share|cite|improve this answer

answered Mar 13 at 7:08

Mr. GMr. G

635521

$endgroup$

add a comment | 

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0

$begingroup$

This was a while ago, but I guess I'll post the answer for future internet travelers. Observe that

$$
sum_{j neq i} p_{j} v_{-j} = left(sum_{j} p_{j} v_{-j}right) - p_{i} v_{-i} tag{3}
$$

Note that the quantity in brackets is independent of $i$. Therefore, let $y = sum_{j} p_{j} v_{-j}$. Substituting $(3)$ into $(1)$ along with the definition of $y$ gives

$$
v_i + p_i v_{i} v_{-i} = p_i + v_i y tag{4}
$$

Interchanging $i$ for $-i$ gives

$$
v_{-i} + p_{-i} v_{i} v_{-i} = p_{-i} + v_{-i} y tag{4}
$$

Now observe that $(4)$ and $(5)$ form a quadratic system for $v_i$ and $v_{-i}$. Solving and taking the positive root gives

$$
v_{i} = frac{1}{2 p_{-i}}left[-(1-y) + sqrt{(1-y)^2 + 4 p_i p_{-i}} right]
$$

Now, let $x = 1 - y$. We have

begin{align}
x &= 1 - sum_{j} p_{j} v_{-j} \
&= 1 - sum_{j = -m}^{m} frac{1}{2}left[-x + sqrt{x^2 + 4 p_j p_{-j}} right] \
&= 1 - sum_{j = 1}^{m}left[-x + sqrt{x^2 + 4 p_j p_{-j}} right] \
&= 1 + m x - sum_{j = 1}^{m}sqrt{x^2 + 4 p_j p_{-j}} \
0 &= 1 + (m - 1)x - sum_{j = 1}^{m}sqrt{x^2 + 4 p_j p_{-j}}
end{align}

as required.

share|cite|improve this answer

answered Mar 13 at 7:08

Mr. GMr. G

635521

$endgroup$

add a comment | 

0

$begingroup$

This was a while ago, but I guess I'll post the answer for future internet travelers. Observe that

$$
sum_{j neq i} p_{j} v_{-j} = left(sum_{j} p_{j} v_{-j}right) - p_{i} v_{-i} tag{3}
$$

Note that the quantity in brackets is independent of $i$. Therefore, let $y = sum_{j} p_{j} v_{-j}$. Substituting $(3)$ into $(1)$ along with the definition of $y$ gives

$$
v_i + p_i v_{i} v_{-i} = p_i + v_i y tag{4}
$$

Interchanging $i$ for $-i$ gives

$$
v_{-i} + p_{-i} v_{i} v_{-i} = p_{-i} + v_{-i} y tag{4}
$$

Now observe that $(4)$ and $(5)$ form a quadratic system for $v_i$ and $v_{-i}$. Solving and taking the positive root gives

$$
v_{i} = frac{1}{2 p_{-i}}left[-(1-y) + sqrt{(1-y)^2 + 4 p_i p_{-i}} right]
$$

Now, let $x = 1 - y$. We have

begin{align}
x &= 1 - sum_{j} p_{j} v_{-j} \
&= 1 - sum_{j = -m}^{m} frac{1}{2}left[-x + sqrt{x^2 + 4 p_j p_{-j}} right] \
&= 1 - sum_{j = 1}^{m}left[-x + sqrt{x^2 + 4 p_j p_{-j}} right] \
&= 1 + m x - sum_{j = 1}^{m}sqrt{x^2 + 4 p_j p_{-j}} \
0 &= 1 + (m - 1)x - sum_{j = 1}^{m}sqrt{x^2 + 4 p_j p_{-j}}
end{align}

as required.

share|cite|improve this answer

answered Mar 13 at 7:08

Mr. GMr. G

635521

$endgroup$

add a comment | 

$begingroup$

This was a while ago, but I guess I'll post the answer for future internet travelers. Observe that

$$
sum_{j neq i} p_{j} v_{-j} = left(sum_{j} p_{j} v_{-j}right) - p_{i} v_{-i} tag{3}
$$

Note that the quantity in brackets is independent of $i$. Therefore, let $y = sum_{j} p_{j} v_{-j}$. Substituting $(3)$ into $(1)$ along with the definition of $y$ gives

$$
v_i + p_i v_{i} v_{-i} = p_i + v_i y tag{4}
$$

Interchanging $i$ for $-i$ gives

$$
v_{-i} + p_{-i} v_{i} v_{-i} = p_{-i} + v_{-i} y tag{4}
$$

Now observe that $(4)$ and $(5)$ form a quadratic system for $v_i$ and $v_{-i}$. Solving and taking the positive root gives

$$
v_{i} = frac{1}{2 p_{-i}}left[-(1-y) + sqrt{(1-y)^2 + 4 p_i p_{-i}} right]
$$

Now, let $x = 1 - y$. We have

begin{align}
x &= 1 - sum_{j} p_{j} v_{-j} \
&= 1 - sum_{j = -m}^{m} frac{1}{2}left[-x + sqrt{x^2 + 4 p_j p_{-j}} right] \
&= 1 - sum_{j = 1}^{m}left[-x + sqrt{x^2 + 4 p_j p_{-j}} right] \
&= 1 + m x - sum_{j = 1}^{m}sqrt{x^2 + 4 p_j p_{-j}} \
0 &= 1 + (m - 1)x - sum_{j = 1}^{m}sqrt{x^2 + 4 p_j p_{-j}}
end{align}

as required.

share|cite|improve this answer

answered Mar 13 at 7:08

Mr. GMr. G

635521

$endgroup$

share|cite|improve this answer

answered Mar 13 at 7:08

Mr. GMr. G

635521

answered Mar 13 at 7:08

Mr. GMr. G

635521

answered Mar 13 at 7:08

Mr. GMr. G

635521