Mixing
Solve the recurrence relation $a_n=2a_{n-1}+15a_{n-2}+8$ for $ngeq2$, $a_0=0$, $a_1=1$
$begingroup$
My task:
$a_n=2a_{n-1}+15a_{n-2}+8$ for $ngeq2$, $a_0=0$, $a_1=1$
My solution
$x^{2}-2x-15$
$Delta=64$
$x1=-3 $
$x2=5$
So I am gonna use following formula:
$a_n=ar^{n}+br^{n}$
$a_n=a*(-3)^{n}+b*5^{n}$
$+8$ is the problem, so I am looking for $c$ that $b_n:=a_n+cimplies b_n=2b_{n-1}+15b_{n-2}$
$$b_n=2(b_{n-1}-c)+15(b_{n-2}-c)+8+c=2b_{n-1}+15b_{n-2}+8-16c$$ I am setting $c=frac{1}{2}$ so
$$b_n=2b_{n-1}+15b_{n-2}impliesexists a,,b:,b_n=a*(-3)^{n}+b*5^{n}.$$From $b_0=frac{1}{2},,b_1=-frac{1}{2}$, after finding $a,,b$. Then $a_n=b_n-frac{1}{2}$.
$$a=frac{3}{8}$$
$$b=frac{1}{8}$$
$b_2=frac{52}{8}$
$a_2=frac{52}{8}-frac{1}{2}=6$
Actual $a_2=10, a_3$=43
So $a_2$ from $b_n$ method is not equal to actual $a_n$.
It means I am doing something wrong here, could anyone point out the mistake?
discrete-mathematics recurrence-relations
asked Jan 27 at 19:20
GorossoGorosso
315
$endgroup$
add a comment |
$begingroup$
My task:
$a_n=2a_{n-1}+15a_{n-2}+8$ for $ngeq2$, $a_0=0$, $a_1=1$
My solution
$x^{2}-2x-15$
$Delta=64$
$x1=-3 $
$x2=5$
So I am gonna use following formula:
$a_n=ar^{n}+br^{n}$
$a_n=a*(-3)^{n}+b*5^{n}$
$+8$ is the problem, so I am looking for $c$ that $b_n:=a_n+cimplies b_n=2b_{n-1}+15b_{n-2}$
$$b_n=2(b_{n-1}-c)+15(b_{n-2}-c)+8+c=2b_{n-1}+15b_{n-2}+8-16c$$ I am setting $c=frac{1}{2}$ so
$$b_n=2b_{n-1}+15b_{n-2}impliesexists a,,b:,b_n=a*(-3)^{n}+b*5^{n}.$$From $b_0=frac{1}{2},,b_1=-frac{1}{2}$, after finding $a,,b$. Then $a_n=b_n-frac{1}{2}$.
$$a=frac{3}{8}$$
$$b=frac{1}{8}$$
$b_2=frac{52}{8}$
$a_2=frac{52}{8}-frac{1}{2}=6$
Actual $a_2=10, a_3$=43
So $a_2$ from $b_n$ method is not equal to actual $a_n$.
It means I am doing something wrong here, could anyone point out the mistake?
discrete-mathematics recurrence-relations
asked Jan 27 at 19:20
GorossoGorosso
315
$endgroup$
1
$begingroup$
check your $a,b$ again. Does not hold when $n=1.$
$endgroup$
– dezdichado
Jan 27 at 19:23
1
$begingroup$
now your $b_1$ is incorrect.
$endgroup$
– dezdichado
Jan 27 at 19:43
$begingroup$
@dezdichado how do I find $b_1$ then? I used method located in comments from this post:(link at the end) So I thought $b_1$ is the same as $b_0$ but with '$-$' math.stackexchange.com/questions/3089757/…
$endgroup$
– Gorosso
Jan 27 at 19:47
1
$begingroup$
$b_1 = a_1+frac 12 = frac 32.$
$endgroup$
– dezdichado
Jan 27 at 19:52
2
$begingroup$
I think you should start with a solution that looks like $a r_1^n + b r_2^n + c$, where $r_1$ and $r_2$ are roots of your characteristic polynomial (which you have already computed).
