Limit of sequence given by $a_n = frac{1}{3}(a_{n-1} + a_{n-2} + a_{n-3})$?
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I have the sequence modeled by the equation $$a_n = dfrac{1}{3}(a_{n-1} + a_{n-2} + a_{n-3}), :a_1 = 1,: a_2 = 2,: a_3 = 3.$$ Simply by manually calculating the sequence I know it converges around $2.3$, but I do not know how to express it in terms of a limit. Since it relies on the previous $3$ values of the sequence I do not know how to express this formulaically.
sequences-and-series limits recurrence-relations
edited Nov 17 at 8:11
user21820
38k541150
asked Nov 17 at 2:15
Elijah
334
|
show 1 more comment
up vote
2
down vote
favorite
I have the sequence modeled by the equation $$a_n = dfrac{1}{3}(a_{n-1} + a_{n-2} + a_{n-3}), :a_1 = 1,: a_2 = 2,: a_3 = 3.$$ Simply by manually calculating the sequence I know it converges around $2.3$, but I do not know how to express it in terms of a limit. Since it relies on the previous $3$ values of the sequence I do not know how to express this formulaically.
sequences-and-series limits recurrence-relations
edited Nov 17 at 8:11
user21820
38k541150
asked Nov 17 at 2:15
Elijah
334
2
Hint: If $displaystylelim_{nrightarrowinfty}a_n=l$ then what happens when you take limit on both sides?
– Yadati Kiran
Nov 17 at 2:22
2
There is a very standard way of solving linear recurrence relation. In this case, if $alpha_k$'s are zeros of $3x^3-x^2-x-1=0$, then $$a_n = sum_{k=1}^{3}c_k alpha_k^n$$ for some constants $c_k$'s, which can be determined by the initial values.
– Sangchul Lee
Nov 17 at 2:33
1
@SangchulLee And one of $alpha_k=1$ in this problem.
– i707107
Nov 17 at 2:35
2
@YadatiKiran I'm not sure it will help here because all you will get is $l=frac{1}{3}[l+l+l]=l$.
– Anurag A
Nov 17 at 3:01
1
@AnuragA: Initially the problem was $a_n = dfrac{1}{3(a_{n-1} + a_{n-2} + a_{n-3})}$. I guess the OP has edited the post.
– Yadati Kiran
Nov 17 at 3:36
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the sequence modeled by the equation $$a_n = dfrac{1}{3}(a_{n-1} + a_{n-2} + a_{n-3}), :a_1 = 1,: a_2 = 2,: a_3 = 3.$$ Simply by manually calculating the sequence I know it converges around $2.3$, but I do not know how to express it in terms of a limit. Since it relies on the previous $3$ values of the sequence I do not know how to express this formulaically.
sequences-and-series limits recurrence-relations
edited Nov 17 at 8:11
user21820
38k541150
asked Nov 17 at 2:15
Elijah
334
I have the sequence modeled by the equation $$a_n = dfrac{1}{3}(a_{n-1} + a_{n-2} + a_{n-3}), :a_1 = 1,: a_2 = 2,: a_3 = 3.$$ Simply by manually calculating the sequence I know it converges around $2.3$, but I do not know how to express it in terms of a limit. Since it relies on the previous $3$ values of the sequence I do not know how to express this formulaically.
sequences-and-series limits recurrence-relations
sequences-and-series limits recurrence-relations
edited Nov 17 at 8:11
user21820
38k541150
asked Nov 17 at 2:15
Elijah
334
edited Nov 17 at 8:11
user21820
38k541150
asked Nov 17 at 2:15
Elijah
334
edited Nov 17 at 8:11
user21820
38k541150
edited Nov 17 at 8:11
user21820
38k541150
edited Nov 17 at 8:11
user21820
38k541150
38k541150
asked Nov 17 at 2:15
Elijah
334
asked Nov 17 at 2:15
Elijah
334
asked Nov 17 at 2:15
Elijah
334
334
2
Hint: If $displaystylelim_{nrightarrowinfty}a_n=l$ then what happens when you take limit on both sides?
