Mixing
How to solve $V(n) = 2 cdot V(n-1) + 2 cdot n$
$begingroup$
How to solve $V(n) = 2 cdot V(n-1) + 2cdot n$? I've tried using telescoping, but I'm not able to get correct solution. The textbook has a solution with homogeneous and particular solution and then gets the final solution but I don't know how to do that. Can I apply telescoping here and what is the proper way to do it?
recurrence-relations telescopic-series
edited Jan 20 at 11:07
user289143
1,002313
asked Jan 20 at 10:59
ponikoliponikoli
416
$endgroup$
add a comment |
$begingroup$
How to solve $V(n) = 2 cdot V(n-1) + 2cdot n$? I've tried using telescoping, but I'm not able to get correct solution. The textbook has a solution with homogeneous and particular solution and then gets the final solution but I don't know how to do that. Can I apply telescoping here and what is the proper way to do it?
recurrence-relations telescopic-series
edited Jan 20 at 11:07
user289143
1,002313
asked Jan 20 at 10:59
ponikoliponikoli
416
$endgroup$
$begingroup$
Can you solve W(n) = W(n-1) + (2n)/2^n ? Then relate the Ws to the Vs and you are done...
$endgroup$
– Did
Jan 20 at 11:03
$begingroup$
@user623855 2n ≠ 2^n.
$endgroup$
– Did
Jan 20 at 11:04
add a comment |
$begingroup$
How to solve $V(n) = 2 cdot V(n-1) + 2cdot n$? I've tried using telescoping, but I'm not able to get correct solution. The textbook has a solution with homogeneous and particular solution and then gets the final solution but I don't know how to do that. Can I apply telescoping here and what is the proper way to do it?
recurrence-relations telescopic-series
edited Jan 20 at 11:07
user289143
1,002313
asked Jan 20 at 10:59
ponikoliponikoli
416
$endgroup$
How to solve $V(n) = 2 cdot V(n-1) + 2cdot n$? I've tried using telescoping, but I'm not able to get correct solution. The textbook has a solution with homogeneous and particular solution and then gets the final solution but I don't know how to do that. Can I apply telescoping here and what is the proper way to do it?
recurrence-relations telescopic-series
recurrence-relations telescopic-series
edited Jan 20 at 11:07
user289143
1,002313
asked Jan 20 at 10:59
ponikoliponikoli
416
edited Jan 20 at 11:07
user289143
1,002313
asked Jan 20 at 10:59
ponikoliponikoli
416
edited Jan 20 at 11:07
user289143
1,002313
edited Jan 20 at 11:07
user289143
1,002313
edited Jan 20 at 11:07
user289143
1,002313
1,002313
asked Jan 20 at 10:59
ponikoliponikoli
416
asked Jan 20 at 10:59
ponikoliponikoli
416
asked Jan 20 at 10:59
ponikoliponikoli
416
416
$begingroup$
Can you solve W(n) = W(n-1) + (2n)/2^n ? Then relate the Ws to the Vs and you are done...
$endgroup$
– Did
Jan 20 at 11:03
$begingroup$
@user623855 2n ≠ 2^n.
$endgroup$
– Did
Jan 20 at 11:04
add a comment |
$begingroup$
Can you solve W(n) = W(n-1) + (2n)/2^n ? Then relate the Ws to the Vs and you are done...
$endgroup$
– Did
Jan 20 at 11:03
$begingroup$
@user623855 2n ≠ 2^n.
$endgroup$
– Did
Jan 20 at 11:04
$begingroup$
Can you solve W(n) = W(n-1) + (2n)/2^n ? Then relate the Ws to the Vs and you are done...
$endgroup$
– Did
Jan 20 at 11:03
$begingroup$
Can you solve W(n) = W(n-1) + (2n)/2^n ? Then relate the Ws to the Vs and you are done...
$endgroup$
– Did
Jan 20 at 11:03
$begingroup$
@user623855 2n ≠ 2^n.
$endgroup$
– Did
Jan 20 at 11:04
$begingroup$
@user623855 2n ≠ 2^n.
