Solve recurrence relation: $T(n) = frac{n}{n+1}T(n-1) + 1$
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I am not able to solve this recurrence relation by substitution and variable change method.
$$T(n) = frac{n}{n+1}T(n-1) + 1; T(1) = 1 $$
recurrence-relations
edited Jan 18 at 18:59
jameselmore
4,39432035
asked Jan 18 at 18:16
KivtasKivtas
275
$endgroup$
add a comment |
$begingroup$
I am not able to solve this recurrence relation by substitution and variable change method.
$$T(n) = frac{n}{n+1}T(n-1) + 1; T(1) = 1 $$
recurrence-relations
edited Jan 18 at 18:59
jameselmore
4,39432035
asked Jan 18 at 18:16
KivtasKivtas
275
$endgroup$
$begingroup$
Is this what you mean? $$begin{cases} T_n=frac{n}{n+1} cdot T_{n-1}+1 \ T_1=1end{cases}$$
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– AmateurMathPirate
Jan 18 at 18:30
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They probably want you to do the substitution $S(n) = (n+1)T(n)$, then solve for $S(n)$.
$endgroup$
– DanielV
Jan 18 at 19:41
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yes @AmateurMathPirate
$endgroup$
– Kivtas
Jan 19 at 4:36
add a comment |
$begingroup$
I am not able to solve this recurrence relation by substitution and variable change method.
$$T(n) = frac{n}{n+1}T(n-1) + 1; T(1) = 1 $$
recurrence-relations
edited Jan 18 at 18:59
jameselmore
4,39432035
asked Jan 18 at 18:16
KivtasKivtas
275
$endgroup$
I am not able to solve this recurrence relation by substitution and variable change method.
$$T(n) = frac{n}{n+1}T(n-1) + 1; T(1) = 1 $$
recurrence-relations
recurrence-relations
edited Jan 18 at 18:59
jameselmore
4,39432035
asked Jan 18 at 18:16
KivtasKivtas
275
edited Jan 18 at 18:59
jameselmore
4,39432035
asked Jan 18 at 18:16
KivtasKivtas
275
edited Jan 18 at 18:59
jameselmore
4,39432035
edited Jan 18 at 18:59
jameselmore
4,39432035
edited Jan 18 at 18:59
jameselmore
4,39432035
4,39432035
asked Jan 18 at 18:16
KivtasKivtas
275
asked Jan 18 at 18:16
KivtasKivtas
275
asked Jan 18 at 18:16
KivtasKivtas
275
275
$begingroup$
Is this what you mean? $$begin{cases} T_n=frac{n}{n+1} cdot T_{n-1}+1 \ T_1=1end{cases}$$
$endgroup$
– AmateurMathPirate
Jan 18 at 18:30
$begingroup$
They probably want you to do the substitution $S(n) = (n+1)T(n)$, then solve for $S(n)$.
$endgroup$
– DanielV
Jan 18 at 19:41
$begingroup$
yes @AmateurMathPirate
$endgroup$
– Kivtas
Jan 19 at 4:36
add a comment |
$begingroup$
Is this what you mean? $$begin{cases} T_n=frac{n}{n+1} cdot T_{n-1}+1 \ T_1=1end{cases}$$
$endgroup$
– AmateurMathPirate
Jan 18 at 18:30
$begingroup$
They probably want you to do the substitution $S(n) = (n+1)T(n)$, then solve for $S(n)$.
$endgroup$
– DanielV
Jan 18 at 19:41
$begingroup$
yes @AmateurMathPirate
$endgroup$
– Kivtas
Jan 19 at 4:36
$begingroup$
Is this what you mean? $$begin{cases} T_n=frac{n}{n+1} cdot T_{n-1}+1 \ T_1=1end{cases}$$
$endgroup$
– AmateurMathPirate
Jan 18 at 18:30
$begingroup$
Is this what you mean? $$begin{cases} T_n=frac{n}{n+1} cdot T_{n-1}+1 \ T_1=1end{cases}$$
$endgroup$
– AmateurMathPirate
Jan 18 at 18:30
$begingroup$
They probably want you to do the substitution $S(n) = (n+1)T(n)$, then solve for $S(n)$.
$endgroup$
– DanielV
Jan 18 at 19:41
$begingroup$
They probably want you to do the substitution $S(n) = (n+1)T(n)$, then solve for $S(n)$.