$endgroup$
– Aditya Dua
Jan 27 at 20:02
add a comment |
$begingroup$
My task:
$a_n=2a_{n-1}+15a_{n-2}+8$ for $ngeq2$, $a_0=0$, $a_1=1$
My solution
$x^{2}-2x-15$
$Delta=64$
$x1=-3 $
$x2=5$
So I am gonna use following formula:
$a_n=ar^{n}+br^{n}$
$a_n=a*(-3)^{n}+b*5^{n}$
$+8$ is the problem, so I am looking for $c$ that $b_n:=a_n+cimplies b_n=2b_{n-1}+15b_{n-2}$
$$b_n=2(b_{n-1}-c)+15(b_{n-2}-c)+8+c=2b_{n-1}+15b_{n-2}+8-16c$$ I am setting $c=frac{1}{2}$ so
$$b_n=2b_{n-1}+15b_{n-2}impliesexists a,,b:,b_n=a*(-3)^{n}+b*5^{n}.$$From $b_0=frac{1}{2},,b_1=-frac{1}{2}$, after finding $a,,b$. Then $a_n=b_n-frac{1}{2}$.
$$a=frac{3}{8}$$
$$b=frac{1}{8}$$
$b_2=frac{52}{8}$
$a_2=frac{52}{8}-frac{1}{2}=6$
Actual $a_2=10, a_3$=43
So $a_2$ from $b_n$ method is not equal to actual $a_n$.
It means I am doing something wrong here, could anyone point out the mistake?
discrete-mathematics recurrence-relations
asked Jan 27 at 19:20
GorossoGorosso
315
$endgroup$
My task:
$a_n=2a_{n-1}+15a_{n-2}+8$ for $ngeq2$, $a_0=0$, $a_1=1$
My solution
$x^{2}-2x-15$
$Delta=64$
$x1=-3 $
$x2=5$
So I am gonna use following formula:
$a_n=ar^{n}+br^{n}$
$a_n=a*(-3)^{n}+b*5^{n}$
$+8$ is the problem, so I am looking for $c$ that $b_n:=a_n+cimplies b_n=2b_{n-1}+15b_{n-2}$
$$b_n=2(b_{n-1}-c)+15(b_{n-2}-c)+8+c=2b_{n-1}+15b_{n-2}+8-16c$$ I am setting $c=frac{1}{2}$ so
$$b_n=2b_{n-1}+15b_{n-2}impliesexists a,,b:,b_n=a*(-3)^{n}+b*5^{n}.$$From $b_0=frac{1}{2},,b_1=-frac{1}{2}$, after finding $a,,b$. Then $a_n=b_n-frac{1}{2}$.
$$a=frac{3}{8}$$
$$b=frac{1}{8}$$
$b_2=frac{52}{8}$
$a_2=frac{52}{8}-frac{1}{2}=6$
Actual $a_2=10, a_3$=43
So $a_2$ from $b_n$ method is not equal to actual $a_n$.
It means I am doing something wrong here, could anyone point out the mistake?
discrete-mathematics recurrence-relations
discrete-mathematics recurrence-relations
asked Jan 27 at 19:20
GorossoGorosso
315
asked Jan 27 at 19:20
GorossoGorosso
315
edited Jan 27 at 19:33
Gorosso
asked Jan 27 at 19:20
GorossoGorosso
315
asked Jan 27 at 19:20
GorossoGorosso
315
asked Jan 27 at 19:20
GorossoGorosso
315
315
1
$begingroup$
check your $a,b$ again. Does not hold when $n=1.$
$endgroup$
– dezdichado
Jan 27 at 19:23
1
$begingroup$
now your $b_1$ is incorrect.
