Mixing
Difference Equation Initial Value Problem
$begingroup$
Solve the IVP:
$$y^2_{k+2}-4y^2_{k+1}+mcdot y^2_k=k, m in mathbb{R},$$
$y_0=1, y_1=2, y_2=sqrt{13}$
I started by taking $k=0$
$y^2_2-4y^2_1+my^2_0=k$ $Rightarrow13-16+m=k$ $Rightarrow m=k+3$
But I don't know how to proceed any help?
ordinary-differential-equations dynamical-systems
asked Feb 3 at 12:06
VakiPitsiVakiPitsi
31719
$endgroup$
add a comment |
$begingroup$
Solve the IVP:
$$y^2_{k+2}-4y^2_{k+1}+mcdot y^2_k=k, m in mathbb{R},$$
$y_0=1, y_1=2, y_2=sqrt{13}$
I started by taking $k=0$
$y^2_2-4y^2_1+my^2_0=k$ $Rightarrow13-16+m=k$ $Rightarrow m=k+3$
But I don't know how to proceed any help?
ordinary-differential-equations dynamical-systems
asked Feb 3 at 12:06
VakiPitsiVakiPitsi
31719
$endgroup$
$begingroup$
Set $x_k=y_k^2.$ Then it becomes a second-order linear inhomogeneous difference equation with constant coefficients. Look here
$endgroup$
– saulspatz
Feb 3 at 12:34
1
$begingroup$
In your equation for $k=0$, you need to set $k=0$ on both sides. Then $m=3$ for the given initial values, and per the theory of linear recursion, the basis solutions are $1$ and $3^k$, so that by undetermined coefficients the general solution has the form $y^2_k=A+B3^k+(Ck+Dk^2)$ with $A,B,C,D$ completely determined by equation and initial values.
$endgroup$
– LutzL
Feb 3 at 17:59
$begingroup$
Hahaha what a silly mistake! Thanks
$endgroup$
– VakiPitsi
Feb 3 at 19:47
$begingroup$
Are you sure that $k$ in the RHS is also the subscript $k$ used in the LHS of the recursion?
$endgroup$
– Did
Feb 5 at 8:08
add a comment |
$begingroup$
Solve the IVP:
$$y^2_{k+2}-4y^2_{k+1}+mcdot y^2_k=k, m in mathbb{R},$$
$y_0=1, y_1=2, y_2=sqrt{13}$
I started by taking $k=0$
$y^2_2-4y^2_1+my^2_0=k$ $Rightarrow13-16+m=k$ $Rightarrow m=k+3$
But I don't know how to proceed any help?
ordinary-differential-equations dynamical-systems
asked Feb 3 at 12:06
VakiPitsiVakiPitsi
31719
$endgroup$
Solve the IVP:
$$y^2_{k+2}-4y^2_{k+1}+mcdot y^2_k=k, m in mathbb{R},$$
$y_0=1, y_1=2, y_2=sqrt{13}$
I started by taking $k=0$
$y^2_2-4y^2_1+my^2_0=k$ $Rightarrow13-16+m=k$ $Rightarrow m=k+3$
But I don't know how to proceed any help?
ordinary-differential-equations dynamical-systems
ordinary-differential-equations dynamical-systems
asked Feb 3 at 12:06
VakiPitsiVakiPitsi
31719
asked Feb 3 at 12:06
VakiPitsiVakiPitsi
31719
asked Feb 3 at 12:06
VakiPitsiVakiPitsi
31719
asked Feb 3 at 12:06
VakiPitsiVakiPitsi
31719
asked Feb 3 at 12:06
VakiPitsiVakiPitsi
31719
31719
$begingroup$
Set $x_k=y_k^2.$ Then it becomes a second-order linear inhomogeneous difference equation with constant coefficients. Look here
$endgroup$
– saulspatz
Feb 3 at 12:34
1
$begingroup$
In your equation for $k=0$, you need to set $k=0$ on both sides. Then $m=3$ for the given initial values, and per the theory of linear recursion, the basis solutions are $1$ and $3^k$, so that by undetermined coefficients the general solution has the form $y^2_k=A+B3^k+(Ck+Dk^2)$ with $A,B,C,D$ completely determined by equation and initial values.
