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Does this sum go infinity?

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5

3

$begingroup$

Consider $F(x)$ that maps from $Bbb N$ to $pm 1$, such that if $x$ is odd, then $F(x)$ = $$(-1)^(fracx-12)$$, and if $x$ is even, then $F(x)=F(y)$, where $y$ is the odd number obtained after dividing $x$ by $2$ until it is odd.

Does $S_n = sum_p=1^n F(p)$ have an explicit formula?
And if $n$ tends to $infty$ does the sum alternates or will it lie between an interval or does it tend to $pm infty$ ?

sequences-and-series algebra-precalculus

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edited 1 hour ago

Hari Krishna P

asked 1 hour ago

Hari Krishna PHari Krishna P

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  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Max thanks for the edit....this was my first question on mathstackexchange!
  $endgroup$
  – Hari Krishna P
  1 hour ago
  

  
  
  
  

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5

3

$begingroup$

Consider $F(x)$ that maps from $Bbb N$ to $pm 1$, such that if $x$ is odd, then $F(x)$ = $$(-1)^(fracx-12)$$, and if $x$ is even, then $F(x)=F(y)$, where $y$ is the odd number obtained after dividing $x$ by $2$ until it is odd.

Does $S_n = sum_p=1^n F(p)$ have an explicit formula?
And if $n$ tends to $infty$ does the sum alternates or will it lie between an interval or does it tend to $pm infty$ ?

sequences-and-series algebra-precalculus

share|cite|improve this question

edited 1 hour ago

Hari Krishna P

asked 1 hour ago

Hari Krishna PHari Krishna P

264

New contributor

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

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Max thanks for the edit....this was my first question on mathstackexchange!
  $endgroup$
  – Hari Krishna P
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider $F(x)$ that maps from $Bbb N$ to $pm 1$, such that if $x$ is odd, then $F(x)$ = $$(-1)^(fracx-12)$$, and if $x$ is even, then $F(x)=F(y)$, where $y$ is the odd number obtained after dividing $x$ by $2$ until it is odd.

Does $S_n = sum_p=1^n F(p)$ have an explicit formula?
And if $n$ tends to $infty$ does the sum alternates or will it lie between an interval or does it tend to $pm infty$ ?

sequences-and-series algebra-precalculus

share|cite|improve this question

edited 1 hour ago

Hari Krishna P

asked 1 hour ago

Hari Krishna PHari Krishna P

264

New contributor

Hari Krishna P 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

edited 1 hour ago

Hari Krishna P

asked 1 hour ago

Hari Krishna PHari Krishna P

264

New contributor

Hari Krishna P 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

edited 1 hour ago

Hari Krishna P

asked 1 hour ago

Hari Krishna PHari Krishna P

264

New contributor

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

asked 1 hour ago

Hari Krishna PHari Krishna P

264

New contributor

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

asked 1 hour ago

Hari Krishna PHari Krishna P

264

1 Answer
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$begingroup$

Let $T(n) = sum_ktext is odd, kle nF(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.

Then, $T(lfloor n/2rfloor) = sum_ktext is odd, kle lfloor n/2rfloorF(k)=sum_ktext is odd, 2kle nF(2k)$, $T(lfloor n/4rfloor) = sum_ktext is odd, 4kle nF(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_m=0^lfloor log_2 nrfloorT(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.

Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^m-1$ place, is always there) or as high as $m$ for $n=lfloor 2^m+1/3rfloor$.

So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.

share|cite|improve this answer

answered 50 mins ago

jmerryjmerry

14.3k1629

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome! Thank you!!
  $endgroup$
  – Hari Krishna P
  18 mins ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

Let $T(n) = sum_ktext is odd, kle nF(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.

Then, $T(lfloor n/2rfloor) = sum_ktext is odd, kle lfloor n/2rfloorF(k)=sum_ktext is odd, 2kle nF(2k)$, $T(lfloor n/4rfloor) = sum_ktext is odd, 4kle nF(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_m=0^lfloor log_2 nrfloorT(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.

Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^m-1$ place, is always there) or as high as $m$ for $n=lfloor 2^m+1/3rfloor$.

So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.

share|cite|improve this answer

answered 50 mins ago

jmerryjmerry

14.3k1629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome! Thank you!!
  $endgroup$
  – Hari Krishna P
  18 mins ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Let $T(n) = sum_ktext is odd, kle nF(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.

Then, $T(lfloor n/2rfloor) = sum_ktext is odd, kle lfloor n/2rfloorF(k)=sum_ktext is odd, 2kle nF(2k)$, $T(lfloor n/4rfloor) = sum_ktext is odd, 4kle nF(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_m=0^lfloor log_2 nrfloorT(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.

Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^m-1$ place, is always there) or as high as $m$ for $n=lfloor 2^m+1/3rfloor$.

So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.

share|cite|improve this answer

answered 50 mins ago

jmerryjmerry

14.3k1629

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Awesome! Thank you!!
  $endgroup$
  – Hari Krishna P
  18 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $T(n) = sum_ktext is odd, kle nF(k)$. We can easily see that $T(0)=0$, $T(1)=1$, $T(2)=1$, $T(3)=0$, and $T(n+4)=T(n)$ for all $nge 0$. This is $1$ if the last two bits of $n$ (when expressed in binary) are different, and $0$ if they are the same.

Then, $T(lfloor n/2rfloor) = sum_ktext is odd, kle lfloor n/2rfloorF(k)=sum_ktext is odd, 2kle nF(2k)$, $T(lfloor n/4rfloor) = sum_ktext is odd, 4kle nF(4k)$, and so on. Every integer is equal to $2^m k$ for some nonnegative $m$ and odd $k$. As such, we can take a sum, and get
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_m=0^lfloor log_2 nrfloorT(lfloor n/2^mrfloor)$$
Each term is $1$ if two particular adjacent bits of $n$ are different and zero if they're equal - the $1$ bit and the $2$ bit for $T(n)$, the $2$ bit and the $4$ bit for $T(lfloor n/2rfloor)$, the $4$ bit and the $8$ bit for $T(lfloor n/4rfloor)$, and so on.

Sum them up, and $S_n$ is the number of times the sequence of bits switches between $0$ and $1$. Among $m$-bit numbers, this can be as low as $1$ for $n=2^m-1$ (the first switch, from $0$ in the $2^m$ place to $1$ in the $2^m-1$ place, is always there) or as high as $m$ for $n=lfloor 2^m+1/3rfloor$.

So, there it is - an explicit form for $S_n$, and a sequence $1,2,5,10,21,42,85,dots$ for which $S_n$ goes to $infty$.

share|cite|improve this answer

answered 50 mins ago

jmerryjmerry

14.3k1629

$endgroup$

share|cite|improve this answer

answered 50 mins ago

jmerryjmerry

14.3k1629

answered 50 mins ago

jmerryjmerry

14.3k1629

answered 50 mins ago

jmerryjmerry

14.3k1629