Is the following Strong Induction conclusion true?
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can someone verify if this is true?
I assume that P(i) holds for i < n, but then I would believe that this means the conclusion should be 3^n-1 and not 3^n.
However, the correct answer given by my teacher says 3^n is correct.
induction
edited Jan 7 at 11:58
Arthur
112k7107190
asked Jan 7 at 11:48
NortonNorton
113
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add a comment |
$begingroup$
can someone verify if this is true?
I assume that P(i) holds for i < n, but then I would believe that this means the conclusion should be 3^n-1 and not 3^n.
However, the correct answer given by my teacher says 3^n is correct.
induction
edited Jan 7 at 11:58
Arthur
112k7107190
asked Jan 7 at 11:48
NortonNorton
113
New contributor
Norton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
please pick up mathjax to type maths on the site. Start by surrounding mathy objects with dollar signs
$endgroup$
– Siong Thye Goh
Jan 7 at 11:56
$begingroup$
$f(n) = 1$ is also a solution. Just so you know. The given induction proof doesn't have base cases (and it can't, since we aren't given any base values for $f$), and therefore hasn't really proven that $f(n) = 3^n$ just yet.
$endgroup$
– Arthur
Jan 7 at 12:00
add a comment |
$begingroup$
can someone verify if this is true?
I assume that P(i) holds for i < n, but then I would believe that this means the conclusion should be 3^n-1 and not 3^n.
However, the correct answer given by my teacher says 3^n is correct.
induction
edited Jan 7 at 11:58
Arthur
112k7107190
asked Jan 7 at 11:48
NortonNorton
113
New contributor
Norton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
can someone verify if this is true?
I assume that P(i) holds for i < n, but then I would believe that this means the conclusion should be 3^n-1 and not 3^n.
However, the correct answer given by my teacher says 3^n is correct.
induction
induction
edited Jan 7 at 11:58
Arthur
112k7107190
asked Jan 7 at 11:48
NortonNorton
113
New contributor
Norton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 7 at 11:58
Arthur
112k7107190
asked Jan 7 at 11:48
NortonNorton
113
New contributor
Norton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 7 at 11:58
Arthur
112k7107190
edited Jan 7 at 11:58
Arthur
112k7107190
edited Jan 7 at 11:58
Arthur
112k7107190
112k7107190
asked Jan 7 at 11:48
NortonNorton
113
New contributor
Norton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 7 at 11:48
NortonNorton
113
asked Jan 7 at 11:48
NortonNorton
113
113
New contributor
Norton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Norton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Norton is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
please pick up mathjax to type maths on the site. Start by surrounding mathy objects with dollar signs
$endgroup$
– Siong Thye Goh
Jan 7 at 11:56
$begingroup$
$f(n) = 1$ is also a solution. Just so you know. The given induction proof doesn't have base cases (and it can't, since we aren't given any base values for $f$), and therefore hasn't really proven that $f(n) = 3^n$ just yet.
$endgroup$
– Arthur
Jan 7 at 12:00
add a comment |
$begingroup$
please pick up mathjax to type maths on the site. Start by surrounding mathy objects with dollar signs
$endgroup$
– Siong Thye Goh
Jan 7 at 11:56
$begingroup$
$f(n) = 1$ is also a solution. Just so you know. The given induction proof doesn't have base cases (and it can't, since we aren't given any base values for $f$), and therefore hasn't really proven that $f(n) = 3^n$ just yet.
$endgroup$
– Arthur
Jan 7 at 12:00
$begingroup$
please pick up mathjax to type maths on the site. Start by surrounding mathy objects with dollar signs
$endgroup$
– Siong Thye Goh
Jan 7 at 11:56
$begingroup$
please pick up mathjax to type maths on the site. Start by surrounding mathy objects with dollar signs
$endgroup$
– Siong Thye Goh
Jan 7 at 11:56
$begingroup$
$f(n) = 1$ is also a solution. Just so you know. The given induction proof doesn't have base cases (and it can't, since we aren't given any base values for $f$), and therefore hasn't really proven that $f(n) = 3^n$ just yet.
$endgroup$
– Arthur
Jan 7 at 12:00
$begingroup$
$f(n) = 1$ is also a solution. Just so you know. The given induction proof doesn't have base cases (and it can't, since we aren't given any base values for $f$), and therefore hasn't really proven that $f(n) = 3^n$ just yet.
$endgroup$
– Arthur
Jan 7 at 12:00
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
By assuming $P(i)$ holds for $i<n$, clearly, we have $f(n-1)=3^{n-1}$ from this assumption.
Our goal for the induction step is to argue that from properties $$P(i) text{ holds for } i<ntag{1}$$ and $$f(n)=4cdot f(n-1)-3f(n-2)tag{2},$$ we can conclude $P(i)$ holds for $i=n$, that is $f(n)=3^n$.
answered Jan 7 at 11:55
Siong Thye GohSiong Thye Goh
100k1465117
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add a comment |
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1 Answer
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oldest
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votes
$begingroup$
By assuming $P(i)$ holds for $i<n$, clearly, we have $f(n-1)=3^{n-1}$ from this assumption.
Our goal for the induction step is to argue that from properties $$P(i) text{ holds for } i<ntag{1}$$ and $$f(n)=4cdot f(n-1)-3f(n-2)tag{2},$$ we can conclude $P(i)$ holds for $i=n$, that is $f(n)=3^n$.
answered Jan 7 at 11:55
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
By assuming $P(i)$ holds for $i<n$, clearly, we have $f(n-1)=3^{n-1}$ from this assumption.
Our goal for the induction step is to argue that from properties $$P(i) text{ holds for } i<ntag{1}$$ and $$f(n)=4cdot f(n-1)-3f(n-2)tag{2},$$ we can conclude $P(i)$ holds for $i=n$, that is $f(n)=3^n$.
answered Jan 7 at 11:55
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
add a comment |
$begingroup$
By assuming $P(i)$ holds for $i<n$, clearly, we have $f(n-1)=3^{n-1}$ from this assumption.
Our goal for the induction step is to argue that from properties $$P(i) text{ holds for } i<ntag{1}$$ and $$f(n)=4cdot f(n-1)-3f(n-2)tag{2},$$ we can conclude $P(i)$ holds for $i=n$, that is $f(n)=3^n$.
answered Jan 7 at 11:55
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
By assuming $P(i)$ holds for $i<n$, clearly, we have $f(n-1)=3^{n-1}$ from this assumption.
Our goal for the induction step is to argue that from properties $$P(i) text{ holds for } i<ntag{1}$$ and $$f(n)=4cdot f(n-1)-3f(n-2)tag{2},$$ we can conclude $P(i)$ holds for $i=n$, that is $f(n)=3^n$.
answered Jan 7 at 11:55
Siong Thye GohSiong Thye Goh
100k1465117
edited Jan 7 at 12:03
answered Jan 7 at 11:55
Siong Thye GohSiong Thye Goh
100k1465117
answered Jan 7 at 11:55
Siong Thye GohSiong Thye Goh
100k1465117
answered Jan 7 at 11:55
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
add a comment |
add a comment |
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$begingroup$
please pick up mathjax to type maths on the site. Start by surrounding mathy objects with dollar signs
$endgroup$
– Siong Thye Goh
Jan 7 at 11:56
$begingroup$
$f(n) = 1$ is also a solution. Just so you know. The given induction proof doesn't have base cases (and it can't, since we aren't given any base values for $f$), and therefore hasn't really proven that $f(n) = 3^n$ just yet.
$endgroup$
– Arthur
Jan 7 at 12:00