$endgroup$
– DanielV
Jan 18 at 19:41
$begingroup$
yes @AmateurMathPirate
$endgroup$
– Kivtas
Jan 19 at 4:36
$begingroup$
yes @AmateurMathPirate
$endgroup$
– Kivtas
Jan 19 at 4:36
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
This is a linear difference equation so
$$
T(n)=T_h(n)+T_p(n)
$$
where
$$
T_h(n)-frac{n}{n+1}T_h(n-1) = 0\
T_p(n)-frac{n}{n+1}T_p(n-1) = 1\
$$
For the homogeneous solution is is clear that making
$$
T_h(n) = frac{C_0}{n+1}
$$
we have the solution. Now for the particular making $T_p(n)= frac{C_0(n)}{n+1}$ we have that
$$
C_0(n)-C_0(n-1)=n+1
$$
hence
$$
C_0(n) = frac 12(n+1)(n+2)
$$
and finally
$$
T(n) = frac{C_0}{n+1}+frac{n+2}{2}
$$
answered Jan 18 at 19:32
CesareoCesareo
8,8293516
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add a comment |
$begingroup$
Let $$S_n=(n+1)T_n$$
Then
$$S_n=S_{n-1}+n+1$$
So
$$S_n=sum_{k=3}^n k +S_1$$
You should be able to continue and compute $S_n$ and therefore $T_n$.
answered Jan 18 at 19:36
Stefan LafonStefan Lafon
1,89618
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add a comment |
$begingroup$
(n+1)T(n) = nT(n-1) + n+1
Let
S(n) = (n+1)T(n) S(0) = T(0) = c
S(n) = S(n-1) + n+1
Since, S(n-1) = S(n-2) + n
therefore, S(n) = S(n-2) + n+1 + n
for k terms...
S(n) = S(n-k) + (n+1) +n +.....+ (n-k+2)
k = n
S(n) = S(0) + (n+1) + n + ...+ (n-k+2)
= c + (n+1) + n + (n-1) + ......+ 2 + 1 - 1
= (n+1)(n+2)/2 + (c-1)
= (n^2 + 3n + 2)/2 + c - 1
= n^2/2 + 3n/2 + c
S(n) = O(n^2)
from our substitution -
T(n)(n+1) = O(n^2)
T(n)=1/(n+1) O(n^2)
We can neglect +1 in n+1
then,
T(n) = O(n)
answered Jan 19 at 14:25
KivtasKivtas
275
$endgroup$
add a comment |
$begingroup$
evaluating a few terms you can find a pattern:
$T_1=1=frac 2 2, T_2=frac{5}{3}, T_3=frac{9}{4}, T(4)=frac {14} 5, ldots$
Namely the pattern is that the numerator of the nth term is the sum of the first (n+1) terms diminished by 1. This can be found most easily by noticing the difference in the numerator between successive terms. The denominator is simply the (n+1)th term. We can then prove it by mathematical induction:
$$T_n=frac{left(sum_{i=1} ^ {n+1} iright) -1}{n+1}= cdots =frac{n+2}{2}-frac 1 {n+1}$$
Proof:
Base case: $n=1$ implies $T_1=frac 3 2 - frac 1 2 =1$ . Good so far.
Inductive step: Does our $T_n implies T_{n+1} $? Letting $nto n+1 $ in the original definition, we multiply our $T_n$ by $frac {n+1} {n+2}$ and add $1$, as this is what is defined by our formula with the index increased by 1....so do we get
$$T_{n+1}=frac{n+3}{2}-frac {1}{n+2}$$ ??
We do..
