Mixing
How can I solve the linear recurrence problem $f(n)=f(n-1)+3 cdot f(n-3)+2n$ using matrix exponentiation when...
$begingroup$
The porblem is $f(n)=f(n-1)+3f(n-3)+2n$. I solved $f(n)=f(n-1)+3f(n-3)$ and adding summation of $2n$ upto $n$. But this is wrong. It requires too much of pre calculation.
I tried this problem based on the concept as given here https://comeoncodeon.wordpress.com/2011/05/08/recurrence-relation-and-matrix-exponentiation/
linear-algebra
edited Jan 20 at 10:23
user289143
1,002313
asked Jan 20 at 9:38
KUNDAN NAYAKKUNDAN NAYAK
11
$endgroup$
add a comment |
$begingroup$
The porblem is $f(n)=f(n-1)+3f(n-3)+2n$. I solved $f(n)=f(n-1)+3f(n-3)$ and adding summation of $2n$ upto $n$. But this is wrong. It requires too much of pre calculation.
I tried this problem based on the concept as given here https://comeoncodeon.wordpress.com/2011/05/08/recurrence-relation-and-matrix-exponentiation/
linear-algebra
edited Jan 20 at 10:23
user289143
1,002313
asked Jan 20 at 9:38
KUNDAN NAYAKKUNDAN NAYAK
11
$endgroup$
$begingroup$
1) The second sentence in your title is not understandable. 2) If you mention "using matrix", it means that you have had already exercises on this subject. What method is advised there ? 3) You do not mention initial conditions $f(1),f(2),f(3)$ ? What are they ? Arbitrary constants ?
$endgroup$
– Jean Marie
Jan 20 at 9:49
$begingroup$
I'm voting to close this question as off-topic because it's taken from a live coding competition, the NIT codathon.
$endgroup$
– lulu
Jan 20 at 12:29
$begingroup$
This problem has shown up a lot lately, see, e.g., this question. Nobody should respond to it until the competition has ended.
$endgroup$
– lulu
Jan 20 at 12:31
add a comment |
$begingroup$
The porblem is $f(n)=f(n-1)+3f(n-3)+2n$. I solved $f(n)=f(n-1)+3f(n-3)$ and adding summation of $2n$ upto $n$. But this is wrong. It requires too much of pre calculation.
I tried this problem based on the concept as given here https://comeoncodeon.wordpress.com/2011/05/08/recurrence-relation-and-matrix-exponentiation/
linear-algebra
edited Jan 20 at 10:23
user289143
1,002313
asked Jan 20 at 9:38
KUNDAN NAYAKKUNDAN NAYAK
11
$endgroup$
The porblem is $f(n)=f(n-1)+3f(n-3)+2n$. I solved $f(n)=f(n-1)+3f(n-3)$ and adding summation of $2n$ upto $n$. But this is wrong. It requires too much of pre calculation.
I tried this problem based on the concept as given here https://comeoncodeon.wordpress.com/2011/05/08/recurrence-relation-and-matrix-exponentiation/
linear-algebra
linear-algebra
edited Jan 20 at 10:23
user289143
1,002313
asked Jan 20 at 9:38
KUNDAN NAYAKKUNDAN NAYAK
11
edited Jan 20 at 10:23
user289143
1,002313
asked Jan 20 at 9:38
KUNDAN NAYAKKUNDAN NAYAK
11
edited Jan 20 at 10:23
user289143
1,002313
edited Jan 20 at 10:23
user289143
1,002313
edited Jan 20 at 10:23
user289143
1,002313
1,002313
asked Jan 20 at 9:38
KUNDAN NAYAKKUNDAN NAYAK
11
asked Jan 20 at 9:38
KUNDAN NAYAKKUNDAN NAYAK
11
asked Jan 20 at 9:38
KUNDAN NAYAKKUNDAN NAYAK
11
11
$begingroup$
1) The second sentence in your title is not understandable. 2) If you mention "using matrix", it means that you have had already exercises on this subject. What method is advised there ? 3) You do not mention initial conditions $f(1),f(2),f(3)$ ? What are they ? Arbitrary constants ?
$endgroup$
– Jean Marie
Jan 20 at 9:49
$begingroup$
I'm voting to close this question as off-topic because it's taken from a live coding competition, the NIT codathon.