$endgroup$
– dezdichado
Jan 27 at 19:43
$begingroup$
@dezdichado how do I find $b_1$ then? I used method located in comments from this post:(link at the end) So I thought $b_1$ is the same as $b_0$ but with '$-$' math.stackexchange.com/questions/3089757/…
$endgroup$
– Gorosso
Jan 27 at 19:47
1
$begingroup$
$b_1 = a_1+frac 12 = frac 32.$
$endgroup$
– dezdichado
Jan 27 at 19:52
2
$begingroup$
I think you should start with a solution that looks like $a r_1^n + b r_2^n + c$, where $r_1$ and $r_2$ are roots of your characteristic polynomial (which you have already computed).
$endgroup$
– Aditya Dua
Jan 27 at 20:02
add a comment |
1
$begingroup$
check your $a,b$ again. Does not hold when $n=1.$
$endgroup$
– dezdichado
Jan 27 at 19:23
1
$begingroup$
now your $b_1$ is incorrect.
$endgroup$
– dezdichado
Jan 27 at 19:43
$begingroup$
@dezdichado how do I find $b_1$ then? I used method located in comments from this post:(link at the end) So I thought $b_1$ is the same as $b_0$ but with '$-$' math.stackexchange.com/questions/3089757/…
$endgroup$
– Gorosso
Jan 27 at 19:47
1
$begingroup$
$b_1 = a_1+frac 12 = frac 32.$
$endgroup$
– dezdichado
Jan 27 at 19:52
2
$begingroup$
I think you should start with a solution that looks like $a r_1^n + b r_2^n + c$, where $r_1$ and $r_2$ are roots of your characteristic polynomial (which you have already computed).
$endgroup$
– Aditya Dua
Jan 27 at 20:02
1
1
$begingroup$
check your $a,b$ again. Does not hold when $n=1.$
$endgroup$
– dezdichado
Jan 27 at 19:23
$begingroup$
check your $a,b$ again. Does not hold when $n=1.$
$endgroup$
– dezdichado
Jan 27 at 19:23
1
1
$begingroup$
now your $b_1$ is incorrect.
$endgroup$
– dezdichado
Jan 27 at 19:43
$begingroup$
now your $b_1$ is incorrect.
$endgroup$
– dezdichado
Jan 27 at 19:43
$begingroup$
@dezdichado how do I find $b_1$ then? I used method located in comments from this post:(link at the end) So I thought $b_1$ is the same as $b_0$ but with '$-$' math.stackexchange.com/questions/3089757/…
$endgroup$
– Gorosso
Jan 27 at 19:47
$begingroup$
@dezdichado how do I find $b_1$ then? I used method located in comments from this post:(link at the end) So I thought $b_1$ is the same as $b_0$ but with '$-$' math.stackexchange.com/questions/3089757/…
$endgroup$
– Gorosso
Jan 27 at 19:47
1
1
$begingroup$
$b_1 = a_1+frac 12 = frac 32.$
$endgroup$
– dezdichado
Jan 27 at 19:52
$begingroup$
$b_1 = a_1+frac 12 = frac 32.$
$endgroup$
– dezdichado
Jan 27 at 19:52
2
2
$begingroup$
I think you should start with a solution that looks like $a r_1^n + b r_2^n + c$, where $r_1$ and $r_2$ are roots of your characteristic polynomial (which you have already computed).
$endgroup$
– Aditya Dua
Jan 27 at 20:02
$begingroup$
I think you should start with a solution that looks like $a r_1^n + b r_2^n + c$, where $r_1$ and $r_2$ are roots of your characteristic polynomial (which you have already computed).
$endgroup$
– Aditya Dua
Jan 27 at 20:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Define $$b_n:=a_n+c=2(b_{n-1}-c)+15(b_{n-2}-c)+8+c=2b_{n-1}+15b_{n-2}+8-16c.$$Choosing $c=frac12$, $$b_n=2b_{n-1}+15b_{n-2}impliesexists a,,b:,b_n=a(-3)^n+b5^n.$$You an obtain $a,,b$ from $b_0=frac12,,b_1=frac{3}{2}$. Then $a_n=b_n-frac12$.
answered Jan 27 at 19:56
J.G.J.G.