– Yadati Kiran
Nov 17 at 2:22
2
There is a very standard way of solving linear recurrence relation. In this case, if $alpha_k$'s are zeros of $3x^3-x^2-x-1=0$, then $$a_n = sum_{k=1}^{3}c_k alpha_k^n$$ for some constants $c_k$'s, which can be determined by the initial values.
– Sangchul Lee
Nov 17 at 2:33
1
@SangchulLee And one of $alpha_k=1$ in this problem.
– i707107
Nov 17 at 2:35
2
@YadatiKiran I'm not sure it will help here because all you will get is $l=frac{1}{3}[l+l+l]=l$.
– Anurag A
Nov 17 at 3:01
1
@AnuragA: Initially the problem was $a_n = dfrac{1}{3(a_{n-1} + a_{n-2} + a_{n-3})}$. I guess the OP has edited the post.
– Yadati Kiran
Nov 17 at 3:36
|
show 1 more comment
2
Hint: If $displaystylelim_{nrightarrowinfty}a_n=l$ then what happens when you take limit on both sides?
– Yadati Kiran
Nov 17 at 2:22
2
There is a very standard way of solving linear recurrence relation. In this case, if $alpha_k$'s are zeros of $3x^3-x^2-x-1=0$, then $$a_n = sum_{k=1}^{3}c_k alpha_k^n$$ for some constants $c_k$'s, which can be determined by the initial values.
– Sangchul Lee
Nov 17 at 2:33
1
@SangchulLee And one of $alpha_k=1$ in this problem.
– i707107
Nov 17 at 2:35
2
@YadatiKiran I'm not sure it will help here because all you will get is $l=frac{1}{3}[l+l+l]=l$.
– Anurag A
Nov 17 at 3:01
1
@AnuragA: Initially the problem was $a_n = dfrac{1}{3(a_{n-1} + a_{n-2} + a_{n-3})}$. I guess the OP has edited the post.
– Yadati Kiran
Nov 17 at 3:36
2
2
Hint: If $displaystylelim_{nrightarrowinfty}a_n=l$ then what happens when you take limit on both sides?
– Yadati Kiran
Nov 17 at 2:22
Hint: If $displaystylelim_{nrightarrowinfty}a_n=l$ then what happens when you take limit on both sides?
– Yadati Kiran
Nov 17 at 2:22
2
2
There is a very standard way of solving linear recurrence relation. In this case, if $alpha_k$'s are zeros of $3x^3-x^2-x-1=0$, then $$a_n = sum_{k=1}^{3}c_k alpha_k^n$$ for some constants $c_k$'s, which can be determined by the initial values.
– Sangchul Lee
Nov 17 at 2:33
There is a very standard way of solving linear recurrence relation. In this case, if $alpha_k$'s are zeros of $3x^3-x^2-x-1=0$, then $$a_n = sum_{k=1}^{3}c_k alpha_k^n$$ for some constants $c_k$'s, which can be determined by the initial values.
– Sangchul Lee
Nov 17 at 2:33
1
1
@SangchulLee And one of $alpha_k=1$ in this problem.
– i707107
Nov 17 at 2:35
@SangchulLee And one of $alpha_k=1$ in this problem.
– i707107
Nov 17 at 2:35
2
2
@YadatiKiran I'm not sure it will help here because all you will get is $l=frac{1}{3}[l+l+l]=l$.
– Anurag A
Nov 17 at 3:01
@YadatiKiran I'm not sure it will help here because all you will get is $l=frac{1}{3}[l+l+l]=l$.
– Anurag A
Nov 17 at 3:01
1
1
@AnuragA: Initially the problem was $a_n = dfrac{1}{3(a_{n-1} + a_{n-2} + a_{n-3})}$. I guess the OP has edited the post.
– Yadati Kiran
Nov 17 at 3:36
@AnuragA: Initially the problem was $a_n = dfrac{1}{3(a_{n-1} + a_{n-2} + a_{n-3})}$. I guess the OP has edited the post.