$endgroup$
– lulu
Jan 20 at 12:29
$begingroup$
This problem has shown up a lot lately, see, e.g., this question. Nobody should respond to it until the competition has ended.
$endgroup$
– lulu
Jan 20 at 12:31
add a comment |
$begingroup$
1) The second sentence in your title is not understandable. 2) If you mention "using matrix", it means that you have had already exercises on this subject. What method is advised there ? 3) You do not mention initial conditions $f(1),f(2),f(3)$ ? What are they ? Arbitrary constants ?
$endgroup$
– Jean Marie
Jan 20 at 9:49
$begingroup$
I'm voting to close this question as off-topic because it's taken from a live coding competition, the NIT codathon.
$endgroup$
– lulu
Jan 20 at 12:29
$begingroup$
This problem has shown up a lot lately, see, e.g., this question. Nobody should respond to it until the competition has ended.
$endgroup$
– lulu
Jan 20 at 12:31
$begingroup$
1) The second sentence in your title is not understandable. 2) If you mention "using matrix", it means that you have had already exercises on this subject. What method is advised there ? 3) You do not mention initial conditions $f(1),f(2),f(3)$ ? What are they ? Arbitrary constants ?
$endgroup$
– Jean Marie
Jan 20 at 9:49
$begingroup$
1) The second sentence in your title is not understandable. 2) If you mention "using matrix", it means that you have had already exercises on this subject. What method is advised there ? 3) You do not mention initial conditions $f(1),f(2),f(3)$ ? What are they ? Arbitrary constants ?
$endgroup$
– Jean Marie
Jan 20 at 9:49
$begingroup$
I'm voting to close this question as off-topic because it's taken from a live coding competition, the NIT codathon.
$endgroup$
– lulu
Jan 20 at 12:29
$begingroup$
I'm voting to close this question as off-topic because it's taken from a live coding competition, the NIT codathon.
$endgroup$
– lulu
Jan 20 at 12:29
$begingroup$
This problem has shown up a lot lately, see, e.g., this question. Nobody should respond to it until the competition has ended.
$endgroup$
– lulu
Jan 20 at 12:31
$begingroup$
This problem has shown up a lot lately, see, e.g., this question. Nobody should respond to it until the competition has ended.
$endgroup$
– lulu
Jan 20 at 12:31
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $v_n$ be the vector
$$ v_n=begin{pmatrix}1\n\f(n)\f(n-1)\f(n-2)end{pmatrix}.$$
Then apparently we can compute $v_n$ from $v_{n-1}$
by a matrix multiplication:
$$ begin{pmatrix}1\n\f(n)\f(n-1)\f(n-2)end{pmatrix}=begin{pmatrix}
1&0&0&0&0\
1&1&0&0&0\
0&2&1&0&3\
0&0&1&0&0\
0&0&0&1&0end{pmatrix}begin{pmatrix}1\n-1\f(n-1)\f(n-2)\f(n-3)end{pmatrix}.$$
Nevertheless, you might be able to proceed faster if you consider $g(n):=f(n)+an+b$ for suitable constants $a,b$. We find the recurrence relation
$$begin{align}g(n)&=f(n)+an+b\
&=f(n-1)+3f(n-3)+2n+an+b\
&=g(n-1)-a(n-1)-b+3g(n-3)-3a(n-3)-3b+2n+an+b\
&=g(n-1)+3g(n-3)+(2-3a)n+(10a-3b)end{align}$$
and this looks particularly nice if you pick $a,b$ such that $2-3a=10a-3b=0$.
answered Jan 20 at 10:31
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
add a comment |
$begingroup$
Define $g(n):=f(n)+an+b$ for constants $a,,b$ we'll choose later so $g(n)=g(n-1)+3g(n-3)+(2-3a)n+10a-3b$ (you should double-check my arithmetic). With $a=2/3,,b=10a/3=20/9$ this simplifies to $g(n)=g(n-1)+g(n-3)$, which you already know how to solve. Then $f(n)=g(n)-an-b$.
answered Jan 20 at 10:29
J.G.J.G.