32k23250
$endgroup$
add a comment |
$begingroup$
Assuming $a(n) = gamma^n$ and substituting into the homogeneous
$$
gamma^n-2gamma^{n-1}-15gamma^{n-2}=0to gamma^nleft(1-frac{2}{gamma}-frac{15}{gamma^2}right)=0
$$
and solving for $gamma$ we have
$$
a_h(n) = C_1(-3)^n + C_2 5^n
$$
and the particular dictates
$$
a_p(n)-2a_p(n-1)-15a_p(n-2) = 8
$$
so making $a_p(n) = C_0$ and substituting into the particular we have
$$
C_0-2C_0-15C_0 = 8to C_0 = -frac 12
$$
and finally
$$
a(n) = a_h(n)+a_p(n) = C_1(-3)^n + C_2 5^n-frac 12
$$
NOTE
$$
a(0) = C_1+C_2-frac 12 = 0\
a(1) = C_1(-3)+C_25-frac 12 =1
$$
and solving for $C_1, C_2$ gives
$$
a(n) = frac 18left(-4+(-3)^n+3 cdot 5^nright)
$$
answered Jan 27 at 20:36
CesareoCesareo
9,4673517
$endgroup$
$begingroup$
Does C1=a and C2= b? If so, I got $$b=frac{1}{8}$$ and $$a=-frac{1}{8}$$ If I try to calculate $a_2$ from your last line I get results=$$frac{30}{8}$$ which is not equal to 10.
$endgroup$
– Gorosso
Jan 27 at 21:20
$begingroup$
@Gorosso Please. See note attached.
$endgroup$
– Cesareo
Jan 27 at 22:17
$begingroup$
$$C_1=-C_2+frac{1}{2}$$ $$3C_2-frac{3}{2}+5C_2-frac{1}{2}=1$$ $$8C_2=3$$ $$C_2=frac{3}{8}$$ $$frac{1}{8}*(-3)^n+frac{3}{8}*5^n-frac{1}{2}$$ Am I doing something wrong here? Even if I use my version or yours $a_2$ is still not equal to 10
$endgroup$
– Gorosso
Jan 27 at 22:37
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
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%2f3090006%2fsolve-the-recurrence-relation-a-n-2a-n-115a-n-28-for-n-geq2-a-0-0%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
$begingroup$
Define $$b_n:=a_n+c=2(b_{n-1}-c)+15(b_{n-2}-c)+8+c=2b_{n-1}+15b_{n-2}+8-16c.$$Choosing $c=frac12$, $$b_n=2b_{n-1}+15b_{n-2}impliesexists a,,b:,b_n=a(-3)^n+b5^n.$$You an obtain $a,,b$ from $b_0=frac12,,b_1=frac{3}{2}$. Then $a_n=b_n-frac12$.
answered Jan 27 at 19:56
J.G.J.G.
32k23250
$endgroup$
add a comment |
$begingroup$
Define $$b_n:=a_n+c=2(b_{n-1}-c)+15(b_{n-2}-c)+8+c=2b_{n-1}+15b_{n-2}+8-16c.$$Choosing $c=frac12$, $$b_n=2b_{n-1}+15b_{n-2}impliesexists a,,b:,b_n=a(-3)^n+b5^n.$$You an obtain $a,,b$ from $b_0=frac12,,b_1=frac{3}{2}$. Then $a_n=b_n-frac12$.
answered Jan 27 at 19:56
J.G.J.G.
32k23250
$endgroup$
add a comment |
$begingroup$
Define $$b_n:=a_n+c=2(b_{n-1}-c)+15(b_{n-2}-c)+8+c=2b_{n-1}+15b_{n-2}+8-16c.$$Choosing $c=frac12$, $$b_n=2b_{n-1}+15b_{n-2}impliesexists a,,b:,b_n=a(-3)^n+b5^n.$$You an obtain $a,,b$ from $b_0=frac12,,b_1=frac{3}{2}$. Then $a_n=b_n-frac12$.
answered Jan 27 at 19:56
J.G.J.G.
