Does this sum go infinity? [closed] The Next CEO of Stack OverflowIf $a_n$ goes to zero, can...
What's the best way to handle refactoring a big file?
What happens if you roll doubles 3 times then land on "Go to jail?"
MessageLevel in QGIS3
What flight has the highest ratio of time difference to flight time?
Can I run my washing machine drain line into a condensate pump so it drains better?
How did the Bene Gesserit know how to make a Kwisatz Haderach?
Why don't programming languages automatically manage the synchronous/asynchronous problem?
Is it professional to write unrelated content in an almost-empty email?
Anatomically Correct Strange Women In Ponds Distributing Swords
Indicator light circuit
In excess I'm lethal
Is it ever safe to open a suspicious html file (e.g. email attachment)?
Is "for causing autism in X" grammatical?
sp_blitzCache results Memory grants
Novel about a guy who is possessed by the divine essence and the world ends?
How do scammers retract money, while you can’t?
If/When UK leaves the EU, can a future goverment conduct a referendum to join the EU?
"In the right combination" vs "with the right combination"?
Do I need to enable Dev Hub in my PROD Org?
What is the result of assigning to std::vector<T>::begin()?
Would a galaxy be visible from outside, but nearby?
Preparing Indesign booklet with .psd graphics for print
How do I transpose the first and deepest levels of an arbitrarily nested array?
Would this house-rule that treats advantage as a +1 to the roll instead (and disadvantage as -1) and allows them to stack be balanced?
Does this sum go infinity? [closed]
The Next CEO of Stack OverflowIf $a_n$ goes to zero, can we find signs $s_n$ such that $sum s_n a_n$ converges?Asymptotic expansion for harmonic sum in two variablesConvergence of $sum_{k=1}^n(1-k/n)a_k$Does $sum_{n=1}^{infty}frac{cosleft(frac{npi}{2}right)}{sqrt{n}}$ converge?Bivariate infinite series: explicit sum?some infinite sum and $liminf$Sum of the inverses of numbers with $n$ divisors.Why $sum_{n=0}^infty (-1)^nx^{2n}$ converge pointwise?Does this series converge absolutely $sum_{n=1}^{infty}frac{b^{n}_{n}cos(npi)}{n}$Does $sumlimits_{k=1}^infty sumlimits_{n=k}^infty frac{(-1)^{n+k}}{n}$ diverge?
$begingroup$
Consider $F(x)$ that maps from $Bbb N$ to ${pm 1}$, such that if $x$ is odd, then $F(x)$ = $$(-1)^{(frac{x-1}{2})}$$, 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
asked Mar 16 at 18:42
Hari Krishna PHari Krishna P
385
$endgroup$
closed as off-topic by RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel Mar 18 at 6:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider $F(x)$ that maps from $Bbb N$ to ${pm 1}$, such that if $x$ is odd, then $F(x)$ = $$(-1)^{(frac{x-1}{2})}$$, 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
asked Mar 16 at 18:42
Hari Krishna PHari Krishna P
385
$endgroup$
closed as off-topic by RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel Mar 18 at 6:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
Mar 16 at 18:51
add a comment |
$begingroup$
Consider $F(x)$ that maps from $Bbb N$ to ${pm 1}$, such that if $x$ is odd, then $F(x)$ = $$(-1)^{(frac{x-1}{2})}$$, 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
asked Mar 16 at 18:42
Hari Krishna PHari Krishna P
385
$endgroup$
Consider $F(x)$ that maps from $Bbb N$ to ${pm 1}$, such that if $x$ is odd, then $F(x)$ = $$(-1)^{(frac{x-1}{2})}$$, 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
sequences-and-series algebra-precalculus
asked Mar 16 at 18:42
Hari Krishna PHari Krishna P
385
asked Mar 16 at 18:42
Hari Krishna PHari Krishna P
385
edited Mar 16 at 18:55
Hari Krishna P
asked Mar 16 at 18:42
Hari Krishna PHari Krishna P
385
asked Mar 16 at 18:42
Hari Krishna PHari Krishna P
385
asked Mar 16 at 18:42
Hari Krishna PHari Krishna P
385
385
closed as off-topic by RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel Mar 18 at 6:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel Mar 18 at 6:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – RRL, Thomas Shelby, Cesareo, José Carlos Santos, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
Mar 16 at 18:51
add a comment |
1
$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
Mar 16 at 18:51
1
1
$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
Mar 16 at 18:51
$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
Mar 16 at 18:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $T(n) = sum_{ktext{ is odd, } kle n}F(k)$. 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, $T(lfloor n/2rfloor) = sum_{ktext{ is odd, } kle lfloor n/2rfloor}F(k)=sum_{ktext{ is odd, } 2kle n}F(2k)$, $T(lfloor n/4rfloor) = sum_{ktext{ is odd, } 4kle n}F(4k)$, 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
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_{m=0}^{lfloor log_2 nrfloor}T(lfloor n/2^mrfloor)$$
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 $infty$.
