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Extended conjecture for $f(m)=sumlimits_n = 1 ^ m (-1)^n sinleft(sumlimits_k = 1 ^infty fraca_k pib_kP_k(n) right)$
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I asked a question that is related to this question and claimed that
Generalized Conjecture:
$$f(m)=sumlimits_n = 1 ^ m (-1)^n sinleft(P_r(n) fraca pibright) tag 1 $$
I have a conjecture that if $P_r(n)=sumlimits_k = 1 ^ n k^2r$ where r is a positive integer,
$f(m)$ function is periodic function when $a,b,m$ positive integers and
$ sumlimits_k = 1 ^T f(k)=0 $
where ($T$) is the period value.
(You may see a proof for (Conjecture (1) above) by @FabioLucchini)
I have found out an example and It shows my generalized conjecture can be extended more. I have tested with many numerical values that It supports my extended conjecture below.
$$f(m)=sumlimits_n = 1 ^ m (-1)^n sinleft(G(n) fraca pibright) tag 2 $$
$$G(n)=fracn(n+1)(2n+1)(3n^2+3n-4)30$$ It also satisfies my generalized conjecture $(1)$ above. $G(n)$ can be written as:
$$G(n)=fracn(n+1)(2n+1)(3n^2+3n-1)30-frac35fracn(n+1)(2n+1)6=sumlimits_k = 1 ^ n k^4-frac35sumlimits_k = 1 ^ n k^2=P_2(n)-frac35P_1(n)$$
Test link for $a=2,b=5$ and $m=500$
The numerical values and my works on the subject estimates the extension of the conjecture above. It is just strong sense without proof that it must be true.
More generalized conjecture can be written:
Extended Conjecture:
$$f(m)=sumlimits_n = 1 ^ m (-1)^n sinleft(sumlimits_k = 1 ^infty fraca_k pib_kP_k(n) right) tag 3 $$
More extended conjecture claims that if $P_r(n)=sumlimits_k = 1 ^ n k^2r$ where r is a positive integer,
$f(m)$ function is periodic function when $a_k$ is any integers, $b_k$ is non-zero integers and $m$ positive integers.$sumlimits_k = 1 ^T f(k)=0 $
where ($T$) is the period value for all possible $a_k,b_k,m$ integers.
Can the extended conjecture be proven? What is the period formula for it?
I also have been still looking for $G(n)$ polynomials that is different from $G(n)=sumlimits_k = 1 ^infty fraca_kb_kP_k(n)$ satisfies $ sumlimits_k = 1 ^T f(k)=0 $ $tag4$ for all $a,b,m$ positive integers
(Please consider Equation $(2)$)
Finally, I claim that there is no any polynomials different from $G(n)=sumlimits_k = 1 ^infty fraca_kb_kP_k(n)$ that satisfy $ sumlimits_k = 1 ^T f(k)=0 $. Please inform me if you find a counter-example.
Thanks for answers
sequences-and-series trigonometric-series
asked Jul 16 at 9:15
Mathlover
6,26222265
add a comment |Â
up vote
3
down vote
favorite
I asked a question that is related to this question and claimed that
Generalized Conjecture:
$$f(m)=sumlimits_n = 1 ^ m (-1)^n sinleft(P_r(n) fraca pibright) tag 1 $$
I have a conjecture that if $P_r(n)=sumlimits_k = 1 ^ n k^2r$ where r is a positive integer,
$f(m)$ function is periodic function when $a,b,m$ positive integers and
$ sumlimits_k = 1 ^T f(k)=0 $
where ($T$) is the period value.
(You may see a proof for (Conjecture (1) above) by @FabioLucchini)
I have found out an example and It shows my generalized conjecture can be extended more. I have tested with many numerical values that It supports my extended conjecture below.
$$f(m)=sumlimits_n = 1 ^ m (-1)^n sinleft(G(n) fraca pibright) tag 2 $$
$$G(n)=fracn(n+1)(2n+1)(3n^2+3n-4)30$$ It also satisfies my generalized conjecture $(1)$ above. $G(n)$ can be written as:
$$G(n)=fracn(n+1)(2n+1)(3n^2+3n-1)30-frac35fracn(n+1)(2n+1)6=sumlimits_k = 1 ^ n k^4-frac35sumlimits_k = 1 ^ n k^2=P_2(n)-frac35P_1(n)$$
Test link for $a=2,b=5$ and $m=500$
The numerical values and my works on the subject estimates the extension of the conjecture above. It is just strong sense without proof that it must be true.