28.6k22845
$endgroup$
add a comment |
$begingroup$
Your problem looks to read
$$
left{ matrix{
f(1) = a hfill cr
f(2) = b hfill cr
f(3) = c hfill cr
f(n) = f(n - 1) + 3f(n - 3) + 2nquad left| {;4 le n} right. hfill cr} right.
$$
that is
$$
eqalign{
& f(n) - f(n - 1) - 3f(n - 3) = cr
& + 2nleft[ {4 le n} right] + aleft[ {1 = n} right] + bleft[ {2 = n} right] + cleft[ {3 = n} right] cr}
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
If you want to use the matrix notation, then
indicating with $bf E$ the shift matrix
$$
{bf E} = left( {matrix{
0 & 0 & 0 & 0 & cdots cr
1 & 0 & 0 & 0 & cdots cr
0 & 1 & 0 & 0 & cdots cr
0 & 0 & 1 & 0 & cdots cr
vdots & vdots & vdots & ddots & ddots cr
} } right)
$$
then the equation is
$$
left( {{bf I} - {bf E} - 3,{bf E}^{,{bf 3}} } right){bf f} = {bf v} = left( {a,b,c,8,10, cdots } right)^T
$$
which gives
$$
eqalign{
& {bf f} = {{bf I} over {{bf I} - {bf E} - 3,{bf E}^{,{bf 3}} }}{bf v} = cr
& = left( {sumlimits_{0, le ,n} {{bf E}^{,{bf n}} left( {{bf I}
+ sqrt 3 ,{bf E}} right)^{,{bf n}} left( {{bf I} - sqrt 3 ,{bf E}} right)^{,{bf n}} } } right){bf v} = cr
& = quad cdots cr}
$$
answered Jan 20 at 11:52
G CabG Cab
19.8k31340
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $v_n$ be the vector
$$ v_n=begin{pmatrix}1\n\f(n)\f(n-1)\f(n-2)end{pmatrix}.$$
Then apparently we can compute $v_n$ from $v_{n-1}$
by a matrix multiplication:
$$ begin{pmatrix}1\n\f(n)\f(n-1)\f(n-2)end{pmatrix}=begin{pmatrix}
1&0&0&0&0\
1&1&0&0&0\
0&2&1&0&3\
0&0&1&0&0\
0&0&0&1&0end{pmatrix}begin{pmatrix}1\n-1\f(n-1)\f(n-2)\f(n-3)end{pmatrix}.$$
Nevertheless, you might be able to proceed faster if you consider $g(n):=f(n)+an+b$ for suitable constants $a,b$. We find the recurrence relation
$$begin{align}g(n)&=f(n)+an+b\
&=f(n-1)+3f(n-3)+2n+an+b\
&=g(n-1)-a(n-1)-b+3g(n-3)-3a(n-3)-3b+2n+an+b\
&=g(n-1)+3g(n-3)+(2-3a)n+(10a-3b)end{align}$$
and this looks particularly nice if you pick $a,b$ such that $2-3a=10a-3b=0$.
answered Jan 20 at 10:31
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
add a comment |
$begingroup$
Let $v_n$ be the vector
$$ v_n=begin{pmatrix}1\n\f(n)\f(n-1)\f(n-2)end{pmatrix}.$$
Then apparently we can compute $v_n$ from $v_{n-1}$
by a matrix multiplication:
$$ begin{pmatrix}1\n\f(n)\f(n-1)\f(n-2)end{pmatrix}=begin{pmatrix}
1&0&0&0&0\
1&1&0&0&0\
0&2&1&0&3\
0&0&1&0&0\
0&0&0&1&0end{pmatrix}begin{pmatrix}1\n-1\f(n-1)\f(n-2)\f(n-3)end{pmatrix}.$$
Nevertheless, you might be able to proceed faster if you consider $g(n):=f(n)+an+b$ for suitable constants $a,b$. We find the recurrence relation
$$begin{align}g(n)&=f(n)+an+b\
&=f(n-1)+3f(n-3)+2n+an+b\
&=g(n-1)-a(n-1)-b+3g(n-3)-3a(n-3)-3b+2n+an+b\
&=g(n-1)+3g(n-3)+(2-3a)n+(10a-3b)end{align}$$
and this looks particularly nice if you pick $a,b$ such that $2-3a=10a-3b=0$.
