Find formula for function of $n$ returning $0$ if $n$ is composite and $1$ if $n$ is prime
Sample problem:
Find an equation $theta(n)$ for which $theta(n)=left{ begin{array} &0, text{when } nin text{Composed} \ n, text{when } nin text{Prime} end{array} right.$
This problem is from the International Youth Math Challenge $2018$ and since they do not return marked sheets, I am unsure if my solution was correct.
My final answer was: $$theta (n)=n-ncdot text{sgn} left(prod_{i=1}^{infty} |n-p_i|right)$$ where $p_i$ is the $i^{text{th}}$ prime number. This is all I could come up with and to be honest, I am not too happy with it, because I feel like I have basically chosen something that will only give the answer I want. Is this solution correct, mathematically? Is there a better solution?
Note: $text {sgn}(n)$ is the $text{sign}$ or $text{signum}$ function and $$text{sgn}(n)=left{ begin{array} &-1; nlt 0\ 0; n=0\ 1; ngt 0 end{array} right.$$
prime-numbers contest-math
edited Dec 21 at 11:02
Did
246k23220454
asked Dec 21 at 5:50
Mohammad Zuhair Khan
1,4452525
add a comment |
Sample problem:
Find an equation $theta(n)$ for which $theta(n)=left{ begin{array} &0, text{when } nin text{Composed} \ n, text{when } nin text{Prime} end{array} right.$
This problem is from the International Youth Math Challenge $2018$ and since they do not return marked sheets, I am unsure if my solution was correct.
My final answer was: $$theta (n)=n-ncdot text{sgn} left(prod_{i=1}^{infty} |n-p_i|right)$$ where $p_i$ is the $i^{text{th}}$ prime number. This is all I could come up with and to be honest, I am not too happy with it, because I feel like I have basically chosen something that will only give the answer I want. Is this solution correct, mathematically? Is there a better solution?
Note: $text {sgn}(n)$ is the $text{sign}$ or $text{signum}$ function and $$text{sgn}(n)=left{ begin{array} &-1; nlt 0\ 0; n=0\ 1; ngt 0 end{array} right.$$
prime-numbers contest-math
edited Dec 21 at 11:02
Did
246k23220454
asked Dec 21 at 5:50
Mohammad Zuhair Khan
1,4452525
Are you sure you're giving the complete statement of the problem? I don't think it makes sense without some sort of restriction as to what kind of functions/primitives you are allowed to use in a solution.
– Evangelos Bampas
Dec 21 at 10:20
The infinite product formula you suggest requires to compute an infinite product, which cannot be done in practice (and to define the signum function at $+infty$, which is not so usual), and to know the full set of primes, which seems rather impractical as well. Easy remedies to these two defects are suggested below.
– Did
Dec 21 at 10:59
@Did which is what my title wonders. I only get the results I want, I cannot make further assumptions.
– Mohammad Zuhair Khan
Dec 21 at 15:33
add a comment |
Sample problem:
Find an equation $theta(n)$ for which $theta(n)=left{ begin{array} &0, text{when } nin text{Composed} \ n, text{when } nin text{Prime} end{array} right.$
This problem is from the International Youth Math Challenge $2018$ and since they do not return marked sheets, I am unsure if my solution was correct.
My final answer was: $$theta (n)=n-ncdot text{sgn} left(prod_{i=1}^{infty} |n-p_i|right)$$ where $p_i$ is the $i^{text{th}}$ prime number. This is all I could come up with and to be honest, I am not too happy with it, because I feel like I have basically chosen something that will only give the answer I want. Is this solution correct, mathematically? Is there a better solution?
Note: $text {sgn}(n)$ is the $text{sign}$ or $text{signum}$ function and $$text{sgn}(n)=left{ begin{array} &-1; nlt 0\ 0; n=0\ 1; ngt 0 end{array} right.$$
prime-numbers contest-math
edited Dec 21 at 11:02
Did
246k23220454
asked Dec 21 at 5:50
Mohammad Zuhair Khan
1,4452525
Sample problem:
Find an equation $theta(n)$ for which $theta(n)=left{ begin{array} &0, text{when } nin text{Composed} \ n, text{when } nin text{Prime} end{array} right.$
This problem is from the International Youth Math Challenge $2018$ and since they do not return marked sheets, I am unsure if my solution was correct.