$endgroup$
– LutzL
Feb 3 at 17:59
$begingroup$
Hahaha what a silly mistake! Thanks
$endgroup$
– VakiPitsi
Feb 3 at 19:47
$begingroup$
Are you sure that $k$ in the RHS is also the subscript $k$ used in the LHS of the recursion?
$endgroup$
– Did
Feb 5 at 8:08
add a comment |
$begingroup$
Set $x_k=y_k^2.$ Then it becomes a second-order linear inhomogeneous difference equation with constant coefficients. Look here
$endgroup$
– saulspatz
Feb 3 at 12:34
1
$begingroup$
In your equation for $k=0$, you need to set $k=0$ on both sides. Then $m=3$ for the given initial values, and per the theory of linear recursion, the basis solutions are $1$ and $3^k$, so that by undetermined coefficients the general solution has the form $y^2_k=A+B3^k+(Ck+Dk^2)$ with $A,B,C,D$ completely determined by equation and initial values.
$endgroup$
– LutzL
Feb 3 at 17:59
$begingroup$
Hahaha what a silly mistake! Thanks
$endgroup$
– VakiPitsi
Feb 3 at 19:47
$begingroup$
Are you sure that $k$ in the RHS is also the subscript $k$ used in the LHS of the recursion?
$endgroup$
– Did
Feb 5 at 8:08
$begingroup$
Set $x_k=y_k^2.$ Then it becomes a second-order linear inhomogeneous difference equation with constant coefficients. Look here
$endgroup$
– saulspatz
Feb 3 at 12:34
$begingroup$
Set $x_k=y_k^2.$ Then it becomes a second-order linear inhomogeneous difference equation with constant coefficients. Look here
$endgroup$
– saulspatz
Feb 3 at 12:34
1
1
$begingroup$
In your equation for $k=0$, you need to set $k=0$ on both sides. Then $m=3$ for the given initial values, and per the theory of linear recursion, the basis solutions are $1$ and $3^k$, so that by undetermined coefficients the general solution has the form $y^2_k=A+B3^k+(Ck+Dk^2)$ with $A,B,C,D$ completely determined by equation and initial values.
$endgroup$
– LutzL
Feb 3 at 17:59
$begingroup$
In your equation for $k=0$, you need to set $k=0$ on both sides. Then $m=3$ for the given initial values, and per the theory of linear recursion, the basis solutions are $1$ and $3^k$, so that by undetermined coefficients the general solution has the form $y^2_k=A+B3^k+(Ck+Dk^2)$ with $A,B,C,D$ completely determined by equation and initial values.
$endgroup$
– LutzL
Feb 3 at 17:59
$begingroup$
Hahaha what a silly mistake! Thanks
$endgroup$
– VakiPitsi
Feb 3 at 19:47
$begingroup$
Hahaha what a silly mistake! Thanks
$endgroup$
– VakiPitsi
Feb 3 at 19:47
$begingroup$
Are you sure that $k$ in the RHS is also the subscript $k$ used in the LHS of the recursion?
$endgroup$
– Did
Feb 5 at 8:08
$begingroup$
Are you sure that $k$ in the RHS is also the subscript $k$ used in the LHS of the recursion?
$endgroup$
– Did
Feb 5 at 8:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I realized that the source I cited in my comment doesn't have an example quite like this one, so I'll sketch the procedure, leaving the details for you.
First, as I said, set $x_k=y_k^2$ giving $$x_{k+2}-4x_{k+1}+mx_k=ktag{1}$$ with initial data
$$begin{align}
x_0&=1\
x_1&=4\
x_2&=13end{align}$$
The general solution of $(1)$ is the general solution of the associated homogeneous equation $$x_{k+2}-4x_{k+1}+mx_k=0tag{2}$$ plus any particular solution of $(1).$
The general solution of $(2)$ is $$x_k=c_1r_1^k+c_2r_2^k$$ where $r_1$ and $r_2$ are the roots of the characteristic polynomial $$x^2-4x+m$$ If there is a double root, which will happen if $m=4,$ then the general solution is a bit different. When $m=4, 2$ is a double root of the characteristic polynomial, and the general solution of $(2)$ becomes $$x_k=c_12^k+c_2k2^k$$
Now we have to guess a particular solution to $(1)$. Since the right-hand side is a linear polynomial in $k$ we guess that a solution in the form of a linear polynomial exists $$x_k=ak+b$$ and we substitute this guess into $(1)$ to solve for $a$ and $b$. When I did did, I found that it worked except in the case $m=3.$ If $m=3$ you should guess that there is a quadratic solution. I didn't actually do this, but it should work.