– Yadati Kiran
Nov 17 at 3:36
|
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
5
down vote
You will find that the limit is $frac16 a_1 + frac13 a_2 + frac12 a_3$
which with $a_1 = 1,: a_2 = 2,: a_3 = 3$ gives a limit of $frac73 approx 2.3333$
More generally, if $a_n$ is the average of the previous $k$ terms then the limit is $$frac{2}{k(k+1)}(a_1+2a_2+3a_3+ldots + ka_k)$$
answered Nov 17 at 2:28
Henry
96.7k474154
add a comment |
up vote
5
down vote
Let
$$b_n=a_n+dfrac23 a_{n-1}+dfrac 13 a_{n-2},tag{*}$$
we have
$$b_n=b_{n-1},qquad b_3=a_3+dfrac23 a_{2}+dfrac 13 a_{1}=frac{14}3.$$
Obviously $b_n$ converges to $b_3=frac{14}3$. If $a_n$ has a limit $l$, by taking limits on both sides of $(*)$,
$$frac{14}3=l+frac 23 l+frac 13 limplies l=frac 73.$$
answered Nov 17 at 2:55
Tianlalu
2,709632
1
If I may ask, how did you come up with general term $b_n$?
– Yadati Kiran
Nov 17 at 3:57
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
You will find that the limit is $frac16 a_1 + frac13 a_2 + frac12 a_3$
which with $a_1 = 1,: a_2 = 2,: a_3 = 3$ gives a limit of $frac73 approx 2.3333$
More generally, if $a_n$ is the average of the previous $k$ terms then the limit is $$frac{2}{k(k+1)}(a_1+2a_2+3a_3+ldots + ka_k)$$
answered Nov 17 at 2:28
Henry
96.7k474154
add a comment |
up vote
5
down vote
You will find that the limit is $frac16 a_1 + frac13 a_2 + frac12 a_3$
which with $a_1 = 1,: a_2 = 2,: a_3 = 3$ gives a limit of $frac73 approx 2.3333$
More generally, if $a_n$ is the average of the previous $k$ terms then the limit is $$frac{2}{k(k+1)}(a_1+2a_2+3a_3+ldots + ka_k)$$
answered Nov 17 at 2:28
Henry
96.7k474154
add a comment |
up vote
5
down vote
up vote
5
down vote
You will find that the limit is $frac16 a_1 + frac13 a_2 + frac12 a_3$
which with $a_1 = 1,: a_2 = 2,: a_3 = 3$ gives a limit of $frac73 approx 2.3333$
More generally, if $a_n$ is the average of the previous $k$ terms then the limit is $$frac{2}{k(k+1)}(a_1+2a_2+3a_3+ldots + ka_k)$$
answered Nov 17 at 2:28
Henry
96.7k474154
You will find that the limit is $frac16 a_1 + frac13 a_2 + frac12 a_3$
which with $a_1 = 1,: a_2 = 2,: a_3 = 3$ gives a limit of $frac73 approx 2.3333$
More generally, if $a_n$ is the average of the previous $k$ terms then the limit is $$frac{2}{k(k+1)}(a_1+2a_2+3a_3+ldots + ka_k)$$
answered Nov 17 at 2:28
Henry
96.7k474154
edited Nov 17 at 3:08
answered Nov 17 at 2:28
Henry
96.7k474154
answered Nov 17 at 2:28
Henry
96.7k474154
answered Nov 17 at 2:28
Henry
96.7k474154
96.7k474154
add a comment |
add a comment |
up vote
5
down vote
Let
$$b_n=a_n+dfrac23 a_{n-1}+dfrac 13 a_{n-2},tag{*}$$
we have
$$b_n=b_{n-1},qquad b_3=a_3+dfrac23 a_{2}+dfrac 13 a_{1}=frac{14}3.$$
Obviously $b_n$ converges to $b_3=frac{14}3$. If $a_n$ has a limit $l$, by taking limits on both sides of $(*)$,
$$frac{14}3=l+frac 23 l+frac 13 limplies l=frac 73.$$
answered Nov 17 at 2:55
Tianlalu
2,709632
1
If I may ask, how did you come up with general term $b_n$?
