Mixing
Limits and While Loops
Clash Royale CLAN TAG#URR8PPP
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1
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Question: Consider the following program. Does $f(1)=0$?
beginalign*
f(i):=&|textwhile frac1i>0\
&||ileftarrow i+1\
&|i
endalign*
I would say that $f(1)=0$ is a true statement. The program does not terminate, but one could consider the sequence of points $x_i_iin mathbbN$ given by $x_i:=frac1i$, so that $x_i_iin mathbbN$ converges to the limit $0$ which makes $underbracefrac1i_=0>0$ false which means $f(1)=0$. However, I could be wrong.
limits programming
edited 24 mins ago
lioness99a
3,6382527
asked 33 mins ago
W. G.
5131416
add a comment |Â
up vote
1
down vote
favorite
Question: Consider the following program. Does $f(1)=0$?
beginalign*
f(i):=&|textwhile frac1i>0\
&||ileftarrow i+1\
&|i
endalign*
I would say that $f(1)=0$ is a true statement. The program does not terminate, but one could consider the sequence of points $x_i_iin mathbbN$ given by $x_i:=frac1i$, so that $x_i_iin mathbbN$ converges to the limit $0$ which makes $underbracefrac1i_=0>0$ false which means $f(1)=0$. However, I could be wrong.
limits programming
edited 24 mins ago
lioness99a
3,6382527
asked 33 mins ago
W. G.
5131416
The line $||ileftarrow i+1$ add 1 to the original variable $i$?
– gimusi
27 mins ago
1
Yes, that is exactly what that means. So, $frac11>0$ is true in the beginning so then we add $1$ to $i$ which looking at the precondition we know $frac12>0$ is true. This loop continues over and over until the statement is false.
– W. G.
24 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Question: Consider the following program. Does $f(1)=0$?
beginalign*
f(i):=&|textwhile frac1i>0\
&||ileftarrow i+1\
&|i
endalign*
I would say that $f(1)=0$ is a true statement. The program does not terminate, but one could consider the sequence of points $x_i_iin mathbbN$ given by $x_i:=frac1i$, so that $x_i_iin mathbbN$ converges to the limit $0$ which makes $underbracefrac1i_=0>0$ false which means $f(1)=0$. However, I could be wrong.
limits programming
edited 24 mins ago
lioness99a
3,6382527
asked 33 mins ago
W. G.
5131416
Question: Consider the following program. Does $f(1)=0$?
beginalign*
f(i):=&|textwhile frac1i>0\
&||ileftarrow i+1\
&|i
endalign*
I would say that $f(1)=0$ is a true statement. The program does not terminate, but one could consider the sequence of points $x_i_iin mathbbN$ given by $x_i:=frac1i$, so that $x_i_iin mathbbN$ converges to the limit $0$ which makes $underbracefrac1i_=0>0$ false which means $f(1)=0$. However, I could be wrong.
limits programming
limits programming
edited 24 mins ago
lioness99a
3,6382527
asked 33 mins ago
W. G.
5131416
edited 24 mins ago
lioness99a
3,6382527
asked 33 mins ago
W. G.
5131416
edited 24 mins ago
lioness99a
3,6382527
edited 24 mins ago
lioness99a
3,6382527
edited 24 mins ago
lioness99a
3,6382527
3,6382527
asked 33 mins ago
W. G.
5131416
asked 33 mins ago
W. G.
5131416
asked 33 mins ago
W. G.
5131416
5131416
The line $||ileftarrow i+1$ add 1 to the original variable $i$?
– gimusi
27 mins ago
1
Yes, that is exactly what that means. So, $frac11>0$ is true in the beginning so then we add $1$ to $i$ which looking at the precondition we know $frac12>0$ is true. This loop continues over and over until the statement is false.
– W. G.
24 mins ago
add a comment |Â
The line $||ileftarrow i+1$ add 1 to the original variable $i$?
– gimusi
27 mins ago
1
Yes, that is exactly what that means. So, $frac11>0$ is true in the beginning so then we add $1$ to $i$ which looking at the precondition we know $frac12>0$ is true. This loop continues over and over until the statement is false.
– W. G.
24 mins ago
The line $||ileftarrow i+1$ add 1 to the original variable $i$?
– gimusi
27 mins ago
The line $||ileftarrow i+1$ add 1 to the original variable $i$?
– gimusi
27 mins ago
1
1
Yes, that is exactly what that means. So, $frac11>0$ is true in the beginning so then we add $1$ to $i$ which looking at the precondition we know $frac12>0$ is true. This loop continues over and over until the statement is false.