Now you have three cases, $m=3,$ $m=4,$ and $mneq 3,4,$ and in each case you have a slightly different formula for the general solution of $(1).$ In each case, you can substitute the initial data to find the undetermined coefficients.
EDIT
As LutzL has pointed out, setting $k=0$ and substituting the initial values in $(1)$ gives $m=3,$ so there is really only one case to consider.
answered Feb 3 at 13:17
saulspatzsaulspatz
17.3k31435
$endgroup$
$begingroup$
Thank you for your answer! Very helpfull
$endgroup$
– VakiPitsi
Feb 3 at 13:31
$begingroup$
When $m>4$, the roots of the characteristic polynomial are complex and the solution of the homogeneous equation takes a different form. More precisely, if $r = rho e^{i theta}$ is one of the roots, you get $x_k = rho^k (c_1 cos (theta k) + c_2 sin (theta k))$. Although this can be recovered from the mentioned formula for $k<4$, it is more direct to have this expression.
$endgroup$
– PierreCarre
Feb 3 at 14:10
2
$begingroup$
With the given 3 initial values for an order-2 recursion equation, the value of $m$ is fixed as $m=3$.
$endgroup$
– LutzL
Feb 3 at 17:07
$begingroup$
@LutzL Good point. I didn't even think about that, though I admit I wondered why they gave three initial values. I feel so stupid.
$endgroup$
– saulspatz
Feb 3 at 17:14
add a comment |
$begingroup$
Substituting $w_k = y_k^2$ you get the linear difference equation
$$
w_{k+2}-4 w_{k+1}+m w_k = m.
$$
The general solution of this linear equation can be written as $w_K = x_k + x_k^*$, where $x_k$ is the general solution of the homogeneous equation (set rhs to zero) and $x_k^*$ is a particular solution of the difference equation.
The characteristic polynomial is $p(lambda) = lambda^2-4 lambda + m$, whose roots are $2 pm sqrt{4-m}$, and so depending on whether $m<4$, $m>4$ or $m=4$ you have the expression for $x_k$.
The particular solution $x_k^*$ can be obtained assuming that it is similar to the rhs and plugging into the equation. The choice of particular solution may also depend on $m$.
answered Feb 3 at 12:46
PierreCarrePierreCarre
2,178215
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3098495%2fdifference-equation-initial-value-problem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I realized that the source I cited in my comment doesn't have an example quite like this one, so I'll sketch the procedure, leaving the details for you.
First, as I said, set $x_k=y_k^2$ giving $$x_{k+2}-4x_{k+1}+mx_k=ktag{1}$$ with initial data
$$begin{align}
x_0&=1\
x_1&=4\
x_2&=13end{align}$$
The general solution of $(1)$ is the general solution of the associated homogeneous equation $$x_{k+2}-4x_{k+1}+mx_k=0tag{2}$$ plus any particular solution of $(1).$
The general solution of $(2)$ is $$x_k=c_1r_1^k+c_2r_2^k$$ where $r_1$ and $r_2$ are the roots of the characteristic polynomial $$x^2-4x+m$$ If there is a double root, which will happen if $m=4,$ then the general solution is a bit different. When $m=4, 2$ is a double root of the characteristic polynomial, and the general solution of $(2)$ becomes $$x_k=c_12^k+c_2k2^k$$
Now we have to guess a particular solution to $(1)$. Since the right-hand side is a linear polynomial in $k$ we guess that a solution in the form of a linear polynomial exists $$x_k=ak+b$$ and we substitute this guess into $(1)$ to solve for $a$ and $b$. When I did did, I found that it worked except in the case $m=3.$ If $m=3$ you should guess that there is a quadratic solution. I didn't actually do this, but it should work.