answered Jan 20 at 10:31
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
add a comment |
$begingroup$
Let $v_n$ be the vector
$$ v_n=begin{pmatrix}1\n\f(n)\f(n-1)\f(n-2)end{pmatrix}.$$
Then apparently we can compute $v_n$ from $v_{n-1}$
by a matrix multiplication:
$$ begin{pmatrix}1\n\f(n)\f(n-1)\f(n-2)end{pmatrix}=begin{pmatrix}
1&0&0&0&0\
1&1&0&0&0\
0&2&1&0&3\
0&0&1&0&0\
0&0&0&1&0end{pmatrix}begin{pmatrix}1\n-1\f(n-1)\f(n-2)\f(n-3)end{pmatrix}.$$
Nevertheless, you might be able to proceed faster if you consider $g(n):=f(n)+an+b$ for suitable constants $a,b$. We find the recurrence relation
$$begin{align}g(n)&=f(n)+an+b\
&=f(n-1)+3f(n-3)+2n+an+b\
&=g(n-1)-a(n-1)-b+3g(n-3)-3a(n-3)-3b+2n+an+b\
&=g(n-1)+3g(n-3)+(2-3a)n+(10a-3b)end{align}$$
and this looks particularly nice if you pick $a,b$ such that $2-3a=10a-3b=0$.
answered Jan 20 at 10:31
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
Let $v_n$ be the vector
$$ v_n=begin{pmatrix}1\n\f(n)\f(n-1)\f(n-2)end{pmatrix}.$$
Then apparently we can compute $v_n$ from $v_{n-1}$
by a matrix multiplication:
$$ begin{pmatrix}1\n\f(n)\f(n-1)\f(n-2)end{pmatrix}=begin{pmatrix}
1&0&0&0&0\
1&1&0&0&0\
0&2&1&0&3\
0&0&1&0&0\
0&0&0&1&0end{pmatrix}begin{pmatrix}1\n-1\f(n-1)\f(n-2)\f(n-3)end{pmatrix}.$$
Nevertheless, you might be able to proceed faster if you consider $g(n):=f(n)+an+b$ for suitable constants $a,b$. We find the recurrence relation
$$begin{align}g(n)&=f(n)+an+b\
&=f(n-1)+3f(n-3)+2n+an+b\
&=g(n-1)-a(n-1)-b+3g(n-3)-3a(n-3)-3b+2n+an+b\
&=g(n-1)+3g(n-3)+(2-3a)n+(10a-3b)end{align}$$
and this looks particularly nice if you pick $a,b$ such that $2-3a=10a-3b=0$.
answered Jan 20 at 10:31
Hagen von EitzenHagen von Eitzen
282k23272505
answered Jan 20 at 10:31
Hagen von EitzenHagen von Eitzen
282k23272505
answered Jan 20 at 10:31
Hagen von EitzenHagen von Eitzen
282k23272505
answered Jan 20 at 10:31
Hagen von EitzenHagen von Eitzen
282k23272505
282k23272505
add a comment |
add a comment |
$begingroup$
Define $g(n):=f(n)+an+b$ for constants $a,,b$ we'll choose later so $g(n)=g(n-1)+3g(n-3)+(2-3a)n+10a-3b$ (you should double-check my arithmetic). With $a=2/3,,b=10a/3=20/9$ this simplifies to $g(n)=g(n-1)+g(n-3)$, which you already know how to solve. Then $f(n)=g(n)-an-b$.
answered Jan 20 at 10:29
J.G.J.G.
28.6k22845
$endgroup$
add a comment |
$begingroup$
Define $g(n):=f(n)+an+b$ for constants $a,,b$ we'll choose later so $g(n)=g(n-1)+3g(n-3)+(2-3a)n+10a-3b$ (you should double-check my arithmetic). With $a=2/3,,b=10a/3=20/9$ this simplifies to $g(n)=g(n-1)+g(n-3)$, which you already know how to solve. Then $f(n)=g(n)-an-b$.
answered Jan 20 at 10:29
J.G.J.G.
28.6k22845
$endgroup$
add a comment |
$begingroup$
Define $g(n):=f(n)+an+b$ for constants $a,,b$ we'll choose later so $g(n)=g(n-1)+3g(n-3)+(2-3a)n+10a-3b$ (you should double-check my arithmetic). With $a=2/3,,b=10a/3=20/9$ this simplifies to $g(n)=g(n-1)+g(n-3)$, which you already know how to solve. Then $f(n)=g(n)-an-b$.
answered Jan 20 at 10:29
J.G.J.G.