– W. G.
24 mins ago
Yes, that is exactly what that means. So, $frac11>0$ is true in the beginning so then we add $1$ to $i$ which looking at the precondition we know $frac12>0$ is true. This loop continues over and over until the statement is false.
– W. G.
24 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
The only thing we can say here is that $f(1)$ is not defined, or that the program does not terminate for the input $1$.
We cannot say that $f(1)=0$.
answered 27 mins ago
5xum
83.4k384148
If I understand this correctly, the term "total correctness" of the program requires the program to terminate which is often considered in proofs. However, why is that $f(1)$ is not defined in the mathematical sense? Is it required to finite?
– W. G.
16 mins ago
1
@W.G. $f(1)$ is not defined because $f$ does not terminate for the input $1$. It's as simple as that.
– 5xum
11 mins ago
I am not trying to rock the boat here or anything, but I agree that the program does not terminate. Any computer would say that $f(1)$ makes no sense . However, we know that the loop continues on forever. Every iteration, can be proved inductively to show that $frac1i$ is false. Thus, the loop must go on forever. If the loop was finite, the program clearly would not exist as $frac1i>0$ would be false. As the loop goes on to infinity, why can't we assume $i$ goes to infinity?
– W. G.
40 secs ago
add a comment |Â
up vote
1
down vote
It's a bit of a funny question. Looks like the best answer is "no." The number $1$ is not in the domain of $f$, because the program doesn't terminate.
It's true that as a human we can see that the in-memory value of $1/i$ is headed toward $0$, but $f$ was never going to return $1/i$, it was going to return $i$, which is headed off to $infty$. Even if it were going to return $1/i$, the answer still shouldn't be $0$, because the program doesn't terminate and there's no notion of a limiting operation made explicit here.
answered 27 mins ago
hunter
13.3k22337
add a comment |Â
up vote
0
down vote
The function is undefined at $i=1$ since $not exists b$ such that $b=f(1)$.
For a fromal proof, indicating with $i_k$ the value assumed by $i$ at the $k^th$ loop, we can show by induction that $forall k$
- $i_kge 1 implies frac 1i_k>0$
therefore the loop will never end.
More in general $f(i)$ is undefined for any $i>0$ and $f(i)=i$ for any $i< 0$.
For $i=0$ it depends on how the "while" command returns to the operation $1/0$.
answered 22 mins ago
gimusi
74.7k73889
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The only thing we can say here is that $f(1)$ is not defined, or that the program does not terminate for the input $1$.
We cannot say that $f(1)=0$.
answered 27 mins ago
5xum
83.4k384148
If I understand this correctly, the term "total correctness" of the program requires the program to terminate which is often considered in proofs. However, why is that $f(1)$ is not defined in the mathematical sense? Is it required to finite?
– W. G.
16 mins ago
1
@W.G. $f(1)$ is not defined because $f$ does not terminate for the input $1$. It's as simple as that.
– 5xum
11 mins ago
I am not trying to rock the boat here or anything, but I agree that the program does not terminate. Any computer would say that $f(1)$ makes no sense . However, we know that the loop continues on forever. Every iteration, can be proved inductively to show that $frac1i$ is false. Thus, the loop must go on forever. If the loop was finite, the program clearly would not exist as $frac1i>0$ would be false. As the loop goes on to infinity, why can't we assume $i$ goes to infinity?
– W. G.
40 secs ago
add a comment |Â
up vote
2
down vote
The only thing we can say here is that $f(1)$ is not defined, or that the program does not terminate for the input $1$.
We cannot say that $f(1)=0$.
answered 27 mins ago
5xum
83.4k384148
If I understand this correctly, the term "total correctness" of the program requires the program to terminate which is often considered in proofs. However, why is that $f(1)$ is not defined in the mathematical sense? Is it required to finite?
– W. G.
16 mins ago
1
@W.G. $f(1)$ is not defined because $f$ does not terminate for the input $1$. It's as simple as that.
– 5xum
11 mins ago
I am not trying to rock the boat here or anything, but I agree that the program does not terminate. Any computer would say that $f(1)$ makes no sense . However, we know that the loop continues on forever. Every iteration, can be proved inductively to show that $frac1i$ is false. Thus, the loop must go on forever. If the loop was finite, the program clearly would not exist as $frac1i>0$ would be false. As the loop goes on to infinity, why can't we assume $i$ goes to infinity?