My final answer was: $$theta (n)=n-ncdot text{sgn} left(prod_{i=1}^{infty} |n-p_i|right)$$ where $p_i$ is the $i^{text{th}}$ prime number. This is all I could come up with and to be honest, I am not too happy with it, because I feel like I have basically chosen something that will only give the answer I want. Is this solution correct, mathematically? Is there a better solution?
Note: $text {sgn}(n)$ is the $text{sign}$ or $text{signum}$ function and $$text{sgn}(n)=left{ begin{array} &-1; nlt 0\ 0; n=0\ 1; ngt 0 end{array} right.$$
prime-numbers contest-math
prime-numbers contest-math
edited Dec 21 at 11:02
Did
246k23220454
asked Dec 21 at 5:50
Mohammad Zuhair Khan
1,4452525
edited Dec 21 at 11:02
Did
246k23220454
asked Dec 21 at 5:50
Mohammad Zuhair Khan
1,4452525
edited Dec 21 at 11:02
Did
246k23220454
edited Dec 21 at 11:02
Did
246k23220454
edited Dec 21 at 11:02
Did
246k23220454
246k23220454
asked Dec 21 at 5:50
Mohammad Zuhair Khan
1,4452525
asked Dec 21 at 5:50
Mohammad Zuhair Khan
1,4452525
asked Dec 21 at 5:50
Mohammad Zuhair Khan
1,4452525
1,4452525
Are you sure you're giving the complete statement of the problem? I don't think it makes sense without some sort of restriction as to what kind of functions/primitives you are allowed to use in a solution.
– Evangelos Bampas
Dec 21 at 10:20
The infinite product formula you suggest requires to compute an infinite product, which cannot be done in practice (and to define the signum function at $+infty$, which is not so usual), and to know the full set of primes, which seems rather impractical as well. Easy remedies to these two defects are suggested below.
– Did
Dec 21 at 10:59
@Did which is what my title wonders. I only get the results I want, I cannot make further assumptions.
– Mohammad Zuhair Khan
Dec 21 at 15:33
add a comment |
Are you sure you're giving the complete statement of the problem? I don't think it makes sense without some sort of restriction as to what kind of functions/primitives you are allowed to use in a solution.
– Evangelos Bampas
Dec 21 at 10:20
The infinite product formula you suggest requires to compute an infinite product, which cannot be done in practice (and to define the signum function at $+infty$, which is not so usual), and to know the full set of primes, which seems rather impractical as well. Easy remedies to these two defects are suggested below.
– Did
Dec 21 at 10:59
@Did which is what my title wonders. I only get the results I want, I cannot make further assumptions.
– Mohammad Zuhair Khan
Dec 21 at 15:33
Are you sure you're giving the complete statement of the problem? I don't think it makes sense without some sort of restriction as to what kind of functions/primitives you are allowed to use in a solution.
– Evangelos Bampas
Dec 21 at 10:20
Are you sure you're giving the complete statement of the problem? I don't think it makes sense without some sort of restriction as to what kind of functions/primitives you are allowed to use in a solution.
– Evangelos Bampas
Dec 21 at 10:20
The infinite product formula you suggest requires to compute an infinite product, which cannot be done in practice (and to define the signum function at $+infty$, which is not so usual), and to know the full set of primes, which seems rather impractical as well. Easy remedies to these two defects are suggested below.
– Did
Dec 21 at 10:59
The infinite product formula you suggest requires to compute an infinite product, which cannot be done in practice (and to define the signum function at $+infty$, which is not so usual), and to know the full set of primes, which seems rather impractical as well. Easy remedies to these two defects are suggested below.
– Did
Dec 21 at 10:59
@Did which is what my title wonders. I only get the results I want, I cannot make further assumptions.
– Mohammad Zuhair Khan
Dec 21 at 15:33
@Did which is what my title wonders. I only get the results I want, I cannot make further assumptions.
