Mixing
Is this how to prove a recurrence?
$begingroup$
Given:
$U(0)=4$
$U(n+1)=3+2cdot U(n)$ ...recurrence case
$U(n)=7cdot 2^n-3$
My Solution:
$U(n+1)=3+2cdot U(n)$
$frac{U(n+1)-3}{2}=U(n)$
$frac{U(n+1)-3}{2}=7cdot 2^n-3$
Since I know $U(0) =4$, and with that I can solve $U(1)=11$
$n = 0$
$frac{U(0+1)-3}{2}=7cdot2^0-3$
$4=4$ .... is this a proof?
recurrence-relations
edited Jan 16 at 13:09
Bernard
121k740116
asked Jan 16 at 10:53
NoobProgrammerNoobProgrammer
32
$endgroup$
add a comment |
$begingroup$
Given:
$U(0)=4$
$U(n+1)=3+2cdot U(n)$ ...recurrence case
$U(n)=7cdot 2^n-3$
My Solution:
$U(n+1)=3+2cdot U(n)$
$frac{U(n+1)-3}{2}=U(n)$
$frac{U(n+1)-3}{2}=7cdot 2^n-3$
Since I know $U(0) =4$, and with that I can solve $U(1)=11$
$n = 0$
$frac{U(0+1)-3}{2}=7cdot2^0-3$
$4=4$ .... is this a proof?
recurrence-relations
edited Jan 16 at 13:09
Bernard
121k740116
asked Jan 16 at 10:53
NoobProgrammerNoobProgrammer
32
$endgroup$
3
$begingroup$
If you already have U(n), what else do you want to solve?
$endgroup$
– idea
Jan 16 at 10:58
$begingroup$
"How to prove a recurrence" is a bit vague. When you have $$ U_{n+1} = 3 + 2U_{n} $$ this is quite clearly a recurrence. It's not clear what is required ...
$endgroup$
– Matti P.
Jan 16 at 11:01
add a comment |
$begingroup$
Given:
$U(0)=4$
$U(n+1)=3+2cdot U(n)$ ...recurrence case
$U(n)=7cdot 2^n-3$
My Solution:
$U(n+1)=3+2cdot U(n)$
$frac{U(n+1)-3}{2}=U(n)$
$frac{U(n+1)-3}{2}=7cdot 2^n-3$
Since I know $U(0) =4$, and with that I can solve $U(1)=11$
$n = 0$
$frac{U(0+1)-3}{2}=7cdot2^0-3$
$4=4$ .... is this a proof?
recurrence-relations
edited Jan 16 at 13:09
Bernard
121k740116
asked Jan 16 at 10:53
NoobProgrammerNoobProgrammer
32
$endgroup$
Given:
$U(0)=4$
$U(n+1)=3+2cdot U(n)$ ...recurrence case
$U(n)=7cdot 2^n-3$
My Solution:
$U(n+1)=3+2cdot U(n)$
$frac{U(n+1)-3}{2}=U(n)$
$frac{U(n+1)-3}{2}=7cdot 2^n-3$
Since I know $U(0) =4$, and with that I can solve $U(1)=11$
$n = 0$
$frac{U(0+1)-3}{2}=7cdot2^0-3$
$4=4$ .... is this a proof?
recurrence-relations
recurrence-relations
edited Jan 16 at 13:09
Bernard
121k740116
asked Jan 16 at 10:53
NoobProgrammerNoobProgrammer
32
edited Jan 16 at 13:09
Bernard
121k740116
asked Jan 16 at 10:53
NoobProgrammerNoobProgrammer
32
edited Jan 16 at 13:09
Bernard
121k740116
edited Jan 16 at 13:09
Bernard
121k740116
edited Jan 16 at 13:09
Bernard
121k740116
121k740116
asked Jan 16 at 10:53
NoobProgrammerNoobProgrammer
32
asked Jan 16 at 10:53
NoobProgrammerNoobProgrammer
32
asked Jan 16 at 10:53
NoobProgrammerNoobProgrammer
32
32
3
$begingroup$
If you already have U(n), what else do you want to solve?
$endgroup$
– idea
Jan 16 at 10:58
$begingroup$
"How to prove a recurrence" is a bit vague. When you have $$ U_{n+1} = 3 + 2U_{n} $$ this is quite clearly a recurrence. It's not clear what is required ...
$endgroup$
– Matti P.
Jan 16 at 11:01
add a comment |
3
$begingroup$
If you already have U(n), what else do you want to solve?
$endgroup$
– idea
Jan 16 at 10:58
$begingroup$
"How to prove a recurrence" is a bit vague. When you have $$ U_{n+1} = 3 + 2U_{n} $$ this is quite clearly a recurrence. It's not clear what is required ...