32k23250
$endgroup$
Define $$b_n:=a_n+c=2(b_{n-1}-c)+15(b_{n-2}-c)+8+c=2b_{n-1}+15b_{n-2}+8-16c.$$Choosing $c=frac12$, $$b_n=2b_{n-1}+15b_{n-2}impliesexists a,,b:,b_n=a(-3)^n+b5^n.$$You an obtain $a,,b$ from $b_0=frac12,,b_1=frac{3}{2}$. Then $a_n=b_n-frac12$.
answered Jan 27 at 19:56
J.G.J.G.
32k23250
answered Jan 27 at 19:56
J.G.J.G.
32k23250
answered Jan 27 at 19:56
J.G.J.G.
32k23250
answered Jan 27 at 19:56
J.G.J.G.
32k23250
32k23250
add a comment |
add a comment |
$begingroup$
Assuming $a(n) = gamma^n$ and substituting into the homogeneous
$$
gamma^n-2gamma^{n-1}-15gamma^{n-2}=0to gamma^nleft(1-frac{2}{gamma}-frac{15}{gamma^2}right)=0
$$
and solving for $gamma$ we have
$$
a_h(n) = C_1(-3)^n + C_2 5^n
$$
and the particular dictates
$$
a_p(n)-2a_p(n-1)-15a_p(n-2) = 8
$$
so making $a_p(n) = C_0$ and substituting into the particular we have
$$
C_0-2C_0-15C_0 = 8to C_0 = -frac 12
$$
and finally
$$
a(n) = a_h(n)+a_p(n) = C_1(-3)^n + C_2 5^n-frac 12
$$
NOTE
$$
a(0) = C_1+C_2-frac 12 = 0\
a(1) = C_1(-3)+C_25-frac 12 =1
$$
and solving for $C_1, C_2$ gives
$$
a(n) = frac 18left(-4+(-3)^n+3 cdot 5^nright)
$$
answered Jan 27 at 20:36
CesareoCesareo
9,4673517
$endgroup$
$begingroup$
Does C1=a and C2= b? If so, I got $$b=frac{1}{8}$$ and $$a=-frac{1}{8}$$ If I try to calculate $a_2$ from your last line I get results=$$frac{30}{8}$$ which is not equal to 10.
$endgroup$
– Gorosso
Jan 27 at 21:20
$begingroup$
@Gorosso Please. See note attached.
$endgroup$
– Cesareo
Jan 27 at 22:17
$begingroup$
$$C_1=-C_2+frac{1}{2}$$ $$3C_2-frac{3}{2}+5C_2-frac{1}{2}=1$$ $$8C_2=3$$ $$C_2=frac{3}{8}$$ $$frac{1}{8}*(-3)^n+frac{3}{8}*5^n-frac{1}{2}$$ Am I doing something wrong here? Even if I use my version or yours $a_2$ is still not equal to 10
$endgroup$
– Gorosso
Jan 27 at 22:37
add a comment |
$begingroup$
Assuming $a(n) = gamma^n$ and substituting into the homogeneous
$$
gamma^n-2gamma^{n-1}-15gamma^{n-2}=0to gamma^nleft(1-frac{2}{gamma}-frac{15}{gamma^2}right)=0
$$
and solving for $gamma$ we have
$$
a_h(n) = C_1(-3)^n + C_2 5^n
$$
and the particular dictates
$$
a_p(n)-2a_p(n-1)-15a_p(n-2) = 8
$$
so making $a_p(n) = C_0$ and substituting into the particular we have
$$
C_0-2C_0-15C_0 = 8to C_0 = -frac 12
$$
and finally
$$
a(n) = a_h(n)+a_p(n) = C_1(-3)^n + C_2 5^n-frac 12
$$
NOTE
$$
a(0) = C_1+C_2-frac 12 = 0\
a(1) = C_1(-3)+C_25-frac 12 =1
$$
and solving for $C_1, C_2$ gives
$$
a(n) = frac 18left(-4+(-3)^n+3 cdot 5^nright)
$$
answered Jan 27 at 20:36
CesareoCesareo
9,4673517
$endgroup$
$begingroup$
Does C1=a and C2= b? If so, I got $$b=frac{1}{8}$$ and $$a=-frac{1}{8}$$ If I try to calculate $a_2$ from your last line I get results=$$frac{30}{8}$$ which is not equal to 10.