answered Mar 16 at 19:51
jmerryjmerry
16.8k11633
$endgroup$
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
Mar 16 at 20:23
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $T(n) = sum_{ktext{ is odd, } kle n}F(k)$. 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, $T(lfloor n/2rfloor) = sum_{ktext{ is odd, } kle lfloor n/2rfloor}F(k)=sum_{ktext{ is odd, } 2kle n}F(2k)$, $T(lfloor n/4rfloor) = sum_{ktext{ is odd, } 4kle n}F(4k)$, 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
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_{m=0}^{lfloor log_2 nrfloor}T(lfloor n/2^mrfloor)$$
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 $infty$.
answered Mar 16 at 19:51
jmerryjmerry
16.8k11633
$endgroup$
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
Mar 16 at 20:23
add a comment |
$begingroup$
Let $T(n) = sum_{ktext{ is odd, } kle n}F(k)$. 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, $T(lfloor n/2rfloor) = sum_{ktext{ is odd, } kle lfloor n/2rfloor}F(k)=sum_{ktext{ is odd, } 2kle n}F(2k)$, $T(lfloor n/4rfloor) = sum_{ktext{ is odd, } 4kle n}F(4k)$, 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
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_{m=0}^{lfloor log_2 nrfloor}T(lfloor n/2^mrfloor)$$
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 $infty$.
answered Mar 16 at 19:51
jmerryjmerry
16.8k11633
$endgroup$
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
Mar 16 at 20:23
add a comment |
$begingroup$
Let $T(n) = sum_{ktext{ is odd, } kle n}F(k)$. 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, $T(lfloor n/2rfloor) = sum_{ktext{ is odd, } kle lfloor n/2rfloor}F(k)=sum_{ktext{ is odd, } 2kle n}F(2k)$, $T(lfloor n/4rfloor) = sum_{ktext{ is odd, } 4kle n}F(4k)$, 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
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_{m=0}^{lfloor log_2 nrfloor}T(lfloor n/2^mrfloor)$$
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 $infty$.
answered Mar 16 at 19:51
jmerryjmerry
16.8k11633
$endgroup$
Let $T(n) = sum_{ktext{ is odd, } kle n}F(k)$. 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, $T(lfloor n/2rfloor) = sum_{ktext{ is odd, } kle lfloor n/2rfloor}F(k)=sum_{ktext{ is odd, } 2kle n}F(2k)$, $T(lfloor n/4rfloor) = sum_{ktext{ is odd, } 4kle n}F(4k)$, 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
$$S_n = T(n)+T(lfloor n/2rfloor)+T(lfloor n/4rfloor)+cdots = sum_{m=0}^{lfloor log_2 nrfloor}T(lfloor n/2^mrfloor)$$
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 $infty$.
answered Mar 16 at 19:51
jmerryjmerry
16.8k11633
answered Mar 16 at 19:51
jmerryjmerry
16.8k11633
answered Mar 16 at 19:51
jmerryjmerry
16.8k11633
answered Mar 16 at 19:51
jmerryjmerry
16.8k11633
16.8k11633
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
Mar 16 at 20:23
add a comment |
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
Mar 16 at 20:23
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
Mar 16 at 20:23
$begingroup$
Awesome! Thank you!!
$endgroup$
– Hari Krishna P
Mar 16 at 20:23
add a comment |
1
$begingroup$
@Max thanks for the edit....this was my first question on mathstackexchange!
$endgroup$
– Hari Krishna P
Mar 16 at 18:51