$endgroup$
– Matti P.
Jan 16 at 11:01
3
3
$begingroup$
If you already have U(n), what else do you want to solve?
$endgroup$
– idea
Jan 16 at 10:58
$begingroup$
If you already have U(n), what else do you want to solve?
$endgroup$
– idea
Jan 16 at 10:58
$begingroup$
"How to prove a recurrence" is a bit vague. When you have $$ U_{n+1} = 3 + 2U_{n} $$ this is quite clearly a recurrence. It's not clear what is required ...
$endgroup$
– Matti P.
Jan 16 at 11:01
$begingroup$
"How to prove a recurrence" is a bit vague. When you have $$ U_{n+1} = 3 + 2U_{n} $$ this is quite clearly a recurrence. It's not clear what is required ...
$endgroup$
– Matti P.
Jan 16 at 11:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Not really a proof, but it contains the idea.
What you're doing should be formalized with induction.
Given: $U(0)=4$, $U(n+1)=3+2U(n)$
Conjecture: $U(n)=7cdot 2^n-3$
Base of induction: $7cdot2^0-3=4=U(0)$. Verified true.
Induction step: Assume $U(n)=7cdot2^n-3$. Then
$$
U(n+1)=3+2U(n)=3+2(7cdot2^n-3)=2cdot7cdot2^{n}+3-6=7cdot2^{n+1}-3.
$$
End.
answered Jan 16 at 11:00
egregegreg
182k1486204
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Not really a proof, but it contains the idea.
What you're doing should be formalized with induction.
Given: $U(0)=4$, $U(n+1)=3+2U(n)$
Conjecture: $U(n)=7cdot 2^n-3$
Base of induction: $7cdot2^0-3=4=U(0)$. Verified true.
Induction step: Assume $U(n)=7cdot2^n-3$. Then
$$
U(n+1)=3+2U(n)=3+2(7cdot2^n-3)=2cdot7cdot2^{n}+3-6=7cdot2^{n+1}-3.
$$
End.
answered Jan 16 at 11:00
egregegreg
182k1486204
$endgroup$
add a comment |
$begingroup$
Not really a proof, but it contains the idea.
What you're doing should be formalized with induction.
Given: $U(0)=4$, $U(n+1)=3+2U(n)$
Conjecture: $U(n)=7cdot 2^n-3$
Base of induction: $7cdot2^0-3=4=U(0)$. Verified true.
Induction step: Assume $U(n)=7cdot2^n-3$. Then
$$
U(n+1)=3+2U(n)=3+2(7cdot2^n-3)=2cdot7cdot2^{n}+3-6=7cdot2^{n+1}-3.
$$
End.
answered Jan 16 at 11:00
egregegreg
182k1486204
$endgroup$
add a comment |
$begingroup$
Not really a proof, but it contains the idea.
What you're doing should be formalized with induction.
Given: $U(0)=4$, $U(n+1)=3+2U(n)$
Conjecture: $U(n)=7cdot 2^n-3$
Base of induction: $7cdot2^0-3=4=U(0)$. Verified true.
Induction step: Assume $U(n)=7cdot2^n-3$. Then
$$
U(n+1)=3+2U(n)=3+2(7cdot2^n-3)=2cdot7cdot2^{n}+3-6=7cdot2^{n+1}-3.
$$
End.
answered Jan 16 at 11:00
egregegreg
182k1486204
$endgroup$
Not really a proof, but it contains the idea.
What you're doing should be formalized with induction.
Given: $U(0)=4$, $U(n+1)=3+2U(n)$
Conjecture: $U(n)=7cdot 2^n-3$
Base of induction: $7cdot2^0-3=4=U(0)$. Verified true.
Induction step: Assume $U(n)=7cdot2^n-3$. Then
$$
U(n+1)=3+2U(n)=3+2(7cdot2^n-3)=2cdot7cdot2^{n}+3-6=7cdot2^{n+1}-3.
$$
End.
answered Jan 16 at 11:00
egregegreg
182k1486204
answered Jan 16 at 11:00
egregegreg
182k1486204
answered Jan 16 at 11:00
egregegreg
182k1486204
answered Jan 16 at 11:00
egregegreg
182k1486204
182k1486204
add a comment |
add a comment |
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3
$begingroup$
If you already have U(n), what else do you want to solve?
$endgroup$
– idea
Jan 16 at 10:58
$begingroup$
"How to prove a recurrence" is a bit vague. When you have $$ U_{n+1} = 3 + 2U_{n} $$ this is quite clearly a recurrence. It's not clear what is required ...
$endgroup$
– Matti P.
Jan 16 at 11:01