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Function problem with Sigma summation
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As a programmer learning how to write math-notation, I want to be able to exit when some function inside the Sigma summation has reached a particular value, but keep the value that the Sigma has computed. In programming this is easy, but not as math notation (unless I have missed something).
Say I have $xinmathbbN$ as bitstring: $x = 1100100110110101_2$ $(=51637_10)$ and I check each pair from right to left: $01$ $01$ $11$ $10$ $01$ $10$ $00$ $11$ using $pmod 4$ I need to make a "counter" function, such that when the pair reaches one of the bitvalues: $11$, $10$ or $00$ the counter stops else increases by $1$.
In decimal this would be equivalent of saying that when $51637 pmod 4$ equals: $3$,$2$ or $0$ the counter should stop else add $1$ and continue.
And on the next iteration we divide $51637$ by $4$ to get $12909$ so: $12909 pmod 4$ equals: $3$,$2$ or 0 counter should stop else add $1$ and continue.
and iterate this procedure until $x = 0$.
So we have the function:
$S(n)=begincases
0 & nequiv 0pmod4 \
1 & nequiv 1pmod4\
0 & nequiv 2pmod4\
0 & nequiv 3pmod4endcases $
The iterative procedure should stop at the first instance of one of these zeros, but keep the value that $S(n)$ has computed along the way.
$S(n)$ has a drawback with this, it does not return after hitting the case: zero. For me it feels like a general barrier (if one could call it that) in all functions though. Or could one use two-dimensional functions to overcome this?
There is no exit-case in sigma-symbol as far as i can tell. The problem is that the "counter" function continues after reaching $3$,$2$ or $0$, but i don't want that.
Are there any solutions to this problem? I also dont know if capital pi notation can be used in the solution or not?
Example
If $c$ is the bitstring length divided by $2$, and $n=51637$, an example would be:
$$Z(n)=sum_i=0^c-1 fracS(n)4^i$$
$Z(n)$ should return the value $3$. But it does not because Sigma has to finish.
Don Fuchs solution seems to give the right answer, if this table is correct (See image below):
functions discrete-mathematics natural-numbers
asked 8 hours ago
Natural Number Guy
366214
add a comment |Â
up vote
0
down vote
favorite
As a programmer learning how to write math-notation, I want to be able to exit when some function inside the Sigma summation has reached a particular value, but keep the value that the Sigma has computed. In programming this is easy, but not as math notation (unless I have missed something).
Say I have $xinmathbbN$ as bitstring: $x = 1100100110110101_2$ $(=51637_10)$ and I check each pair from right to left: $01$ $01$ $11$ $10$ $01$ $10$ $00$ $11$ using $pmod 4$ I need to make a "counter" function, such that when the pair reaches one of the bitvalues: $11$, $10$ or $00$ the counter stops else increases by $1$.
In decimal this would be equivalent of saying that when $51637 pmod 4$ equals: $3$,$2$ or $0$ the counter should stop else add $1$ and continue.
And on the next iteration we divide $51637$ by $4$ to get $12909$ so: $12909 pmod 4$ equals: $3$,$2$ or 0 counter should stop else add $1$ and continue.
and iterate this procedure until $x = 0$.
So we have the function:
$S(n)=begincases
0 & nequiv 0pmod4 \
1 & nequiv 1pmod4\
0 & nequiv 2pmod4\
0 & nequiv 3pmod4endcases $
The iterative procedure should stop at the first instance of one of these zeros, but keep the value that $S(n)$ has computed along the way.
$S(n)$ has a drawback with this, it does not return after hitting the case: zero. For me it feels like a general barrier (if one could call it that) in all functions though. Or could one use two-dimensional functions to overcome this?
There is no exit-case in sigma-symbol as far as i can tell. The problem is that the "counter" function continues after reaching $3$,$2$ or $0$, but i don't want that.
Are there any solutions to this problem? I also dont know if capital pi notation can be used in the solution or not?