answered Jan 18 at 19:20
AmateurMathPirateAmateurMathPirate
1,336621
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is a linear difference equation so
$$
T(n)=T_h(n)+T_p(n)
$$
where
$$
T_h(n)-frac{n}{n+1}T_h(n-1) = 0\
T_p(n)-frac{n}{n+1}T_p(n-1) = 1\
$$
For the homogeneous solution is is clear that making
$$
T_h(n) = frac{C_0}{n+1}
$$
we have the solution. Now for the particular making $T_p(n)= frac{C_0(n)}{n+1}$ we have that
$$
C_0(n)-C_0(n-1)=n+1
$$
hence
$$
C_0(n) = frac 12(n+1)(n+2)
$$
and finally
$$
T(n) = frac{C_0}{n+1}+frac{n+2}{2}
$$
answered Jan 18 at 19:32
CesareoCesareo
8,8293516
$endgroup$
add a comment |
$begingroup$
This is a linear difference equation so
$$
T(n)=T_h(n)+T_p(n)
$$
where
$$
T_h(n)-frac{n}{n+1}T_h(n-1) = 0\
T_p(n)-frac{n}{n+1}T_p(n-1) = 1\
$$
For the homogeneous solution is is clear that making
$$
T_h(n) = frac{C_0}{n+1}
$$
we have the solution. Now for the particular making $T_p(n)= frac{C_0(n)}{n+1}$ we have that
$$
C_0(n)-C_0(n-1)=n+1
$$
hence
$$
C_0(n) = frac 12(n+1)(n+2)
$$
and finally
$$
T(n) = frac{C_0}{n+1}+frac{n+2}{2}
$$
answered Jan 18 at 19:32
CesareoCesareo
8,8293516
$endgroup$
add a comment |
$begingroup$
This is a linear difference equation so
$$
T(n)=T_h(n)+T_p(n)
$$
where
$$
T_h(n)-frac{n}{n+1}T_h(n-1) = 0\
T_p(n)-frac{n}{n+1}T_p(n-1) = 1\
$$
For the homogeneous solution is is clear that making
$$
T_h(n) = frac{C_0}{n+1}
$$
we have the solution. Now for the particular making $T_p(n)= frac{C_0(n)}{n+1}$ we have that
$$
C_0(n)-C_0(n-1)=n+1
$$
hence
$$
C_0(n) = frac 12(n+1)(n+2)
$$
and finally
$$
T(n) = frac{C_0}{n+1}+frac{n+2}{2}
$$
answered Jan 18 at 19:32
CesareoCesareo
8,8293516
$endgroup$
This is a linear difference equation so
$$
T(n)=T_h(n)+T_p(n)
$$
where
$$
T_h(n)-frac{n}{n+1}T_h(n-1) = 0\
T_p(n)-frac{n}{n+1}T_p(n-1) = 1\
$$
For the homogeneous solution is is clear that making
$$
T_h(n) = frac{C_0}{n+1}
$$
we have the solution. Now for the particular making $T_p(n)= frac{C_0(n)}{n+1}$ we have that
$$
C_0(n)-C_0(n-1)=n+1
$$
hence
$$
C_0(n) = frac 12(n+1)(n+2)
$$
and finally
$$
T(n) = frac{C_0}{n+1}+frac{n+2}{2}
$$
answered Jan 18 at 19:32
CesareoCesareo
8,8293516
answered Jan 18 at 19:32
CesareoCesareo
8,8293516
answered Jan 18 at 19:32
CesareoCesareo
8,8293516
answered Jan 18 at 19:32
CesareoCesareo
8,8293516
8,8293516
add a comment |
add a comment |
$begingroup$
Let $$S_n=(n+1)T_n$$
Then
$$S_n=S_{n-1}+n+1$$
So
$$S_n=sum_{k=3}^n k +S_1$$
You should be able to continue and compute $S_n$ and therefore $T_n$.
answered Jan 18 at 19:36
Stefan LafonStefan Lafon
1,89618
$endgroup$
add a comment |
$begingroup$
Let $$S_n=(n+1)T_n$$
Then
$$S_n=S_{n-1}+n+1$$
So
$$S_n=sum_{k=3}^n k +S_1$$
You should be able to continue and compute $S_n$ and therefore $T_n$.
answered Jan 18 at 19:36
Stefan LafonStefan Lafon
1,89618
$endgroup$
add a comment |
$begingroup$
Let $$S_n=(n+1)T_n$$
Then
$$S_n=S_{n-1}+n+1$$
So
$$S_n=sum_{k=3}^n k +S_1$$
You should be able to continue and compute $S_n$ and therefore $T_n$.
answered Jan 18 at 19:36
Stefan LafonStefan Lafon
1,89618
$endgroup$
Let $$S_n=(n+1)T_n$$
Then
$$S_n=S_{n-1}+n+1$$
So
$$S_n=sum_{k=3}^n k +S_1$$
You should be able to continue and compute $S_n$ and therefore $T_n$.
answered Jan 18 at 19:36
Stefan LafonStefan Lafon
1,89618
answered Jan 18 at 19:36
Stefan LafonStefan Lafon
1,89618
answered Jan 18 at 19:36
Stefan LafonStefan Lafon
1,89618
answered Jan 18 at 19:36
Stefan LafonStefan Lafon
1,89618
1,89618
add a comment |
add a comment |
$begingroup$
(n+1)T(n) = nT(n-1) + n+1
Let
S(n) = (n+1)T(n) S(0) = T(0) = c
S(n) = S(n-1) + n+1
Since, S(n-1) = S(n-2) + n
therefore, S(n) = S(n-2) + n+1 + n
for k terms...