– Mohammad Zuhair Khan
Dec 21 at 15:33
add a comment |
3 Answers
3
active
oldest
votes
I'm not sure what kind of functions are allowed but here is a similar one (might be equivalent after some small changes), the differences being it's finite and doesn't require ability to select primes:
For any positive integer $p$, define this function
$$
f(n,p):= leftlceil frac{n-plfloor frac{n}{p}rfloor}{n} rightrceil
$$
If $p$ divides $n$ then $f(n,p)=0$, otherwise $n-plfloor n/prfloorneq 0$ so $f(n,p)=1$.
You can then use this to make the following:
$$
theta(n):= n - nprod_{p=2}^{n-1}f(n,p)
$$
If $n$ is composite then one of the $p$'s will make the product $0$ and hence $theta(n)=n$. Otherwise $n$ is prime and the product is $1$, giving $theta(n)=0$.
answered Dec 21 at 6:48
Yong Hao Ng
3,2341220
add a comment |
Good try; but there's something that needs fixing. When $n$ is composite, the product diverges to $infty$; so you should define $text {sgn} (infty) = 1$.
answered Dec 21 at 6:04
Ovi
12.4k1038111
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
Dec 21 at 6:05
2
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
Dec 21 at 6:09
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
Dec 21 at 6:11
@MohammadZuhairKhan I am thinking about it
– Ovi
Dec 21 at 6:12
@Ovi I would definitely not assume we are working in $mathbb{R}$, $mathbb{N}$ might be a better choice (nitpicks). I also don't see any reason to include such a definition, I would much prefer simpy choosing and noting which set does $n$ belong to. I would go as far as to consider writing $text{sgn}(infty)$ a mistake without well explaining why are you doing that and what do you mean by that.
– J.E
Dec 21 at 11:10
|
show 1 more comment
Instead of the sign function, you could potentially use the Kronecker delta, which is defined as
$$delta_{mn}=begin{cases}
1 & text{if }n=m\
0 & text{if }nneq m
end{cases}$$
It basically compares two numbers and gives $1$ if there is a match and $0$ otherwise. By summing over such Kronecker delta's, you could build:
$$theta(n)=nsum_{i=1}^{infty}delta_{np_i}$$
(where $p_i$ is the $i$-th prime number). However, I'm wondering if this would be accepted because it kind of bypasses the question of "checking if $n$ is prime". Both our formulas are just a nice rewording of the "text-form" formula given in the question, so I'm not certain this was the kind of answer that was expected.
answered Dec 21 at 6:39
orion2112
461210
1
To be honest, the entire problem seems to be a test of a student's thought process.
– Mohammad Zuhair Khan
Dec 21 at 15:36
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
I'm not sure what kind of functions are allowed but here is a similar one (might be equivalent after some small changes), the differences being it's finite and doesn't require ability to select primes:
For any positive integer $p$, define this function
$$
f(n,p):= leftlceil frac{n-plfloor frac{n}{p}rfloor}{n} rightrceil
$$
If $p$ divides $n$ then $f(n,p)=0$, otherwise $n-plfloor n/prfloorneq 0$ so $f(n,p)=1$.
You can then use this to make the following:
$$
theta(n):= n - nprod_{p=2}^{n-1}f(n,p)
$$
If $n$ is composite then one of the $p$'s will make the product $0$ and hence $theta(n)=n$. Otherwise $n$ is prime and the product is $1$, giving $theta(n)=0$.
answered Dec 21 at 6:48
Yong Hao Ng
3,2341220
add a comment |
I'm not sure what kind of functions are allowed but here is a similar one (might be equivalent after some small changes), the differences being it's finite and doesn't require ability to select primes:
For any positive integer $p$, define this function
$$
f(n,p):= leftlceil frac{n-plfloor frac{n}{p}rfloor}{n} rightrceil
$$
If $p$ divides $n$ then $f(n,p)=0$, otherwise $n-plfloor n/prfloorneq 0$ so $f(n,p)=1$.