Now you have three cases, $m=3,$ $m=4,$ and $mneq 3,4,$ and in each case you have a slightly different formula for the general solution of $(1).$ In each case, you can substitute the initial data to find the undetermined coefficients.
EDIT
As LutzL has pointed out, setting $k=0$ and substituting the initial values in $(1)$ gives $m=3,$ so there is really only one case to consider.
answered Feb 3 at 13:17
saulspatzsaulspatz
17.3k31435
$endgroup$
$begingroup$
Thank you for your answer! Very helpfull
$endgroup$
– VakiPitsi
Feb 3 at 13:31
$begingroup$
When $m>4$, the roots of the characteristic polynomial are complex and the solution of the homogeneous equation takes a different form. More precisely, if $r = rho e^{i theta}$ is one of the roots, you get $x_k = rho^k (c_1 cos (theta k) + c_2 sin (theta k))$. Although this can be recovered from the mentioned formula for $k<4$, it is more direct to have this expression.
$endgroup$
– PierreCarre
Feb 3 at 14:10
2
$begingroup$
With the given 3 initial values for an order-2 recursion equation, the value of $m$ is fixed as $m=3$.
$endgroup$
– LutzL
Feb 3 at 17:07
$begingroup$
@LutzL Good point. I didn't even think about that, though I admit I wondered why they gave three initial values. I feel so stupid.
$endgroup$
– saulspatz
Feb 3 at 17:14
add a comment |
$begingroup$
I realized that the source I cited in my comment doesn't have an example quite like this one, so I'll sketch the procedure, leaving the details for you.
First, as I said, set $x_k=y_k^2$ giving $$x_{k+2}-4x_{k+1}+mx_k=ktag{1}$$ with initial data
$$begin{align}
x_0&=1\
x_1&=4\
x_2&=13end{align}$$
The general solution of $(1)$ is the general solution of the associated homogeneous equation $$x_{k+2}-4x_{k+1}+mx_k=0tag{2}$$ plus any particular solution of $(1).$
The general solution of $(2)$ is $$x_k=c_1r_1^k+c_2r_2^k$$ where $r_1$ and $r_2$ are the roots of the characteristic polynomial $$x^2-4x+m$$ If there is a double root, which will happen if $m=4,$ then the general solution is a bit different. When $m=4, 2$ is a double root of the characteristic polynomial, and the general solution of $(2)$ becomes $$x_k=c_12^k+c_2k2^k$$
Now we have to guess a particular solution to $(1)$. Since the right-hand side is a linear polynomial in $k$ we guess that a solution in the form of a linear polynomial exists $$x_k=ak+b$$ and we substitute this guess into $(1)$ to solve for $a$ and $b$. When I did did, I found that it worked except in the case $m=3.$ If $m=3$ you should guess that there is a quadratic solution. I didn't actually do this, but it should work.
Now you have three cases, $m=3,$ $m=4,$ and $mneq 3,4,$ and in each case you have a slightly different formula for the general solution of $(1).$ In each case, you can substitute the initial data to find the undetermined coefficients.
EDIT
As LutzL has pointed out, setting $k=0$ and substituting the initial values in $(1)$ gives $m=3,$ so there is really only one case to consider.
answered Feb 3 at 13:17
saulspatzsaulspatz
17.3k31435
$endgroup$
$begingroup$
Thank you for your answer! Very helpfull
$endgroup$
– VakiPitsi
Feb 3 at 13:31
$begingroup$
When $m>4$, the roots of the characteristic polynomial are complex and the solution of the homogeneous equation takes a different form. More precisely, if $r = rho e^{i theta}$ is one of the roots, you get $x_k = rho^k (c_1 cos (theta k) + c_2 sin (theta k))$. Although this can be recovered from the mentioned formula for $k<4$, it is more direct to have this expression.
$endgroup$
– PierreCarre
Feb 3 at 14:10
2
$begingroup$
With the given 3 initial values for an order-2 recursion equation, the value of $m$ is fixed as $m=3$.
$endgroup$
– LutzL
Feb 3 at 17:07
$begingroup$
@LutzL Good point. I didn't even think about that, though I admit I wondered why they gave three initial values. I feel so stupid.