28.6k22845
$endgroup$
Define $g(n):=f(n)+an+b$ for constants $a,,b$ we'll choose later so $g(n)=g(n-1)+3g(n-3)+(2-3a)n+10a-3b$ (you should double-check my arithmetic). With $a=2/3,,b=10a/3=20/9$ this simplifies to $g(n)=g(n-1)+g(n-3)$, which you already know how to solve. Then $f(n)=g(n)-an-b$.
answered Jan 20 at 10:29
J.G.J.G.
28.6k22845
answered Jan 20 at 10:29
J.G.J.G.
28.6k22845
answered Jan 20 at 10:29
J.G.J.G.
28.6k22845
answered Jan 20 at 10:29
J.G.J.G.
28.6k22845
28.6k22845
add a comment |
add a comment |
$begingroup$
Your problem looks to read
$$
left{ matrix{
f(1) = a hfill cr
f(2) = b hfill cr
f(3) = c hfill cr
f(n) = f(n - 1) + 3f(n - 3) + 2nquad left| {;4 le n} right. hfill cr} right.
$$
that is
$$
eqalign{
& f(n) - f(n - 1) - 3f(n - 3) = cr
& + 2nleft[ {4 le n} right] + aleft[ {1 = n} right] + bleft[ {2 = n} right] + cleft[ {3 = n} right] cr}
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
If you want to use the matrix notation, then
indicating with $bf E$ the shift matrix
$$
{bf E} = left( {matrix{
0 & 0 & 0 & 0 & cdots cr
1 & 0 & 0 & 0 & cdots cr
0 & 1 & 0 & 0 & cdots cr
0 & 0 & 1 & 0 & cdots cr
vdots & vdots & vdots & ddots & ddots cr
} } right)
$$
then the equation is
$$
left( {{bf I} - {bf E} - 3,{bf E}^{,{bf 3}} } right){bf f} = {bf v} = left( {a,b,c,8,10, cdots } right)^T
$$
which gives
$$
eqalign{
& {bf f} = {{bf I} over {{bf I} - {bf E} - 3,{bf E}^{,{bf 3}} }}{bf v} = cr
& = left( {sumlimits_{0, le ,n} {{bf E}^{,{bf n}} left( {{bf I}
+ sqrt 3 ,{bf E}} right)^{,{bf n}} left( {{bf I} - sqrt 3 ,{bf E}} right)^{,{bf n}} } } right){bf v} = cr
& = quad cdots cr}
$$
answered Jan 20 at 11:52
G CabG Cab
19.8k31340
$endgroup$
add a comment |
$begingroup$
Your problem looks to read
$$
left{ matrix{
f(1) = a hfill cr
f(2) = b hfill cr
f(3) = c hfill cr
f(n) = f(n - 1) + 3f(n - 3) + 2nquad left| {;4 le n} right. hfill cr} right.
$$
that is
$$
eqalign{
& f(n) - f(n - 1) - 3f(n - 3) = cr
& + 2nleft[ {4 le n} right] + aleft[ {1 = n} right] + bleft[ {2 = n} right] + cleft[ {3 = n} right] cr}
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
If you want to use the matrix notation, then
indicating with $bf E$ the shift matrix
$$
{bf E} = left( {matrix{
0 & 0 & 0 & 0 & cdots cr
1 & 0 & 0 & 0 & cdots cr
0 & 1 & 0 & 0 & cdots cr
0 & 0 & 1 & 0 & cdots cr
vdots & vdots & vdots & ddots & ddots cr
} } right)
$$
then the equation is
$$
left( {{bf I} - {bf E} - 3,{bf E}^{,{bf 3}} } right){bf f} = {bf v} = left( {a,b,c,8,10, cdots } right)^T
$$
which gives
$$
eqalign{
& {bf f} = {{bf I} over {{bf I} - {bf E} - 3,{bf E}^{,{bf 3}} }}{bf v} = cr
& = left( {sumlimits_{0, le ,n} {{bf E}^{,{bf n}} left( {{bf I}
+ sqrt 3 ,{bf E}} right)^{,{bf n}} left( {{bf I} - sqrt 3 ,{bf E}} right)^{,{bf n}} } } right){bf v} = cr
& = quad cdots cr}
$$
answered Jan 20 at 11:52
G CabG Cab
19.8k31340
$endgroup$
add a comment |
$begingroup$
Your problem looks to read
$$
left{ matrix{
f(1) = a hfill cr
f(2) = b hfill cr
f(3) = c hfill cr
f(n) = f(n - 1) + 3f(n - 3) + 2nquad left| {;4 le n} right. hfill cr} right.
