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gerund and noun applications

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How to understand this recursive definition?

How to solve this recursive equation?Solve this recursionRecursive definition of sequences inverse/equivalency.Prove by induction of recursive sequenceRecursive Definition of a SeriesFactorial Series Written As Recursive DefinitionNot sure if minimization is needed for definition of function 'half', or if primitive recursive definition possibleRecursive sequence is sequence of integersHow to write a recursive definition in a mathematical notation?Recursive definition of Wallis product.

0

$begingroup$

My question is:

Now know through my Discrete-Mathematics course what recursion is, but this notation confuses me.

$$
mathbf a_n =
begin{cases}
n, & text{for $ 0le n le 2 $ } \
a_{n-3}+a_{n-2}+a_{n-1}, & text{for $ n ge 3 $}
end{cases}
$$

So this is an example of my problem.

discrete-mathematics recursion

share|cite|improve this question

asked Mar 11 at 20:06

bite_Over3idebite_Over3ide

31

New contributor

bite_Over3ide is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What exactly is your problem? Can you say what $a_2$ is? What about $a_4?$
  $endgroup$
  – saulspatz
  Mar 11 at 20:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, I understand it now....
  $endgroup$
  – bite_Over3ide
  Mar 11 at 21:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Glad to hear it. Now that I read the accepted answer, I see that your problem was just with the notation. I didn't realize that before.
  $endgroup$
  – saulspatz
  Mar 11 at 21:52
  

  
  
  
  

add a comment | 

0

$begingroup$

My question is:

Now know through my Discrete-Mathematics course what recursion is, but this notation confuses me.

$$
mathbf a_n =
begin{cases}
n, & text{for $ 0le n le 2 $ } \
a_{n-3}+a_{n-2}+a_{n-1}, & text{for $ n ge 3 $}
end{cases}
$$

So this is an example of my problem.

discrete-mathematics recursion

share|cite|improve this question

asked Mar 11 at 20:06

bite_Over3idebite_Over3ide

31

New contributor

bite_Over3ide is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What exactly is your problem? Can you say what $a_2$ is? What about $a_4?$
  $endgroup$
  – saulspatz
  Mar 11 at 20:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, I understand it now....
  $endgroup$
  – bite_Over3ide
  Mar 11 at 21:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Glad to hear it. Now that I read the accepted answer, I see that your problem was just with the notation. I didn't realize that before.
  $endgroup$
  – saulspatz
  Mar 11 at 21:52
  

  
  
  
  

add a comment | 

$begingroup$

My question is:

Now know through my Discrete-Mathematics course what recursion is, but this notation confuses me.

$$
mathbf a_n =
begin{cases}
n, & text{for $ 0le n le 2 $ } \
a_{n-3}+a_{n-2}+a_{n-1}, & text{for $ n ge 3 $}
end{cases}
$$

So this is an example of my problem.

discrete-mathematics recursion

share|cite|improve this question

asked Mar 11 at 20:06

bite_Over3idebite_Over3ide

31

New contributor

bite_Over3ide is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Mar 11 at 20:06

bite_Over3idebite_Over3ide

31

New contributor

bite_Over3ide is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Mar 11 at 20:06

bite_Over3idebite_Over3ide

31

New contributor

bite_Over3ide is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 11 at 20:06

bite_Over3idebite_Over3ide

31

New contributor

bite_Over3ide is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 11 at 20:06

bite_Over3idebite_Over3ide

31

3 Answers
3

active

oldest

votes

1

$begingroup$

When an author uses the left-brace and conditions sort of notation you present here, the way to interpret it is to look at the right-hand column first. In this case, the first line says $0leq nleq 2$ so whatever it is saying applies to only $n$ values in that range, namely (if $n$ is assumed to be an integer) to $n=0$, $n=1$ and $n=2$. So the first line tells you that
$$
a_0 = 0
\ a_1 = 1
\a_2 = 2
$$
but it tells you nothing about $a_3$ or $a_4$ or $a_n$ for any other values of the index $n$.

The second line, on the right, says "Now I'm going to tell you about $n_n$ when $n geq 3$," which fortunately comprises all the other values of $n$ not covered by the first line. So for example,
$$
a_3 = a_2 + a_1 + a_0 = 2+1+0 = 3 \
a_4 = a_3+a_2+a_1 = 3+2+1 = 6
$$
but that second line does not apply if you wanted to find $a_2 = a_1 + a_0 + a_{-1}$. (In fact, $a_{-1}$ is left undefined in this definition of the $a_k$.)

share|cite|improve this answer

answered Mar 11 at 20:16

Mark FischlerMark Fischler

33.5k12452

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, now it make sense...
  $endgroup$
  – bite_Over3ide
  Mar 11 at 21:33
  

  
  
  
  

add a comment | 

1

$begingroup$

That means that $a_0=0,a_1=1,a_2=2$ and then you use the definition $a_n =a_{n-3} +a_{n-2}+a_{n-1}$

share|cite|improve this answer

answered Mar 11 at 20:09

JoseSquareJoseSquare

867116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I understand it now...
  $endgroup$
  – bite_Over3ide
  Mar 11 at 21:34
  

  
  
  
  

add a comment | 

0

$begingroup$

It means what it says. $a_0 = 0; a_1 = 1, a_2 = 2$.