$endgroup$
– saulspatz
Feb 3 at 17:14
add a comment |
$begingroup$
I realized that the source I cited in my comment doesn't have an example quite like this one, so I'll sketch the procedure, leaving the details for you.
First, as I said, set $x_k=y_k^2$ giving $$x_{k+2}-4x_{k+1}+mx_k=ktag{1}$$ with initial data
$$begin{align}
x_0&=1\
x_1&=4\
x_2&=13end{align}$$
The general solution of $(1)$ is the general solution of the associated homogeneous equation $$x_{k+2}-4x_{k+1}+mx_k=0tag{2}$$ plus any particular solution of $(1).$
The general solution of $(2)$ is $$x_k=c_1r_1^k+c_2r_2^k$$ where $r_1$ and $r_2$ are the roots of the characteristic polynomial $$x^2-4x+m$$ If there is a double root, which will happen if $m=4,$ then the general solution is a bit different. When $m=4, 2$ is a double root of the characteristic polynomial, and the general solution of $(2)$ becomes $$x_k=c_12^k+c_2k2^k$$
Now we have to guess a particular solution to $(1)$. Since the right-hand side is a linear polynomial in $k$ we guess that a solution in the form of a linear polynomial exists $$x_k=ak+b$$ and we substitute this guess into $(1)$ to solve for $a$ and $b$. When I did did, I found that it worked except in the case $m=3.$ If $m=3$ you should guess that there is a quadratic solution. I didn't actually do this, but it should work.
Now you have three cases, $m=3,$ $m=4,$ and $mneq 3,4,$ and in each case you have a slightly different formula for the general solution of $(1).$ In each case, you can substitute the initial data to find the undetermined coefficients.
EDIT
As LutzL has pointed out, setting $k=0$ and substituting the initial values in $(1)$ gives $m=3,$ so there is really only one case to consider.
answered Feb 3 at 13:17
saulspatzsaulspatz
17.3k31435
$endgroup$
I realized that the source I cited in my comment doesn't have an example quite like this one, so I'll sketch the procedure, leaving the details for you.
First, as I said, set $x_k=y_k^2$ giving $$x_{k+2}-4x_{k+1}+mx_k=ktag{1}$$ with initial data
$$begin{align}
x_0&=1\
x_1&=4\
x_2&=13end{align}$$
The general solution of $(1)$ is the general solution of the associated homogeneous equation $$x_{k+2}-4x_{k+1}+mx_k=0tag{2}$$ plus any particular solution of $(1).$
The general solution of $(2)$ is $$x_k=c_1r_1^k+c_2r_2^k$$ where $r_1$ and $r_2$ are the roots of the characteristic polynomial $$x^2-4x+m$$ If there is a double root, which will happen if $m=4,$ then the general solution is a bit different. When $m=4, 2$ is a double root of the characteristic polynomial, and the general solution of $(2)$ becomes $$x_k=c_12^k+c_2k2^k$$
Now we have to guess a particular solution to $(1)$. Since the right-hand side is a linear polynomial in $k$ we guess that a solution in the form of a linear polynomial exists $$x_k=ak+b$$ and we substitute this guess into $(1)$ to solve for $a$ and $b$. When I did did, I found that it worked except in the case $m=3.$ If $m=3$ you should guess that there is a quadratic solution. I didn't actually do this, but it should work.
Now you have three cases, $m=3,$ $m=4,$ and $mneq 3,4,$ and in each case you have a slightly different formula for the general solution of $(1).$ In each case, you can substitute the initial data to find the undetermined coefficients.
EDIT
As LutzL has pointed out, setting $k=0$ and substituting the initial values in $(1)$ gives $m=3,$ so there is really only one case to consider.
answered Feb 3 at 13:17
saulspatzsaulspatz
17.3k31435
edited Feb 3 at 18:04
answered Feb 3 at 13:17
saulspatzsaulspatz
17.3k31435
answered Feb 3 at 13:17
saulspatzsaulspatz
17.3k31435
answered Feb 3 at 13:17
saulspatzsaulspatz
17.3k31435
17.3k31435
$begingroup$
Thank you for your answer! Very helpfull
$endgroup$
– VakiPitsi
Feb 3 at 13:31
$begingroup$
When $m>4$, the roots of the characteristic polynomial are complex and the solution of the homogeneous equation takes a different form. More precisely, if $r = rho e^{i theta}$ is one of the roots, you get $x_k = rho^k (c_1 cos (theta k) + c_2 sin (theta k))$. Although this can be recovered from the mentioned formula for $k<4$, it is more direct to have this expression.