$endgroup$
– Gorosso
Jan 27 at 21:20
$begingroup$
@Gorosso Please. See note attached.
$endgroup$
– Cesareo
Jan 27 at 22:17
$begingroup$
$$C_1=-C_2+frac{1}{2}$$ $$3C_2-frac{3}{2}+5C_2-frac{1}{2}=1$$ $$8C_2=3$$ $$C_2=frac{3}{8}$$ $$frac{1}{8}*(-3)^n+frac{3}{8}*5^n-frac{1}{2}$$ Am I doing something wrong here? Even if I use my version or yours $a_2$ is still not equal to 10
$endgroup$
– Gorosso
Jan 27 at 22:37
add a comment |
$begingroup$
Assuming $a(n) = gamma^n$ and substituting into the homogeneous
$$
gamma^n-2gamma^{n-1}-15gamma^{n-2}=0to gamma^nleft(1-frac{2}{gamma}-frac{15}{gamma^2}right)=0
$$
and solving for $gamma$ we have
$$
a_h(n) = C_1(-3)^n + C_2 5^n
$$
and the particular dictates
$$
a_p(n)-2a_p(n-1)-15a_p(n-2) = 8
$$
so making $a_p(n) = C_0$ and substituting into the particular we have
$$
C_0-2C_0-15C_0 = 8to C_0 = -frac 12
$$
and finally
$$
a(n) = a_h(n)+a_p(n) = C_1(-3)^n + C_2 5^n-frac 12
$$
NOTE
$$
a(0) = C_1+C_2-frac 12 = 0\
a(1) = C_1(-3)+C_25-frac 12 =1
$$
and solving for $C_1, C_2$ gives
$$
a(n) = frac 18left(-4+(-3)^n+3 cdot 5^nright)
$$
answered Jan 27 at 20:36
CesareoCesareo
9,4673517
$endgroup$
Assuming $a(n) = gamma^n$ and substituting into the homogeneous
$$
gamma^n-2gamma^{n-1}-15gamma^{n-2}=0to gamma^nleft(1-frac{2}{gamma}-frac{15}{gamma^2}right)=0
$$
and solving for $gamma$ we have
$$
a_h(n) = C_1(-3)^n + C_2 5^n
$$
and the particular dictates
$$
a_p(n)-2a_p(n-1)-15a_p(n-2) = 8
$$
so making $a_p(n) = C_0$ and substituting into the particular we have
$$
C_0-2C_0-15C_0 = 8to C_0 = -frac 12
$$
and finally
$$
a(n) = a_h(n)+a_p(n) = C_1(-3)^n + C_2 5^n-frac 12
$$
NOTE
$$
a(0) = C_1+C_2-frac 12 = 0\
a(1) = C_1(-3)+C_25-frac 12 =1
$$
and solving for $C_1, C_2$ gives
$$
a(n) = frac 18left(-4+(-3)^n+3 cdot 5^nright)
$$
answered Jan 27 at 20:36
CesareoCesareo
9,4673517
edited Jan 27 at 22:17
answered Jan 27 at 20:36
CesareoCesareo
9,4673517
answered Jan 27 at 20:36
CesareoCesareo
9,4673517
answered Jan 27 at 20:36
CesareoCesareo
9,4673517
9,4673517
$begingroup$
Does C1=a and C2= b? If so, I got $$b=frac{1}{8}$$ and $$a=-frac{1}{8}$$ If I try to calculate $a_2$ from your last line I get results=$$frac{30}{8}$$ which is not equal to 10.
$endgroup$
– Gorosso
Jan 27 at 21:20
$begingroup$
@Gorosso Please. See note attached.