You can then use this to make the following:
$$
theta(n):= n - nprod_{p=2}^{n-1}f(n,p)
$$
If $n$ is composite then one of the $p$'s will make the product $0$ and hence $theta(n)=n$. Otherwise $n$ is prime and the product is $1$, giving $theta(n)=0$.
answered Dec 21 at 6:48
Yong Hao Ng
3,2341220
add a comment |
I'm not sure what kind of functions are allowed but here is a similar one (might be equivalent after some small changes), the differences being it's finite and doesn't require ability to select primes:
For any positive integer $p$, define this function
$$
f(n,p):= leftlceil frac{n-plfloor frac{n}{p}rfloor}{n} rightrceil
$$
If $p$ divides $n$ then $f(n,p)=0$, otherwise $n-plfloor n/prfloorneq 0$ so $f(n,p)=1$.
You can then use this to make the following:
$$
theta(n):= n - nprod_{p=2}^{n-1}f(n,p)
$$
If $n$ is composite then one of the $p$'s will make the product $0$ and hence $theta(n)=n$. Otherwise $n$ is prime and the product is $1$, giving $theta(n)=0$.
answered Dec 21 at 6:48
Yong Hao Ng
3,2341220
I'm not sure what kind of functions are allowed but here is a similar one (might be equivalent after some small changes), the differences being it's finite and doesn't require ability to select primes:
For any positive integer $p$, define this function
$$
f(n,p):= leftlceil frac{n-plfloor frac{n}{p}rfloor}{n} rightrceil
$$
If $p$ divides $n$ then $f(n,p)=0$, otherwise $n-plfloor n/prfloorneq 0$ so $f(n,p)=1$.
You can then use this to make the following:
$$
theta(n):= n - nprod_{p=2}^{n-1}f(n,p)
$$
If $n$ is composite then one of the $p$'s will make the product $0$ and hence $theta(n)=n$. Otherwise $n$ is prime and the product is $1$, giving $theta(n)=0$.
answered Dec 21 at 6:48
Yong Hao Ng
3,2341220
answered Dec 21 at 6:48
Yong Hao Ng
3,2341220
answered Dec 21 at 6:48
Yong Hao Ng
3,2341220
answered Dec 21 at 6:48
Yong Hao Ng
3,2341220
3,2341220
add a comment |
add a comment |
Good try; but there's something that needs fixing. When $n$ is composite, the product diverges to $infty$; so you should define $text {sgn} (infty) = 1$.
answered Dec 21 at 6:04
Ovi
12.4k1038111
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
Dec 21 at 6:05
2
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
Dec 21 at 6:09
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
Dec 21 at 6:11
@MohammadZuhairKhan I am thinking about it
– Ovi
Dec 21 at 6:12
@Ovi I would definitely not assume we are working in $mathbb{R}$, $mathbb{N}$ might be a better choice (nitpicks). I also don't see any reason to include such a definition, I would much prefer simpy choosing and noting which set does $n$ belong to. I would go as far as to consider writing $text{sgn}(infty)$ a mistake without well explaining why are you doing that and what do you mean by that.
– J.E
Dec 21 at 11:10
|
show 1 more comment
Good try; but there's something that needs fixing. When $n$ is composite, the product diverges to $infty$; so you should define $text {sgn} (infty) = 1$.
answered Dec 21 at 6:04
Ovi
12.4k1038111
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
Dec 21 at 6:05
2
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
Dec 21 at 6:09
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
Dec 21 at 6:11
@MohammadZuhairKhan I am thinking about it
– Ovi
Dec 21 at 6:12
@Ovi I would definitely not assume we are working in $mathbb{R}$, $mathbb{N}$ might be a better choice (nitpicks). I also don't see any reason to include such a definition, I would much prefer simpy choosing and noting which set does $n$ belong to. I would go as far as to consider writing $text{sgn}(infty)$ a mistake without well explaining why are you doing that and what do you mean by that.