$endgroup$
– PierreCarre
Feb 3 at 14:10
2
$begingroup$
With the given 3 initial values for an order-2 recursion equation, the value of $m$ is fixed as $m=3$.
$endgroup$
– LutzL
Feb 3 at 17:07
$begingroup$
@LutzL Good point. I didn't even think about that, though I admit I wondered why they gave three initial values. I feel so stupid.
$endgroup$
– saulspatz
Feb 3 at 17:14
add a comment |
$begingroup$
Thank you for your answer! Very helpfull
$endgroup$
– VakiPitsi
Feb 3 at 13:31
$begingroup$
When $m>4$, the roots of the characteristic polynomial are complex and the solution of the homogeneous equation takes a different form. More precisely, if $r = rho e^{i theta}$ is one of the roots, you get $x_k = rho^k (c_1 cos (theta k) + c_2 sin (theta k))$. Although this can be recovered from the mentioned formula for $k<4$, it is more direct to have this expression.
$endgroup$
– PierreCarre
Feb 3 at 14:10
2
$begingroup$
With the given 3 initial values for an order-2 recursion equation, the value of $m$ is fixed as $m=3$.
$endgroup$
– LutzL
Feb 3 at 17:07
$begingroup$
@LutzL Good point. I didn't even think about that, though I admit I wondered why they gave three initial values. I feel so stupid.
$endgroup$
– saulspatz
Feb 3 at 17:14
$begingroup$
Thank you for your answer! Very helpfull
$endgroup$
– VakiPitsi
Feb 3 at 13:31
$begingroup$
Thank you for your answer! Very helpfull
$endgroup$
– VakiPitsi
Feb 3 at 13:31
$begingroup$
When $m>4$, the roots of the characteristic polynomial are complex and the solution of the homogeneous equation takes a different form. More precisely, if $r = rho e^{i theta}$ is one of the roots, you get $x_k = rho^k (c_1 cos (theta k) + c_2 sin (theta k))$. Although this can be recovered from the mentioned formula for $k<4$, it is more direct to have this expression.
$endgroup$
– PierreCarre
Feb 3 at 14:10
$begingroup$
When $m>4$, the roots of the characteristic polynomial are complex and the solution of the homogeneous equation takes a different form. More precisely, if $r = rho e^{i theta}$ is one of the roots, you get $x_k = rho^k (c_1 cos (theta k) + c_2 sin (theta k))$. Although this can be recovered from the mentioned formula for $k<4$, it is more direct to have this expression.
$endgroup$
– PierreCarre
Feb 3 at 14:10
2
2
$begingroup$
With the given 3 initial values for an order-2 recursion equation, the value of $m$ is fixed as $m=3$.
$endgroup$
– LutzL
Feb 3 at 17:07
$begingroup$
With the given 3 initial values for an order-2 recursion equation, the value of $m$ is fixed as $m=3$.
$endgroup$
– LutzL
Feb 3 at 17:07
$begingroup$
@LutzL Good point. I didn't even think about that, though I admit I wondered why they gave three initial values. I feel so stupid.
$endgroup$
– saulspatz
Feb 3 at 17:14
$begingroup$
@LutzL Good point. I didn't even think about that, though I admit I wondered why they gave three initial values. I feel so stupid.
$endgroup$
– saulspatz
Feb 3 at 17:14
add a comment |
$begingroup$
Substituting $w_k = y_k^2$ you get the linear difference equation
$$
w_{k+2}-4 w_{k+1}+m w_k = m.
$$
The general solution of this linear equation can be written as $w_K = x_k + x_k^*$, where $x_k$ is the general solution of the homogeneous equation (set rhs to zero) and $x_k^*$ is a particular solution of the difference equation.