– W. G.
40 secs ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The only thing we can say here is that $f(1)$ is not defined, or that the program does not terminate for the input $1$.
We cannot say that $f(1)=0$.
answered 27 mins ago
5xum
83.4k384148
The only thing we can say here is that $f(1)$ is not defined, or that the program does not terminate for the input $1$.
We cannot say that $f(1)=0$.
answered 27 mins ago
5xum
83.4k384148
answered 27 mins ago
5xum
83.4k384148
answered 27 mins ago
5xum
83.4k384148
answered 27 mins ago
5xum
83.4k384148
83.4k384148
If I understand this correctly, the term "total correctness" of the program requires the program to terminate which is often considered in proofs. However, why is that $f(1)$ is not defined in the mathematical sense? Is it required to finite?
– W. G.
16 mins ago
1
@W.G. $f(1)$ is not defined because $f$ does not terminate for the input $1$. It's as simple as that.
– 5xum
11 mins ago
I am not trying to rock the boat here or anything, but I agree that the program does not terminate. Any computer would say that $f(1)$ makes no sense . However, we know that the loop continues on forever. Every iteration, can be proved inductively to show that $frac1i$ is false. Thus, the loop must go on forever. If the loop was finite, the program clearly would not exist as $frac1i>0$ would be false. As the loop goes on to infinity, why can't we assume $i$ goes to infinity?
– W. G.
40 secs ago
add a comment |Â
If I understand this correctly, the term "total correctness" of the program requires the program to terminate which is often considered in proofs. However, why is that $f(1)$ is not defined in the mathematical sense? Is it required to finite?
– W. G.
16 mins ago
1
@W.G. $f(1)$ is not defined because $f$ does not terminate for the input $1$. It's as simple as that.
– 5xum
11 mins ago
I am not trying to rock the boat here or anything, but I agree that the program does not terminate. Any computer would say that $f(1)$ makes no sense . However, we know that the loop continues on forever. Every iteration, can be proved inductively to show that $frac1i$ is false. Thus, the loop must go on forever. If the loop was finite, the program clearly would not exist as $frac1i>0$ would be false. As the loop goes on to infinity, why can't we assume $i$ goes to infinity?
– W. G.
40 secs ago
If I understand this correctly, the term "total correctness" of the program requires the program to terminate which is often considered in proofs. However, why is that $f(1)$ is not defined in the mathematical sense? Is it required to finite?
– W. G.
16 mins ago
If I understand this correctly, the term "total correctness" of the program requires the program to terminate which is often considered in proofs. However, why is that $f(1)$ is not defined in the mathematical sense? Is it required to finite?
– W. G.
16 mins ago
1
1
@W.G. $f(1)$ is not defined because $f$ does not terminate for the input $1$. It's as simple as that.
– 5xum
11 mins ago
@W.G. $f(1)$ is not defined because $f$ does not terminate for the input $1$. It's as simple as that.
– 5xum
11 mins ago
I am not trying to rock the boat here or anything, but I agree that the program does not terminate. Any computer would say that $f(1)$ makes no sense . However, we know that the loop continues on forever. Every iteration, can be proved inductively to show that $frac1i$ is false. Thus, the loop must go on forever. If the loop was finite, the program clearly would not exist as $frac1i>0$ would be false. As the loop goes on to infinity, why can't we assume $i$ goes to infinity?
– W. G.
40 secs ago
I am not trying to rock the boat here or anything, but I agree that the program does not terminate. Any computer would say that $f(1)$ makes no sense . However, we know that the loop continues on forever. Every iteration, can be proved inductively to show that $frac1i$ is false. Thus, the loop must go on forever. If the loop was finite, the program clearly would not exist as $frac1i>0$ would be false. As the loop goes on to infinity, why can't we assume $i$ goes to infinity?
– W. G.
40 secs ago
add a comment |Â
up vote
1
down vote
It's a bit of a funny question. Looks like the best answer is "no." The number $1$ is not in the domain of $f$, because the program doesn't terminate.
It's true that as a human we can see that the in-memory value of $1/i$ is headed toward $0$, but $f$ was never going to return $1/i$, it was going to return $i$, which is headed off to $infty$. Even if it were going to return $1/i$, the answer still shouldn't be $0$, because the program doesn't terminate and there's no notion of a limiting operation made explicit here.
answered 27 mins ago
hunter
13.3k22337
add a comment |Â
up vote
1
down vote
It's a bit of a funny question. Looks like the best answer is "no." The number $1$ is not in the domain of $f$, because the program doesn't terminate.