More generalized conjecture can be written:
Extended Conjecture:
$$f(m)=sumlimits_n = 1 ^ m (-1)^n sinleft(sumlimits_k = 1 ^infty fraca_k pib_kP_k(n) right) tag 3 $$
More extended conjecture claims that if $P_r(n)=sumlimits_k = 1 ^ n k^2r$ where r is a positive integer,
$f(m)$ function is periodic function when $a_k$ is any integers, $b_k$ is non-zero integers and $m$ positive integers.$sumlimits_k = 1 ^T f(k)=0 $
where ($T$) is the period value for all possible $a_k,b_k,m$ integers.
Can the extended conjecture be proven? What is the period formula for it?
I also have been still looking for $G(n)$ polynomials that is different from $G(n)=sumlimits_k = 1 ^infty fraca_kb_kP_k(n)$ satisfies $ sumlimits_k = 1 ^T f(k)=0 $ $tag4$ for all $a,b,m$ positive integers
(Please consider Equation $(2)$)
Finally, I claim that there is no any polynomials different from $G(n)=sumlimits_k = 1 ^infty fraca_kb_kP_k(n)$ that satisfy $ sumlimits_k = 1 ^T f(k)=0 $. Please inform me if you find a counter-example.
Thanks for answers
sequences-and-series trigonometric-series
asked Jul 16 at 9:15
Mathlover
6,26222265
I don't know if this is helpful, but it's possible to put all into a power of $-1$ (and remove the trig function): It can be shown $f(m) = Im[sum_n=1^m (-1)^g(n)]$, where $g(n) = n+ sum_k=1^infty fraca_kb_kP_k(n)$.
– packetpacket
Aug 6 at 14:17
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I asked a question that is related to this question and claimed that
Generalized Conjecture:
$$f(m)=sumlimits_n = 1 ^ m (-1)^n sinleft(P_r(n) fraca pibright) tag 1 $$
I have a conjecture that if $P_r(n)=sumlimits_k = 1 ^ n k^2r$ where r is a positive integer,
$f(m)$ function is periodic function when $a,b,m$ positive integers and
$ sumlimits_k = 1 ^T f(k)=0 $
where ($T$) is the period value.
(You may see a proof for (Conjecture (1) above) by @FabioLucchini)
I have found out an example and It shows my generalized conjecture can be extended more. I have tested with many numerical values that It supports my extended conjecture below.
$$f(m)=sumlimits_n = 1 ^ m (-1)^n sinleft(G(n) fraca pibright) tag 2 $$
$$G(n)=fracn(n+1)(2n+1)(3n^2+3n-4)30$$ It also satisfies my generalized conjecture $(1)$ above. $G(n)$ can be written as:
$$G(n)=fracn(n+1)(2n+1)(3n^2+3n-1)30-frac35fracn(n+1)(2n+1)6=sumlimits_k = 1 ^ n k^4-frac35sumlimits_k = 1 ^ n k^2=P_2(n)-frac35P_1(n)$$
Test link for $a=2,b=5$ and $m=500$
The numerical values and my works on the subject estimates the extension of the conjecture above. It is just strong sense without proof that it must be true.
More generalized conjecture can be written:
Extended Conjecture:
$$f(m)=sumlimits_n = 1 ^ m (-1)^n sinleft(sumlimits_k = 1 ^infty fraca_k pib_kP_k(n) right) tag 3 $$
More extended conjecture claims that if $P_r(n)=sumlimits_k = 1 ^ n k^2r$ where r is a positive integer,
$f(m)$ function is periodic function when $a_k$ is any integers, $b_k$ is non-zero integers and $m$ positive integers.$sumlimits_k = 1 ^T f(k)=0 $
where ($T$) is the period value for all possible $a_k,b_k,m$ integers.
Can the extended conjecture be proven? What is the period formula for it?
I also have been still looking for $G(n)$ polynomials that is different from $G(n)=sumlimits_k = 1 ^infty fraca_kb_kP_k(n)$ satisfies $ sumlimits_k = 1 ^T f(k)=0 $ $tag4$ for all $a,b,m$ positive integers
(Please consider Equation $(2)$)
Finally, I claim that there is no any polynomials different from $G(n)=sumlimits_k = 1 ^infty fraca_kb_kP_k(n)$ that satisfy $ sumlimits_k = 1 ^T f(k)=0 $. Please inform me if you find a counter-example.