The characteristic polynomial is $p(lambda) = lambda^2-4 lambda + m$, whose roots are $2 pm sqrt{4-m}$, and so depending on whether $m<4$, $m>4$ or $m=4$ you have the expression for $x_k$.
The particular solution $x_k^*$ can be obtained assuming that it is similar to the rhs and plugging into the equation. The choice of particular solution may also depend on $m$.
answered Feb 3 at 12:46
PierreCarrePierreCarre
2,178215
$endgroup$
add a comment |
$begingroup$
Substituting $w_k = y_k^2$ you get the linear difference equation
$$
w_{k+2}-4 w_{k+1}+m w_k = m.
$$
The general solution of this linear equation can be written as $w_K = x_k + x_k^*$, where $x_k$ is the general solution of the homogeneous equation (set rhs to zero) and $x_k^*$ is a particular solution of the difference equation.
The characteristic polynomial is $p(lambda) = lambda^2-4 lambda + m$, whose roots are $2 pm sqrt{4-m}$, and so depending on whether $m<4$, $m>4$ or $m=4$ you have the expression for $x_k$.
The particular solution $x_k^*$ can be obtained assuming that it is similar to the rhs and plugging into the equation. The choice of particular solution may also depend on $m$.
answered Feb 3 at 12:46
PierreCarrePierreCarre
2,178215
$endgroup$
add a comment |
$begingroup$
Substituting $w_k = y_k^2$ you get the linear difference equation
$$
w_{k+2}-4 w_{k+1}+m w_k = m.
$$
The general solution of this linear equation can be written as $w_K = x_k + x_k^*$, where $x_k$ is the general solution of the homogeneous equation (set rhs to zero) and $x_k^*$ is a particular solution of the difference equation.
The characteristic polynomial is $p(lambda) = lambda^2-4 lambda + m$, whose roots are $2 pm sqrt{4-m}$, and so depending on whether $m<4$, $m>4$ or $m=4$ you have the expression for $x_k$.
The particular solution $x_k^*$ can be obtained assuming that it is similar to the rhs and plugging into the equation. The choice of particular solution may also depend on $m$.
answered Feb 3 at 12:46
PierreCarrePierreCarre
2,178215
$endgroup$
Substituting $w_k = y_k^2$ you get the linear difference equation
$$
w_{k+2}-4 w_{k+1}+m w_k = m.
$$
The general solution of this linear equation can be written as $w_K = x_k + x_k^*$, where $x_k$ is the general solution of the homogeneous equation (set rhs to zero) and $x_k^*$ is a particular solution of the difference equation.
The characteristic polynomial is $p(lambda) = lambda^2-4 lambda + m$, whose roots are $2 pm sqrt{4-m}$, and so depending on whether $m<4$, $m>4$ or $m=4$ you have the expression for $x_k$.
The particular solution $x_k^*$ can be obtained assuming that it is similar to the rhs and plugging into the equation. The choice of particular solution may also depend on $m$.
answered Feb 3 at 12:46
PierreCarrePierreCarre
2,178215
answered Feb 3 at 12:46
PierreCarrePierreCarre
2,178215
answered Feb 3 at 12:46
PierreCarrePierreCarre
2,178215
answered Feb 3 at 12:46
PierreCarrePierreCarre
2,178215
2,178215
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3098495%2fdifference-equation-initial-value-problem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Set $x_k=y_k^2.$ Then it becomes a second-order linear inhomogeneous difference equation with constant coefficients. Look here
$endgroup$
– saulspatz
Feb 3 at 12:34
1
$begingroup$
In your equation for $k=0$, you need to set $k=0$ on both sides. Then $m=3$ for the given initial values, and per the theory of linear recursion, the basis solutions are $1$ and $3^k$, so that by undetermined coefficients the general solution has the form $y^2_k=A+B3^k+(Ck+Dk^2)$ with $A,B,C,D$ completely determined by equation and initial values.
$endgroup$
– LutzL
Feb 3 at 17:59
$begingroup$
Hahaha what a silly mistake! Thanks
$endgroup$
– VakiPitsi
Feb 3 at 19:47
$begingroup$
Are you sure that $k$ in the RHS is also the subscript $k$ used in the LHS of the recursion?
$endgroup$
– Did
Feb 5 at 8:08