Example
If $c$ is the bitstring length divided by $2$, and $n=51637$, an example would be:
$$Z(n)=sum_i=0^c-1 fracS(n)4^i$$
$Z(n)$ should return the value $3$. But it does not because Sigma has to finish.
Don Fuchs solution seems to give the right answer, if this table is correct (See image below):
functions discrete-mathematics natural-numbers
asked 8 hours ago
Natural Number Guy
366214
Mathematically it seems that "counter stops" is the same as "counter increases by 0."
– John Wayland Bales
8 hours ago
That's true John
– Natural Number Guy
53 mins ago
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
As a programmer learning how to write math-notation, I want to be able to exit when some function inside the Sigma summation has reached a particular value, but keep the value that the Sigma has computed. In programming this is easy, but not as math notation (unless I have missed something).
Say I have $xinmathbbN$ as bitstring: $x = 1100100110110101_2$ $(=51637_10)$ and I check each pair from right to left: $01$ $01$ $11$ $10$ $01$ $10$ $00$ $11$ using $pmod 4$ I need to make a "counter" function, such that when the pair reaches one of the bitvalues: $11$, $10$ or $00$ the counter stops else increases by $1$.
In decimal this would be equivalent of saying that when $51637 pmod 4$ equals: $3$,$2$ or $0$ the counter should stop else add $1$ and continue.
And on the next iteration we divide $51637$ by $4$ to get $12909$ so: $12909 pmod 4$ equals: $3$,$2$ or 0 counter should stop else add $1$ and continue.
and iterate this procedure until $x = 0$.
So we have the function:
$S(n)=begincases
0 & nequiv 0pmod4 \
1 & nequiv 1pmod4\
0 & nequiv 2pmod4\
0 & nequiv 3pmod4endcases $
The iterative procedure should stop at the first instance of one of these zeros, but keep the value that $S(n)$ has computed along the way.
$S(n)$ has a drawback with this, it does not return after hitting the case: zero. For me it feels like a general barrier (if one could call it that) in all functions though. Or could one use two-dimensional functions to overcome this?
There is no exit-case in sigma-symbol as far as i can tell. The problem is that the "counter" function continues after reaching $3$,$2$ or $0$, but i don't want that.
Are there any solutions to this problem? I also dont know if capital pi notation can be used in the solution or not?
Example
If $c$ is the bitstring length divided by $2$, and $n=51637$, an example would be:
$$Z(n)=sum_i=0^c-1 fracS(n)4^i$$
$Z(n)$ should return the value $3$. But it does not because Sigma has to finish.
Don Fuchs solution seems to give the right answer, if this table is correct (See image below):
functions discrete-mathematics natural-numbers
asked 8 hours ago
Natural Number Guy
366214
As a programmer learning how to write math-notation, I want to be able to exit when some function inside the Sigma summation has reached a particular value, but keep the value that the Sigma has computed. In programming this is easy, but not as math notation (unless I have missed something).
Say I have $xinmathbbN$ as bitstring: $x = 1100100110110101_2$ $(=51637_10)$ and I check each pair from right to left: $01$ $01$ $11$ $10$ $01$ $10$ $00$ $11$ using $pmod 4$ I need to make a "counter" function, such that when the pair reaches one of the bitvalues: $11$, $10$ or $00$ the counter stops else increases by $1$.
In decimal this would be equivalent of saying that when $51637 pmod 4$ equals: $3$,$2$ or $0$ the counter should stop else add $1$ and continue.
And on the next iteration we divide $51637$ by $4$ to get $12909$ so: $12909 pmod 4$ equals: $3$,$2$ or 0 counter should stop else add $1$ and continue.
and iterate this procedure until $x = 0$.
So we have the function:
$S(n)=begincases
0 & nequiv 0pmod4 \
1 & nequiv 1pmod4\
0 & nequiv 2pmod4\
0 & nequiv 3pmod4endcases $
The iterative procedure should stop at the first instance of one of these zeros, but keep the value that $S(n)$ has computed along the way.