$endgroup$
– Cesareo
Jan 27 at 22:17
$begingroup$
$$C_1=-C_2+frac{1}{2}$$ $$3C_2-frac{3}{2}+5C_2-frac{1}{2}=1$$ $$8C_2=3$$ $$C_2=frac{3}{8}$$ $$frac{1}{8}*(-3)^n+frac{3}{8}*5^n-frac{1}{2}$$ Am I doing something wrong here? Even if I use my version or yours $a_2$ is still not equal to 10
$endgroup$
– Gorosso
Jan 27 at 22:37
add a comment |
$begingroup$
Does C1=a and C2= b? If so, I got $$b=frac{1}{8}$$ and $$a=-frac{1}{8}$$ If I try to calculate $a_2$ from your last line I get results=$$frac{30}{8}$$ which is not equal to 10.
$endgroup$
– Gorosso
Jan 27 at 21:20
$begingroup$
@Gorosso Please. See note attached.
$endgroup$
– Cesareo
Jan 27 at 22:17
$begingroup$
$$C_1=-C_2+frac{1}{2}$$ $$3C_2-frac{3}{2}+5C_2-frac{1}{2}=1$$ $$8C_2=3$$ $$C_2=frac{3}{8}$$ $$frac{1}{8}*(-3)^n+frac{3}{8}*5^n-frac{1}{2}$$ Am I doing something wrong here? Even if I use my version or yours $a_2$ is still not equal to 10
$endgroup$
– Gorosso
Jan 27 at 22:37
$begingroup$
Does C1=a and C2= b? If so, I got $$b=frac{1}{8}$$ and $$a=-frac{1}{8}$$ If I try to calculate $a_2$ from your last line I get results=$$frac{30}{8}$$ which is not equal to 10.
$endgroup$
– Gorosso
Jan 27 at 21:20
$begingroup$
Does C1=a and C2= b? If so, I got $$b=frac{1}{8}$$ and $$a=-frac{1}{8}$$ If I try to calculate $a_2$ from your last line I get results=$$frac{30}{8}$$ which is not equal to 10.
$endgroup$
– Gorosso
Jan 27 at 21:20
$begingroup$
@Gorosso Please. See note attached.
$endgroup$
– Cesareo
Jan 27 at 22:17
$begingroup$
@Gorosso Please. See note attached.
$endgroup$
– Cesareo
Jan 27 at 22:17
$begingroup$
$$C_1=-C_2+frac{1}{2}$$ $$3C_2-frac{3}{2}+5C_2-frac{1}{2}=1$$ $$8C_2=3$$ $$C_2=frac{3}{8}$$ $$frac{1}{8}*(-3)^n+frac{3}{8}*5^n-frac{1}{2}$$ Am I doing something wrong here? Even if I use my version or yours $a_2$ is still not equal to 10
$endgroup$
– Gorosso
Jan 27 at 22:37
$begingroup$
$$C_1=-C_2+frac{1}{2}$$ $$3C_2-frac{3}{2}+5C_2-frac{1}{2}=1$$ $$8C_2=3$$ $$C_2=frac{3}{8}$$ $$frac{1}{8}*(-3)^n+frac{3}{8}*5^n-frac{1}{2}$$ Am I doing something wrong here? Even if I use my version or yours $a_2$ is still not equal to 10
$endgroup$
– Gorosso
Jan 27 at 22:37
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%2f3090006%2fsolve-the-recurrence-relation-a-n-2a-n-115a-n-28-for-n-geq2-a-0-0%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
1
$begingroup$
check your $a,b$ again. Does not hold when $n=1.$
$endgroup$
– dezdichado
Jan 27 at 19:23
1
$begingroup$
now your $b_1$ is incorrect.
$endgroup$
– dezdichado
Jan 27 at 19:43
$begingroup$
@dezdichado how do I find $b_1$ then? I used method located in comments from this post:(link at the end) So I thought $b_1$ is the same as $b_0$ but with '$-$' math.stackexchange.com/questions/3089757/…
$endgroup$
– Gorosso
Jan 27 at 19:47
1
$begingroup$
$b_1 = a_1+frac 12 = frac 32.$
$endgroup$
– dezdichado
Jan 27 at 19:52
2
$begingroup$
I think you should start with a solution that looks like $a r_1^n + b r_2^n + c$, where $r_1$ and $r_2$ are roots of your characteristic polynomial (which you have already computed).
$endgroup$
– Aditya Dua
Jan 27 at 20:02