$endgroup$
– Did
Jan 20 at 11:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Try to expand and generalize the case (by induction):
$$V(n) = 2times(2V(n-2) + 2(n-1)) + 2n = 2^2 V(n-2) + 2^2(n-1) + 2n = sum_{i=1}^{n}2^i(n-i+1)$$
Suppose $V(1) = 2$ in this case.
answered Jan 20 at 11:05
OmGOmG
2,502822
$endgroup$
add a comment |
$begingroup$
Let $V(m)=f(m)+am+b$
$$2n=V(n)-2V(n-1)=f(n)+an+b-2{f(m-1)+a(n-1)+b}$$
$$2n=f(n)-2f(n-1)-an+2a-b$$
Set $2a-b=0, -an=2$ so that $f(n)=2f(n-1)=2^rf(n-r),0le rle n$
$implies a=-2, b=2(-2)=?$
answered Jan 20 at 12:42
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try to expand and generalize the case (by induction):
$$V(n) = 2times(2V(n-2) + 2(n-1)) + 2n = 2^2 V(n-2) + 2^2(n-1) + 2n = sum_{i=1}^{n}2^i(n-i+1)$$
Suppose $V(1) = 2$ in this case.
answered Jan 20 at 11:05
OmGOmG
2,502822
$endgroup$
add a comment |
$begingroup$
Try to expand and generalize the case (by induction):
$$V(n) = 2times(2V(n-2) + 2(n-1)) + 2n = 2^2 V(n-2) + 2^2(n-1) + 2n = sum_{i=1}^{n}2^i(n-i+1)$$
Suppose $V(1) = 2$ in this case.
answered Jan 20 at 11:05
OmGOmG
2,502822
$endgroup$
add a comment |
$begingroup$
Try to expand and generalize the case (by induction):
$$V(n) = 2times(2V(n-2) + 2(n-1)) + 2n = 2^2 V(n-2) + 2^2(n-1) + 2n = sum_{i=1}^{n}2^i(n-i+1)$$
Suppose $V(1) = 2$ in this case.
answered Jan 20 at 11:05
OmGOmG
2,502822
$endgroup$
Try to expand and generalize the case (by induction):
$$V(n) = 2times(2V(n-2) + 2(n-1)) + 2n = 2^2 V(n-2) + 2^2(n-1) + 2n = sum_{i=1}^{n}2^i(n-i+1)$$
Suppose $V(1) = 2$ in this case.
answered Jan 20 at 11:05
OmGOmG
2,502822
answered Jan 20 at 11:05
OmGOmG
2,502822
answered Jan 20 at 11:05
OmGOmG
2,502822
answered Jan 20 at 11:05
OmGOmG
2,502822
2,502822
add a comment |
add a comment |
$begingroup$
Let $V(m)=f(m)+am+b$
$$2n=V(n)-2V(n-1)=f(n)+an+b-2{f(m-1)+a(n-1)+b}$$
$$2n=f(n)-2f(n-1)-an+2a-b$$
Set $2a-b=0, -an=2$ so that $f(n)=2f(n-1)=2^rf(n-r),0le rle n$
$implies a=-2, b=2(-2)=?$
answered Jan 20 at 12:42
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Let $V(m)=f(m)+am+b$
$$2n=V(n)-2V(n-1)=f(n)+an+b-2{f(m-1)+a(n-1)+b}$$
$$2n=f(n)-2f(n-1)-an+2a-b$$
Set $2a-b=0, -an=2$ so that $f(n)=2f(n-1)=2^rf(n-r),0le rle n$
$implies a=-2, b=2(-2)=?$
answered Jan 20 at 12:42
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
add a comment |
$begingroup$
Let $V(m)=f(m)+am+b$
$$2n=V(n)-2V(n-1)=f(n)+an+b-2{f(m-1)+a(n-1)+b}$$
$$2n=f(n)-2f(n-1)-an+2a-b$$
Set $2a-b=0, -an=2$ so that $f(n)=2f(n-1)=2^rf(n-r),0le rle n$
$implies a=-2, b=2(-2)=?$
answered Jan 20 at 12:42
lab bhattacharjeelab bhattacharjee
226k15157275
$endgroup$
Let $V(m)=f(m)+am+b$
$$2n=V(n)-2V(n-1)=f(n)+an+b-2{f(m-1)+a(n-1)+b}$$
$$2n=f(n)-2f(n-1)-an+2a-b$$
Set $2a-b=0, -an=2$ so that $f(n)=2f(n-1)=2^rf(n-r),0le rle n$
$implies a=-2, b=2(-2)=?$
answered Jan 20 at 12:42
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 20 at 12:42
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 20 at 12:42
lab bhattacharjeelab bhattacharjee
226k15157275
answered Jan 20 at 12:42
lab bhattacharjeelab bhattacharjee
226k15157275
226k15157275
add a comment |
add a comment |
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$begingroup$
Can you solve W(n) = W(n-1) + (2n)/2^n ? Then relate the Ws to the Vs and you are done...
$endgroup$
– Did
Jan 20 at 11:03
$begingroup$
@user623855 2n ≠ 2^n.
$endgroup$
– Did
Jan 20 at 11:04