And $a_3 = a_0 + a_1 + a_2$.

And $a_4 = a_1 + a_2 + a_3$.

And $a_5 = a_2 + a_3 + a_4$.

And $a_6 = a_3 + a_4 + a_5$ and so on.

In general: $a_n = a_{n-3} + a_{n-2} + a_{n-1}$.

Would it have made more sense if we had written

$a_{n+3} =a_{n} + a_{n+1} + a_{n+2}$?

share|cite|improve this answer

answered Mar 11 at 20:10

fleabloodfleablood

72.7k22788

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes both make sense now, thank you...
  $endgroup$
  – bite_Over3ide
  Mar 11 at 21:35
  

  
  
  
  

add a comment | 

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1

$begingroup$

When an author uses the left-brace and conditions sort of notation you present here, the way to interpret it is to look at the right-hand column first. In this case, the first line says $0leq nleq 2$ so whatever it is saying applies to only $n$ values in that range, namely (if $n$ is assumed to be an integer) to $n=0$, $n=1$ and $n=2$. So the first line tells you that
$$
a_0 = 0
\ a_1 = 1
\a_2 = 2
$$
but it tells you nothing about $a_3$ or $a_4$ or $a_n$ for any other values of the index $n$.

The second line, on the right, says "Now I'm going to tell you about $n_n$ when $n geq 3$," which fortunately comprises all the other values of $n$ not covered by the first line. So for example,
$$
a_3 = a_2 + a_1 + a_0 = 2+1+0 = 3 \
a_4 = a_3+a_2+a_1 = 3+2+1 = 6
$$
but that second line does not apply if you wanted to find $a_2 = a_1 + a_0 + a_{-1}$. (In fact, $a_{-1}$ is left undefined in this definition of the $a_k$.)

share|cite|improve this answer

answered Mar 11 at 20:16

Mark FischlerMark Fischler

33.5k12452

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, now it make sense...
  $endgroup$
  – bite_Over3ide
  Mar 11 at 21:33
  

  
  
  
  

add a comment | 

1

$begingroup$

When an author uses the left-brace and conditions sort of notation you present here, the way to interpret it is to look at the right-hand column first. In this case, the first line says $0leq nleq 2$ so whatever it is saying applies to only $n$ values in that range, namely (if $n$ is assumed to be an integer) to $n=0$, $n=1$ and $n=2$. So the first line tells you that
$$
a_0 = 0
\ a_1 = 1
\a_2 = 2
$$
but it tells you nothing about $a_3$ or $a_4$ or $a_n$ for any other values of the index $n$.

The second line, on the right, says "Now I'm going to tell you about $n_n$ when $n geq 3$," which fortunately comprises all the other values of $n$ not covered by the first line. So for example,
$$
a_3 = a_2 + a_1 + a_0 = 2+1+0 = 3 \
a_4 = a_3+a_2+a_1 = 3+2+1 = 6
$$
but that second line does not apply if you wanted to find $a_2 = a_1 + a_0 + a_{-1}$. (In fact, $a_{-1}$ is left undefined in this definition of the $a_k$.)

share|cite|improve this answer

answered Mar 11 at 20:16

Mark FischlerMark Fischler

33.5k12452

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, now it make sense...
  $endgroup$
  – bite_Over3ide
  Mar 11 at 21:33
  

  
  
  
  

add a comment | 

$begingroup$

When an author uses the left-brace and conditions sort of notation you present here, the way to interpret it is to look at the right-hand column first. In this case, the first line says $0leq nleq 2$ so whatever it is saying applies to only $n$ values in that range, namely (if $n$ is assumed to be an integer) to $n=0$, $n=1$ and $n=2$. So the first line tells you that
$$
a_0 = 0
\ a_1 = 1
\a_2 = 2
$$
but it tells you nothing about $a_3$ or $a_4$ or $a_n$ for any other values of the index $n$.