– Yadati Kiran
Nov 17 at 3:57
add a comment |
up vote
5
down vote
Let
$$b_n=a_n+dfrac23 a_{n-1}+dfrac 13 a_{n-2},tag{*}$$
we have
$$b_n=b_{n-1},qquad b_3=a_3+dfrac23 a_{2}+dfrac 13 a_{1}=frac{14}3.$$
Obviously $b_n$ converges to $b_3=frac{14}3$. If $a_n$ has a limit $l$, by taking limits on both sides of $(*)$,
$$frac{14}3=l+frac 23 l+frac 13 limplies l=frac 73.$$
answered Nov 17 at 2:55
Tianlalu
2,709632
1
If I may ask, how did you come up with general term $b_n$?
– Yadati Kiran
Nov 17 at 3:57
add a comment |
up vote
5
down vote
up vote
5
down vote
Let
$$b_n=a_n+dfrac23 a_{n-1}+dfrac 13 a_{n-2},tag{*}$$
we have
$$b_n=b_{n-1},qquad b_3=a_3+dfrac23 a_{2}+dfrac 13 a_{1}=frac{14}3.$$
Obviously $b_n$ converges to $b_3=frac{14}3$. If $a_n$ has a limit $l$, by taking limits on both sides of $(*)$,
$$frac{14}3=l+frac 23 l+frac 13 limplies l=frac 73.$$
answered Nov 17 at 2:55
Tianlalu
2,709632
Let
$$b_n=a_n+dfrac23 a_{n-1}+dfrac 13 a_{n-2},tag{*}$$
we have
$$b_n=b_{n-1},qquad b_3=a_3+dfrac23 a_{2}+dfrac 13 a_{1}=frac{14}3.$$
Obviously $b_n$ converges to $b_3=frac{14}3$. If $a_n$ has a limit $l$, by taking limits on both sides of $(*)$,
$$frac{14}3=l+frac 23 l+frac 13 limplies l=frac 73.$$
answered Nov 17 at 2:55
Tianlalu
2,709632
edited Nov 17 at 3:10
answered Nov 17 at 2:55
Tianlalu
2,709632
answered Nov 17 at 2:55
Tianlalu
2,709632
answered Nov 17 at 2:55
Tianlalu
2,709632
2,709632
1
If I may ask, how did you come up with general term $b_n$?
– Yadati Kiran
Nov 17 at 3:57
add a comment |
1
If I may ask, how did you come up with general term $b_n$?
– Yadati Kiran
Nov 17 at 3:57
1
1
If I may ask, how did you come up with general term $b_n$?
– Yadati Kiran
Nov 17 at 3:57
If I may ask, how did you come up with general term $b_n$?
– Yadati Kiran
Nov 17 at 3:57
add a comment |
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2
Hint: If $displaystylelim_{nrightarrowinfty}a_n=l$ then what happens when you take limit on both sides?
– Yadati Kiran
Nov 17 at 2:22
2
There is a very standard way of solving linear recurrence relation. In this case, if $alpha_k$'s are zeros of $3x^3-x^2-x-1=0$, then $$a_n = sum_{k=1}^{3}c_k alpha_k^n$$ for some constants $c_k$'s, which can be determined by the initial values.
– Sangchul Lee
Nov 17 at 2:33
1
@SangchulLee And one of $alpha_k=1$ in this problem.
– i707107
Nov 17 at 2:35
2
@YadatiKiran I'm not sure it will help here because all you will get is $l=frac{1}{3}[l+l+l]=l$.
– Anurag A
Nov 17 at 3:01
1
@AnuragA: Initially the problem was $a_n = dfrac{1}{3(a_{n-1} + a_{n-2} + a_{n-3})}$. I guess the OP has edited the post.
– Yadati Kiran
Nov 17 at 3:36