$$
that is
$$
eqalign{
& f(n) - f(n - 1) - 3f(n - 3) = cr
& + 2nleft[ {4 le n} right] + aleft[ {1 = n} right] + bleft[ {2 = n} right] + cleft[ {3 = n} right] cr}
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
If you want to use the matrix notation, then
indicating with $bf E$ the shift matrix
$$
{bf E} = left( {matrix{
0 & 0 & 0 & 0 & cdots cr
1 & 0 & 0 & 0 & cdots cr
0 & 1 & 0 & 0 & cdots cr
0 & 0 & 1 & 0 & cdots cr
vdots & vdots & vdots & ddots & ddots cr
} } right)
$$
then the equation is
$$
left( {{bf I} - {bf E} - 3,{bf E}^{,{bf 3}} } right){bf f} = {bf v} = left( {a,b,c,8,10, cdots } right)^T
$$
which gives
$$
eqalign{
& {bf f} = {{bf I} over {{bf I} - {bf E} - 3,{bf E}^{,{bf 3}} }}{bf v} = cr
& = left( {sumlimits_{0, le ,n} {{bf E}^{,{bf n}} left( {{bf I}
+ sqrt 3 ,{bf E}} right)^{,{bf n}} left( {{bf I} - sqrt 3 ,{bf E}} right)^{,{bf n}} } } right){bf v} = cr
& = quad cdots cr}
$$
answered Jan 20 at 11:52
G CabG Cab
19.8k31340
$endgroup$
Your problem looks to read
$$
left{ matrix{
f(1) = a hfill cr
f(2) = b hfill cr
f(3) = c hfill cr
f(n) = f(n - 1) + 3f(n - 3) + 2nquad left| {;4 le n} right. hfill cr} right.
$$
that is
$$
eqalign{
& f(n) - f(n - 1) - 3f(n - 3) = cr
& + 2nleft[ {4 le n} right] + aleft[ {1 = n} right] + bleft[ {2 = n} right] + cleft[ {3 = n} right] cr}
$$
where $[P]$ denotes the Iverson bracket
$$
left[ P right] = left{ {begin{array}{*{20}c}
1 & {P = TRUE} \
0 & {P = FALSE} \
end{array} } right.
$$
If you want to use the matrix notation, then
indicating with $bf E$ the shift matrix
$$
{bf E} = left( {matrix{
0 & 0 & 0 & 0 & cdots cr
1 & 0 & 0 & 0 & cdots cr
0 & 1 & 0 & 0 & cdots cr
0 & 0 & 1 & 0 & cdots cr
vdots & vdots & vdots & ddots & ddots cr
} } right)
$$
then the equation is
$$
left( {{bf I} - {bf E} - 3,{bf E}^{,{bf 3}} } right){bf f} = {bf v} = left( {a,b,c,8,10, cdots } right)^T
$$
which gives
$$
eqalign{
& {bf f} = {{bf I} over {{bf I} - {bf E} - 3,{bf E}^{,{bf 3}} }}{bf v} = cr
& = left( {sumlimits_{0, le ,n} {{bf E}^{,{bf n}} left( {{bf I}
+ sqrt 3 ,{bf E}} right)^{,{bf n}} left( {{bf I} - sqrt 3 ,{bf E}} right)^{,{bf n}} } } right){bf v} = cr
& = quad cdots cr}
$$
answered Jan 20 at 11:52
G CabG Cab
19.8k31340
answered Jan 20 at 11:52
G CabG Cab
19.8k31340
answered Jan 20 at 11:52
G CabG Cab
19.8k31340
answered Jan 20 at 11:52
G CabG Cab
19.8k31340
19.8k31340
add a comment |
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$begingroup$
1) The second sentence in your title is not understandable. 2) If you mention "using matrix", it means that you have had already exercises on this subject. What method is advised there ? 3) You do not mention initial conditions $f(1),f(2),f(3)$ ? What are they ? Arbitrary constants ?
$endgroup$
– Jean Marie
Jan 20 at 9:49
$begingroup$
I'm voting to close this question as off-topic because it's taken from a live coding competition, the NIT codathon.
$endgroup$
– lulu
Jan 20 at 12:29
$begingroup$
This problem has shown up a lot lately, see, e.g., this question. Nobody should respond to it until the competition has ended.
$endgroup$
– lulu
Jan 20 at 12:31