The second line, on the right, says "Now I'm going to tell you about $n_n$ when $n geq 3$," which fortunately comprises all the other values of $n$ not covered by the first line. So for example,
$$
a_3 = a_2 + a_1 + a_0 = 2+1+0 = 3 \
a_4 = a_3+a_2+a_1 = 3+2+1 = 6
$$
but that second line does not apply if you wanted to find $a_2 = a_1 + a_0 + a_{-1}$. (In fact, $a_{-1}$ is left undefined in this definition of the $a_k$.)

share|cite|improve this answer

answered Mar 11 at 20:16

Mark FischlerMark Fischler

33.5k12452

$endgroup$

share|cite|improve this answer

answered Mar 11 at 20:16

Mark FischlerMark Fischler

33.5k12452

answered Mar 11 at 20:16

Mark FischlerMark Fischler

33.5k12452

answered Mar 11 at 20:16

Mark FischlerMark Fischler

33.5k12452

1

$begingroup$

That means that $a_0=0,a_1=1,a_2=2$ and then you use the definition $a_n =a_{n-3} +a_{n-2}+a_{n-1}$

share|cite|improve this answer

answered Mar 11 at 20:09

JoseSquareJoseSquare

867116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I understand it now...
  $endgroup$
  – bite_Over3ide
  Mar 11 at 21:34
  

  
  
  
  

add a comment | 

1

$begingroup$

That means that $a_0=0,a_1=1,a_2=2$ and then you use the definition $a_n =a_{n-3} +a_{n-2}+a_{n-1}$

share|cite|improve this answer

answered Mar 11 at 20:09

JoseSquareJoseSquare

867116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you, I understand it now...
  $endgroup$
  – bite_Over3ide
  Mar 11 at 21:34
  

  
  
  
  

add a comment | 

$begingroup$

That means that $a_0=0,a_1=1,a_2=2$ and then you use the definition $a_n =a_{n-3} +a_{n-2}+a_{n-1}$

share|cite|improve this answer

answered Mar 11 at 20:09

JoseSquareJoseSquare

867116

$endgroup$

share|cite|improve this answer

answered Mar 11 at 20:09

JoseSquareJoseSquare

867116

answered Mar 11 at 20:09

JoseSquareJoseSquare

867116

answered Mar 11 at 20:09

JoseSquareJoseSquare

867116

0

$begingroup$

It means what it says. $a_0 = 0; a_1 = 1, a_2 = 2$.

And $a_3 = a_0 + a_1 + a_2$.

And $a_4 = a_1 + a_2 + a_3$.

And $a_5 = a_2 + a_3 + a_4$.

And $a_6 = a_3 + a_4 + a_5$ and so on.

In general: $a_n = a_{n-3} + a_{n-2} + a_{n-1}$.

Would it have made more sense if we had written

$a_{n+3} =a_{n} + a_{n+1} + a_{n+2}$?

share|cite|improve this answer

answered Mar 11 at 20:10

fleabloodfleablood

72.7k22788

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes both make sense now, thank you...
  $endgroup$
  – bite_Over3ide
  Mar 11 at 21:35
  

  
  
  
  

add a comment | 

0

$begingroup$

It means what it says. $a_0 = 0; a_1 = 1, a_2 = 2$.

And $a_3 = a_0 + a_1 + a_2$.

And $a_4 = a_1 + a_2 + a_3$.

And $a_5 = a_2 + a_3 + a_4$.

And $a_6 = a_3 + a_4 + a_5$ and so on.

In general: $a_n = a_{n-3} + a_{n-2} + a_{n-1}$.

Would it have made more sense if we had written

$a_{n+3} =a_{n} + a_{n+1} + a_{n+2}$?

share|cite|improve this answer

answered Mar 11 at 20:10

fleabloodfleablood

72.7k22788

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes both make sense now, thank you...
  $endgroup$
  – bite_Over3ide
  Mar 11 at 21:35
  

  
  
  
  

add a comment | 

$begingroup$

It means what it says. $a_0 = 0; a_1 = 1, a_2 = 2$.

And $a_3 = a_0 + a_1 + a_2$.

And $a_4 = a_1 + a_2 + a_3$.

And $a_5 = a_2 + a_3 + a_4$.

And $a_6 = a_3 + a_4 + a_5$ and so on.

In general: $a_n = a_{n-3} + a_{n-2} + a_{n-1}$.

Would it have made more sense if we had written

$a_{n+3} =a_{n} + a_{n+1} + a_{n+2}$?

share|cite|improve this answer

answered Mar 11 at 20:10

fleabloodfleablood

72.7k22788

$endgroup$

share|cite|improve this answer

answered Mar 11 at 20:10

fleabloodfleablood

72.7k22788

answered Mar 11 at 20:10

fleabloodfleablood

72.7k22788

answered Mar 11 at 20:10

fleabloodfleablood

72.7k22788