It's true that as a human we can see that the in-memory value of $1/i$ is headed toward $0$, but $f$ was never going to return $1/i$, it was going to return $i$, which is headed off to $infty$. Even if it were going to return $1/i$, the answer still shouldn't be $0$, because the program doesn't terminate and there's no notion of a limiting operation made explicit here.
answered 27 mins ago
hunter
13.3k22337
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It's a bit of a funny question. Looks like the best answer is "no." The number $1$ is not in the domain of $f$, because the program doesn't terminate.
It's true that as a human we can see that the in-memory value of $1/i$ is headed toward $0$, but $f$ was never going to return $1/i$, it was going to return $i$, which is headed off to $infty$. Even if it were going to return $1/i$, the answer still shouldn't be $0$, because the program doesn't terminate and there's no notion of a limiting operation made explicit here.
answered 27 mins ago
hunter
13.3k22337
It's a bit of a funny question. Looks like the best answer is "no." The number $1$ is not in the domain of $f$, because the program doesn't terminate.
It's true that as a human we can see that the in-memory value of $1/i$ is headed toward $0$, but $f$ was never going to return $1/i$, it was going to return $i$, which is headed off to $infty$. Even if it were going to return $1/i$, the answer still shouldn't be $0$, because the program doesn't terminate and there's no notion of a limiting operation made explicit here.
answered 27 mins ago
hunter
13.3k22337
answered 27 mins ago
hunter
13.3k22337
answered 27 mins ago
hunter
13.3k22337
answered 27 mins ago
hunter
13.3k22337
13.3k22337
add a comment |Â
add a comment |Â
up vote
0
down vote
The function is undefined at $i=1$ since $not exists b$ such that $b=f(1)$.
For a fromal proof, indicating with $i_k$ the value assumed by $i$ at the $k^th$ loop, we can show by induction that $forall k$
- $i_kge 1 implies frac 1i_k>0$
therefore the loop will never end.
More in general $f(i)$ is undefined for any $i>0$ and $f(i)=i$ for any $i< 0$.
For $i=0$ it depends on how the "while" command returns to the operation $1/0$.
answered 22 mins ago
gimusi
74.7k73889
add a comment |Â
up vote
0
down vote
The function is undefined at $i=1$ since $not exists b$ such that $b=f(1)$.
For a fromal proof, indicating with $i_k$ the value assumed by $i$ at the $k^th$ loop, we can show by induction that $forall k$
- $i_kge 1 implies frac 1i_k>0$
therefore the loop will never end.
More in general $f(i)$ is undefined for any $i>0$ and $f(i)=i$ for any $i< 0$.
For $i=0$ it depends on how the "while" command returns to the operation $1/0$.
answered 22 mins ago
gimusi
74.7k73889
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The function is undefined at $i=1$ since $not exists b$ such that $b=f(1)$.
For a fromal proof, indicating with $i_k$ the value assumed by $i$ at the $k^th$ loop, we can show by induction that $forall k$
- $i_kge 1 implies frac 1i_k>0$
therefore the loop will never end.
More in general $f(i)$ is undefined for any $i>0$ and $f(i)=i$ for any $i< 0$.
For $i=0$ it depends on how the "while" command returns to the operation $1/0$.
answered 22 mins ago
gimusi
74.7k73889
The function is undefined at $i=1$ since $not exists b$ such that $b=f(1)$.
For a fromal proof, indicating with $i_k$ the value assumed by $i$ at the $k^th$ loop, we can show by induction that $forall k$
- $i_kge 1 implies frac 1i_k>0$
therefore the loop will never end.
More in general $f(i)$ is undefined for any $i>0$ and $f(i)=i$ for any $i< 0$.
For $i=0$ it depends on how the "while" command returns to the operation $1/0$.
answered 22 mins ago
gimusi
74.7k73889
edited 10 mins ago
answered 22 mins ago
gimusi
74.7k73889
answered 22 mins ago
gimusi
74.7k73889
answered 22 mins ago
gimusi
74.7k73889
74.7k73889
add a comment |Â
add a comment |Â
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The line $||ileftarrow i+1$ add 1 to the original variable $i$?
– gimusi
27 mins ago
1
Yes, that is exactly what that means. So, $frac11>0$ is true in the beginning so then we add $1$ to $i$ which looking at the precondition we know $frac12>0$ is true. This loop continues over and over until the statement is false.
– W. G.
24 mins ago