– J.E
Dec 21 at 11:10
|
show 1 more comment
Good try; but there's something that needs fixing. When $n$ is composite, the product diverges to $infty$; so you should define $text {sgn} (infty) = 1$.
answered Dec 21 at 6:04
Ovi
12.4k1038111
Good try; but there's something that needs fixing. When $n$ is composite, the product diverges to $infty$; so you should define $text {sgn} (infty) = 1$.
answered Dec 21 at 6:04
Ovi
12.4k1038111
answered Dec 21 at 6:04
Ovi
12.4k1038111
answered Dec 21 at 6:04
Ovi
12.4k1038111
answered Dec 21 at 6:04
Ovi
12.4k1038111
12.4k1038111
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
Dec 21 at 6:05
2
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
Dec 21 at 6:09
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
Dec 21 at 6:11
@MohammadZuhairKhan I am thinking about it
– Ovi
Dec 21 at 6:12
@Ovi I would definitely not assume we are working in $mathbb{R}$, $mathbb{N}$ might be a better choice (nitpicks). I also don't see any reason to include such a definition, I would much prefer simpy choosing and noting which set does $n$ belong to. I would go as far as to consider writing $text{sgn}(infty)$ a mistake without well explaining why are you doing that and what do you mean by that.
– J.E
Dec 21 at 11:10
|
show 1 more comment
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
Dec 21 at 6:05
2
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
Dec 21 at 6:09
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
Dec 21 at 6:11
@MohammadZuhairKhan I am thinking about it
– Ovi
Dec 21 at 6:12
@Ovi I would definitely not assume we are working in $mathbb{R}$, $mathbb{N}$ might be a better choice (nitpicks). I also don't see any reason to include such a definition, I would much prefer simpy choosing and noting which set does $n$ belong to. I would go as far as to consider writing $text{sgn}(infty)$ a mistake without well explaining why are you doing that and what do you mean by that.
– J.E
Dec 21 at 11:10
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
Dec 21 at 6:05
Thanks for the advice. But, unless I am mistaken, $+infty gt 0?$
– Mohammad Zuhair Khan
Dec 21 at 6:05
2
2
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
Dec 21 at 6:09
@MohammadZuhairKhan Yes; just personally, I would mention that case, since often people assume that youre working in $mathbb{R}$ instead of the extended number system. I would just put it to make sure there's no chance of getting points off.
– Ovi
Dec 21 at 6:09
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
Dec 21 at 6:11
Oh okay thank you. Ofcourse, as it is done already, I can not (and would not) update my answer. But is there any other solution for this problem? I have a suspicion that this might be an open problem, or atleast a sensible version of it will.
– Mohammad Zuhair Khan
Dec 21 at 6:11
@MohammadZuhairKhan I am thinking about it
– Ovi
Dec 21 at 6:12
@MohammadZuhairKhan I am thinking about it
– Ovi
Dec 21 at 6:12
@Ovi I would definitely not assume we are working in $mathbb{R}$, $mathbb{N}$ might be a better choice (nitpicks). I also don't see any reason to include such a definition, I would much prefer simpy choosing and noting which set does $n$ belong to. I would go as far as to consider writing $text{sgn}(infty)$ a mistake without well explaining why are you doing that and what do you mean by that.
– J.E
Dec 21 at 11:10
@Ovi I would definitely not assume we are working in $mathbb{R}$, $mathbb{N}$ might be a better choice (nitpicks). I also don't see any reason to include such a definition, I would much prefer simpy choosing and noting which set does $n$ belong to. I would go as far as to consider writing $text{sgn}(infty)$ a mistake without well explaining why are you doing that and what do you mean by that.
– J.E
Dec 21 at 11:10
|
show 1 more comment
Instead of the sign function, you could potentially use the Kronecker delta, which is defined as
$$delta_{mn}=begin{cases}
1 & text{if }n=m\
0 & text{if }nneq m
end{cases}$$
It basically compares two numbers and gives $1$ if there is a match and $0$ otherwise. By summing over such Kronecker delta's, you could build:
$$theta(n)=nsum_{i=1}^{infty}delta_{np_i}$$
(where $p_i$ is the $i$-th prime number). However, I'm wondering if this would be accepted because it kind of bypasses the question of "checking if $n$ is prime". Both our formulas are just a nice rewording of the "text-form" formula given in the question, so I'm not certain this was the kind of answer that was expected.
answered Dec 21 at 6:39
orion2112
461210
1
To be honest, the entire problem seems to be a test of a student's thought process.