$S(n)$ has a drawback with this, it does not return after hitting the case: zero. For me it feels like a general barrier (if one could call it that) in all functions though. Or could one use two-dimensional functions to overcome this?
There is no exit-case in sigma-symbol as far as i can tell. The problem is that the "counter" function continues after reaching $3$,$2$ or $0$, but i don't want that.
Are there any solutions to this problem? I also dont know if capital pi notation can be used in the solution or not?
Example
If $c$ is the bitstring length divided by $2$, and $n=51637$, an example would be:
$$Z(n)=sum_i=0^c-1 fracS(n)4^i$$
$Z(n)$ should return the value $3$. But it does not because Sigma has to finish.
Don Fuchs solution seems to give the right answer, if this table is correct (See image below):
functions discrete-mathematics natural-numbers
asked 8 hours ago
Natural Number Guy
366214
edited 4 hours ago
asked 8 hours ago
Natural Number Guy
366214
asked 8 hours ago
Natural Number Guy
366214
asked 8 hours ago
Natural Number Guy
366214
366214
Mathematically it seems that "counter stops" is the same as "counter increases by 0."
– John Wayland Bales
8 hours ago
That's true John
– Natural Number Guy
53 mins ago
add a comment |Â
Mathematically it seems that "counter stops" is the same as "counter increases by 0."
– John Wayland Bales
8 hours ago
That's true John
– Natural Number Guy
53 mins ago
Mathematically it seems that "counter stops" is the same as "counter increases by 0."
– John Wayland Bales
8 hours ago
Mathematically it seems that "counter stops" is the same as "counter increases by 0."
– John Wayland Bales
8 hours ago
That's true John
– Natural Number Guy
53 mins ago
That's true John
– Natural Number Guy
53 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
Your idea to use a product isn't too bad; I would try
$$
Z(n)=sum_i=0^inftyleft[Sleft(fracn4^iright)prod_j=0^iSleft(fracn4^jright)right].
$$
With the first $i$ such that $S(n/4^i)$ equals $0$ the inner product contains a $0$ factor and therefore is $0$, and remains $0$ even for higher indices $k$ where $S(n/4^k)$ might equal $1$ (it "has a memory").
Also note the divisor $4^i$ dividing the argument of $S$, not $S(n)$ itself, to get your desired iterative structure (if I got you right there).
answered 7 hours ago
Don Fuchs
234
Can you look at the table I've added to see if I understand your solution is same. Yes I the pi-notation seems correct. I have to be sure I understand it first. And I will accept your answer if so. (And of course I see that I could just add 1 to the whole function to get the answer 3. so thats just a minor problem).
– Natural Number Guy
4 hours ago
And yes, you are right about dividing the argument and not $S(n)$ itself.
– Natural Number Guy
55 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Your idea to use a product isn't too bad; I would try
$$
Z(n)=sum_i=0^inftyleft[Sleft(fracn4^iright)prod_j=0^iSleft(fracn4^jright)right].
$$
With the first $i$ such that $S(n/4^i)$ equals $0$ the inner product contains a $0$ factor and therefore is $0$, and remains $0$ even for higher indices $k$ where $S(n/4^k)$ might equal $1$ (it "has a memory").
Also note the divisor $4^i$ dividing the argument of $S$, not $S(n)$ itself, to get your desired iterative structure (if I got you right there).
answered 7 hours ago
Don Fuchs
234
Can you look at the table I've added to see if I understand your solution is same. Yes I the pi-notation seems correct. I have to be sure I understand it first. And I will accept your answer if so. (And of course I see that I could just add 1 to the whole function to get the answer 3. so thats just a minor problem).
– Natural Number Guy
4 hours ago
And yes, you are right about dividing the argument and not $S(n)$ itself.
– Natural Number Guy
55 mins ago
add a comment |Â
up vote
1
down vote
Your idea to use a product isn't too bad; I would try
$$
Z(n)=sum_i=0^inftyleft[Sleft(fracn4^iright)prod_j=0^iSleft(fracn4^jright)right].
$$
With the first $i$ such that $S(n/4^i)$ equals $0$ the inner product contains a $0$ factor and therefore is $0$, and remains $0$ even for higher indices $k$ where $S(n/4^k)$ might equal $1$ (it "has a memory").