S(n) = S(n-k) + (n+1) +n +.....+ (n-k+2)
k = n
S(n) = S(0) + (n+1) + n + ...+ (n-k+2)
= c + (n+1) + n + (n-1) + ......+ 2 + 1 - 1
= (n+1)(n+2)/2 + (c-1)
= (n^2 + 3n + 2)/2 + c - 1
= n^2/2 + 3n/2 + c
S(n) = O(n^2)
from our substitution -
T(n)(n+1) = O(n^2)
T(n)=1/(n+1) O(n^2)
We can neglect +1 in n+1
then,
T(n) = O(n)
answered Jan 19 at 14:25
KivtasKivtas
275
$endgroup$
add a comment |
$begingroup$
(n+1)T(n) = nT(n-1) + n+1
Let
S(n) = (n+1)T(n) S(0) = T(0) = c
S(n) = S(n-1) + n+1
Since, S(n-1) = S(n-2) + n
therefore, S(n) = S(n-2) + n+1 + n
for k terms...
S(n) = S(n-k) + (n+1) +n +.....+ (n-k+2)
k = n
S(n) = S(0) + (n+1) + n + ...+ (n-k+2)
= c + (n+1) + n + (n-1) + ......+ 2 + 1 - 1
= (n+1)(n+2)/2 + (c-1)
= (n^2 + 3n + 2)/2 + c - 1
= n^2/2 + 3n/2 + c
S(n) = O(n^2)
from our substitution -
T(n)(n+1) = O(n^2)
T(n)=1/(n+1) O(n^2)
We can neglect +1 in n+1
then,
T(n) = O(n)
answered Jan 19 at 14:25
KivtasKivtas
275
$endgroup$
add a comment |
$begingroup$
(n+1)T(n) = nT(n-1) + n+1
Let
S(n) = (n+1)T(n) S(0) = T(0) = c
S(n) = S(n-1) + n+1
Since, S(n-1) = S(n-2) + n
therefore, S(n) = S(n-2) + n+1 + n
for k terms...
S(n) = S(n-k) + (n+1) +n +.....+ (n-k+2)
k = n
S(n) = S(0) + (n+1) + n + ...+ (n-k+2)
= c + (n+1) + n + (n-1) + ......+ 2 + 1 - 1
= (n+1)(n+2)/2 + (c-1)
= (n^2 + 3n + 2)/2 + c - 1
= n^2/2 + 3n/2 + c
S(n) = O(n^2)
from our substitution -
T(n)(n+1) = O(n^2)
T(n)=1/(n+1) O(n^2)
We can neglect +1 in n+1
then,
T(n) = O(n)
answered Jan 19 at 14:25
KivtasKivtas
275
$endgroup$
(n+1)T(n) = nT(n-1) + n+1
Let
S(n) = (n+1)T(n) S(0) = T(0) = c
S(n) = S(n-1) + n+1
Since, S(n-1) = S(n-2) + n
therefore, S(n) = S(n-2) + n+1 + n
for k terms...
S(n) = S(n-k) + (n+1) +n +.....+ (n-k+2)
k = n
S(n) = S(0) + (n+1) + n + ...+ (n-k+2)
= c + (n+1) + n + (n-1) + ......+ 2 + 1 - 1
= (n+1)(n+2)/2 + (c-1)
= (n^2 + 3n + 2)/2 + c - 1
= n^2/2 + 3n/2 + c
S(n) = O(n^2)
from our substitution -
T(n)(n+1) = O(n^2)
T(n)=1/(n+1) O(n^2)
We can neglect +1 in n+1
then,
T(n) = O(n)
answered Jan 19 at 14:25
KivtasKivtas
275
answered Jan 19 at 14:25
KivtasKivtas
275
answered Jan 19 at 14:25
KivtasKivtas
275
answered Jan 19 at 14:25
KivtasKivtas
275
275
add a comment |
add a comment |
$begingroup$
evaluating a few terms you can find a pattern:
$T_1=1=frac 2 2, T_2=frac{5}{3}, T_3=frac{9}{4}, T(4)=frac {14} 5, ldots$
Namely the pattern is that the numerator of the nth term is the sum of the first (n+1) terms diminished by 1. This can be found most easily by noticing the difference in the numerator between successive terms. The denominator is simply the (n+1)th term. We can then prove it by mathematical induction:
$$T_n=frac{left(sum_{i=1} ^ {n+1} iright) -1}{n+1}= cdots =frac{n+2}{2}-frac 1 {n+1}$$
Proof:
Base case: $n=1$ implies $T_1=frac 3 2 - frac 1 2 =1$ . Good so far.
Inductive step: Does our $T_n implies T_{n+1} $? Letting $nto n+1 $ in the original definition, we multiply our $T_n$ by $frac {n+1} {n+2}$ and add $1$, as this is what is defined by our formula with the index increased by 1....so do we get
$$T_{n+1}=frac{n+3}{2}-frac {1}{n+2}$$ ??