– Mohammad Zuhair Khan
Dec 21 at 15:36
add a comment |
Instead of the sign function, you could potentially use the Kronecker delta, which is defined as
$$delta_{mn}=begin{cases}
1 & text{if }n=m\
0 & text{if }nneq m
end{cases}$$
It basically compares two numbers and gives $1$ if there is a match and $0$ otherwise. By summing over such Kronecker delta's, you could build:
$$theta(n)=nsum_{i=1}^{infty}delta_{np_i}$$
(where $p_i$ is the $i$-th prime number). However, I'm wondering if this would be accepted because it kind of bypasses the question of "checking if $n$ is prime". Both our formulas are just a nice rewording of the "text-form" formula given in the question, so I'm not certain this was the kind of answer that was expected.
answered Dec 21 at 6:39
orion2112
461210
1
To be honest, the entire problem seems to be a test of a student's thought process.
– Mohammad Zuhair Khan
Dec 21 at 15:36
add a comment |
Instead of the sign function, you could potentially use the Kronecker delta, which is defined as
$$delta_{mn}=begin{cases}
1 & text{if }n=m\
0 & text{if }nneq m
end{cases}$$
It basically compares two numbers and gives $1$ if there is a match and $0$ otherwise. By summing over such Kronecker delta's, you could build:
$$theta(n)=nsum_{i=1}^{infty}delta_{np_i}$$
(where $p_i$ is the $i$-th prime number). However, I'm wondering if this would be accepted because it kind of bypasses the question of "checking if $n$ is prime". Both our formulas are just a nice rewording of the "text-form" formula given in the question, so I'm not certain this was the kind of answer that was expected.
answered Dec 21 at 6:39
orion2112
461210
Instead of the sign function, you could potentially use the Kronecker delta, which is defined as
$$delta_{mn}=begin{cases}
1 & text{if }n=m\
0 & text{if }nneq m
end{cases}$$
It basically compares two numbers and gives $1$ if there is a match and $0$ otherwise. By summing over such Kronecker delta's, you could build:
$$theta(n)=nsum_{i=1}^{infty}delta_{np_i}$$
(where $p_i$ is the $i$-th prime number). However, I'm wondering if this would be accepted because it kind of bypasses the question of "checking if $n$ is prime". Both our formulas are just a nice rewording of the "text-form" formula given in the question, so I'm not certain this was the kind of answer that was expected.
answered Dec 21 at 6:39
orion2112
461210
answered Dec 21 at 6:39
orion2112
461210
answered Dec 21 at 6:39
orion2112
461210
answered Dec 21 at 6:39
orion2112
461210
461210
1
To be honest, the entire problem seems to be a test of a student's thought process.
– Mohammad Zuhair Khan
Dec 21 at 15:36
add a comment |
1
To be honest, the entire problem seems to be a test of a student's thought process.
– Mohammad Zuhair Khan
Dec 21 at 15:36
1
1
To be honest, the entire problem seems to be a test of a student's thought process.
– Mohammad Zuhair Khan
Dec 21 at 15:36
To be honest, the entire problem seems to be a test of a student's thought process.
– Mohammad Zuhair Khan
Dec 21 at 15:36
add a comment |
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Are you sure you're giving the complete statement of the problem? I don't think it makes sense without some sort of restriction as to what kind of functions/primitives you are allowed to use in a solution.
– Evangelos Bampas
Dec 21 at 10:20
The infinite product formula you suggest requires to compute an infinite product, which cannot be done in practice (and to define the signum function at $+infty$, which is not so usual), and to know the full set of primes, which seems rather impractical as well. Easy remedies to these two defects are suggested below.
– Did
Dec 21 at 10:59
@Did which is what my title wonders. I only get the results I want, I cannot make further assumptions.
– Mohammad Zuhair Khan
Dec 21 at 15:33