Also note the divisor $4^i$ dividing the argument of $S$, not $S(n)$ itself, to get your desired iterative structure (if I got you right there).
answered 7 hours ago
Don Fuchs
234
Can you look at the table I've added to see if I understand your solution is same. Yes I the pi-notation seems correct. I have to be sure I understand it first. And I will accept your answer if so. (And of course I see that I could just add 1 to the whole function to get the answer 3. so thats just a minor problem).
– Natural Number Guy
4 hours ago
And yes, you are right about dividing the argument and not $S(n)$ itself.
– Natural Number Guy
55 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your idea to use a product isn't too bad; I would try
$$
Z(n)=sum_i=0^inftyleft[Sleft(fracn4^iright)prod_j=0^iSleft(fracn4^jright)right].
$$
With the first $i$ such that $S(n/4^i)$ equals $0$ the inner product contains a $0$ factor and therefore is $0$, and remains $0$ even for higher indices $k$ where $S(n/4^k)$ might equal $1$ (it "has a memory").
Also note the divisor $4^i$ dividing the argument of $S$, not $S(n)$ itself, to get your desired iterative structure (if I got you right there).
answered 7 hours ago
Don Fuchs
234
Your idea to use a product isn't too bad; I would try
$$
Z(n)=sum_i=0^inftyleft[Sleft(fracn4^iright)prod_j=0^iSleft(fracn4^jright)right].
$$
With the first $i$ such that $S(n/4^i)$ equals $0$ the inner product contains a $0$ factor and therefore is $0$, and remains $0$ even for higher indices $k$ where $S(n/4^k)$ might equal $1$ (it "has a memory").
Also note the divisor $4^i$ dividing the argument of $S$, not $S(n)$ itself, to get your desired iterative structure (if I got you right there).
answered 7 hours ago
Don Fuchs
234
edited 7 hours ago
answered 7 hours ago
Don Fuchs
234
answered 7 hours ago
Don Fuchs
234
answered 7 hours ago
Don Fuchs
234
234
Can you look at the table I've added to see if I understand your solution is same. Yes I the pi-notation seems correct. I have to be sure I understand it first. And I will accept your answer if so. (And of course I see that I could just add 1 to the whole function to get the answer 3. so thats just a minor problem).
– Natural Number Guy
4 hours ago
And yes, you are right about dividing the argument and not $S(n)$ itself.
– Natural Number Guy
55 mins ago
add a comment |Â
Can you look at the table I've added to see if I understand your solution is same. Yes I the pi-notation seems correct. I have to be sure I understand it first. And I will accept your answer if so. (And of course I see that I could just add 1 to the whole function to get the answer 3. so thats just a minor problem).
– Natural Number Guy
4 hours ago
And yes, you are right about dividing the argument and not $S(n)$ itself.
– Natural Number Guy
55 mins ago
Can you look at the table I've added to see if I understand your solution is same. Yes I the pi-notation seems correct. I have to be sure I understand it first. And I will accept your answer if so. (And of course I see that I could just add 1 to the whole function to get the answer 3. so thats just a minor problem).
– Natural Number Guy
4 hours ago
Can you look at the table I've added to see if I understand your solution is same. Yes I the pi-notation seems correct. I have to be sure I understand it first. And I will accept your answer if so. (And of course I see that I could just add 1 to the whole function to get the answer 3. so thats just a minor problem).
– Natural Number Guy
4 hours ago
And yes, you are right about dividing the argument and not $S(n)$ itself.
– Natural Number Guy
55 mins ago
And yes, you are right about dividing the argument and not $S(n)$ itself.
– Natural Number Guy
55 mins ago
add a comment |Â
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Mathematically it seems that "counter stops" is the same as "counter increases by 0."
– John Wayland Bales
8 hours ago
That's true John
– Natural Number Guy
53 mins ago