We do..
answered Jan 18 at 19:20
AmateurMathPirateAmateurMathPirate
1,336621
$endgroup$
add a comment |
$begingroup$
evaluating a few terms you can find a pattern:
$T_1=1=frac 2 2, T_2=frac{5}{3}, T_3=frac{9}{4}, T(4)=frac {14} 5, ldots$
Namely the pattern is that the numerator of the nth term is the sum of the first (n+1) terms diminished by 1. This can be found most easily by noticing the difference in the numerator between successive terms. The denominator is simply the (n+1)th term. We can then prove it by mathematical induction:
$$T_n=frac{left(sum_{i=1} ^ {n+1} iright) -1}{n+1}= cdots =frac{n+2}{2}-frac 1 {n+1}$$
Proof:
Base case: $n=1$ implies $T_1=frac 3 2 - frac 1 2 =1$ . Good so far.
Inductive step: Does our $T_n implies T_{n+1} $? Letting $nto n+1 $ in the original definition, we multiply our $T_n$ by $frac {n+1} {n+2}$ and add $1$, as this is what is defined by our formula with the index increased by 1....so do we get
$$T_{n+1}=frac{n+3}{2}-frac {1}{n+2}$$ ??
We do..
answered Jan 18 at 19:20
AmateurMathPirateAmateurMathPirate
1,336621
$endgroup$
add a comment |
$begingroup$
evaluating a few terms you can find a pattern:
$T_1=1=frac 2 2, T_2=frac{5}{3}, T_3=frac{9}{4}, T(4)=frac {14} 5, ldots$
Namely the pattern is that the numerator of the nth term is the sum of the first (n+1) terms diminished by 1. This can be found most easily by noticing the difference in the numerator between successive terms. The denominator is simply the (n+1)th term. We can then prove it by mathematical induction:
$$T_n=frac{left(sum_{i=1} ^ {n+1} iright) -1}{n+1}= cdots =frac{n+2}{2}-frac 1 {n+1}$$
Proof:
Base case: $n=1$ implies $T_1=frac 3 2 - frac 1 2 =1$ . Good so far.
Inductive step: Does our $T_n implies T_{n+1} $? Letting $nto n+1 $ in the original definition, we multiply our $T_n$ by $frac {n+1} {n+2}$ and add $1$, as this is what is defined by our formula with the index increased by 1....so do we get
$$T_{n+1}=frac{n+3}{2}-frac {1}{n+2}$$ ??
We do..
answered Jan 18 at 19:20
AmateurMathPirateAmateurMathPirate
1,336621
$endgroup$
evaluating a few terms you can find a pattern:
$T_1=1=frac 2 2, T_2=frac{5}{3}, T_3=frac{9}{4}, T(4)=frac {14} 5, ldots$
Namely the pattern is that the numerator of the nth term is the sum of the first (n+1) terms diminished by 1. This can be found most easily by noticing the difference in the numerator between successive terms. The denominator is simply the (n+1)th term. We can then prove it by mathematical induction:
$$T_n=frac{left(sum_{i=1} ^ {n+1} iright) -1}{n+1}= cdots =frac{n+2}{2}-frac 1 {n+1}$$
Proof:
Base case: $n=1$ implies $T_1=frac 3 2 - frac 1 2 =1$ . Good so far.
Inductive step: Does our $T_n implies T_{n+1} $? Letting $nto n+1 $ in the original definition, we multiply our $T_n$ by $frac {n+1} {n+2}$ and add $1$, as this is what is defined by our formula with the index increased by 1....so do we get
$$T_{n+1}=frac{n+3}{2}-frac {1}{n+2}$$ ??
We do..
answered Jan 18 at 19:20
AmateurMathPirateAmateurMathPirate
1,336621
edited Jan 20 at 16:06
answered Jan 18 at 19:20
AmateurMathPirateAmateurMathPirate
1,336621
answered Jan 18 at 19:20
AmateurMathPirateAmateurMathPirate
1,336621
answered Jan 18 at 19:20
AmateurMathPirateAmateurMathPirate
1,336621
1,336621
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$begingroup$
Is this what you mean? $$begin{cases} T_n=frac{n}{n+1} cdot T_{n-1}+1 \ T_1=1end{cases}$$
$endgroup$
– AmateurMathPirate
Jan 18 at 18:30
$begingroup$
They probably want you to do the substitution $S(n) = (n+1)T(n)$, then solve for $S(n)$.
$endgroup$
– DanielV
Jan 18 at 19:41
$begingroup$
yes @AmateurMathPirate
$endgroup$
– Kivtas
Jan 19 at 4:36