Thanks for answers
sequences-and-series trigonometric-series
asked Jul 16 at 9:15
Mathlover
6,26222265
I asked a question that is related to this question and claimed that
Generalized Conjecture:
$$f(m)=sumlimits_n = 1 ^ m (-1)^n sinleft(P_r(n) fraca pibright) tag 1 $$
I have a conjecture that if $P_r(n)=sumlimits_k = 1 ^ n k^2r$ where r is a positive integer,
$f(m)$ function is periodic function when $a,b,m$ positive integers and
$ sumlimits_k = 1 ^T f(k)=0 $
where ($T$) is the period value.
(You may see a proof for (Conjecture (1) above) by @FabioLucchini)
I have found out an example and It shows my generalized conjecture can be extended more. I have tested with many numerical values that It supports my extended conjecture below.
$$f(m)=sumlimits_n = 1 ^ m (-1)^n sinleft(G(n) fraca pibright) tag 2 $$
$$G(n)=fracn(n+1)(2n+1)(3n^2+3n-4)30$$ It also satisfies my generalized conjecture $(1)$ above. $G(n)$ can be written as:
$$G(n)=fracn(n+1)(2n+1)(3n^2+3n-1)30-frac35fracn(n+1)(2n+1)6=sumlimits_k = 1 ^ n k^4-frac35sumlimits_k = 1 ^ n k^2=P_2(n)-frac35P_1(n)$$
Test link for $a=2,b=5$ and $m=500$
The numerical values and my works on the subject estimates the extension of the conjecture above. It is just strong sense without proof that it must be true.
More generalized conjecture can be written:
Extended Conjecture:
$$f(m)=sumlimits_n = 1 ^ m (-1)^n sinleft(sumlimits_k = 1 ^infty fraca_k pib_kP_k(n) right) tag 3 $$
More extended conjecture claims that if $P_r(n)=sumlimits_k = 1 ^ n k^2r$ where r is a positive integer,
$f(m)$ function is periodic function when $a_k$ is any integers, $b_k$ is non-zero integers and $m$ positive integers.$sumlimits_k = 1 ^T f(k)=0 $
where ($T$) is the period value for all possible $a_k,b_k,m$ integers.
Can the extended conjecture be proven? What is the period formula for it?
I also have been still looking for $G(n)$ polynomials that is different from $G(n)=sumlimits_k = 1 ^infty fraca_kb_kP_k(n)$ satisfies $ sumlimits_k = 1 ^T f(k)=0 $ $tag4$ for all $a,b,m$ positive integers
(Please consider Equation $(2)$)
Finally, I claim that there is no any polynomials different from $G(n)=sumlimits_k = 1 ^infty fraca_kb_kP_k(n)$ that satisfy $ sumlimits_k = 1 ^T f(k)=0 $. Please inform me if you find a counter-example.
Thanks for answers
sequences-and-series trigonometric-series
asked Jul 16 at 9:15
Mathlover
6,26222265
edited Jul 17 at 8:28
asked Jul 16 at 9:15
Mathlover
6,26222265
asked Jul 16 at 9:15
Mathlover
6,26222265
asked Jul 16 at 9:15
Mathlover
6,26222265
6,26222265
I don't know if this is helpful, but it's possible to put all into a power of $-1$ (and remove the trig function): It can be shown $f(m) = Im[sum_n=1^m (-1)^g(n)]$, where $g(n) = n+ sum_k=1^infty fraca_kb_kP_k(n)$.
– packetpacket
Aug 6 at 14:17
add a comment |Â
I don't know if this is helpful, but it's possible to put all into a power of $-1$ (and remove the trig function): It can be shown $f(m) = Im[sum_n=1^m (-1)^g(n)]$, where $g(n) = n+ sum_k=1^infty fraca_kb_kP_k(n)$.
– packetpacket
Aug 6 at 14:17
I don't know if this is helpful, but it's possible to put all into a power of $-1$ (and remove the trig function): It can be shown $f(m) = Im[sum_n=1^m (-1)^g(n)]$, where $g(n) = n+ sum_k=1^infty fraca_kb_kP_k(n)$.
– packetpacket
Aug 6 at 14:17
I don't know if this is helpful, but it's possible to put all into a power of $-1$ (and remove the trig function): It can be shown $f(m) = Im[sum_n=1^m (-1)^g(n)]$, where $g(n) = n+ sum_k=1^infty fraca_kb_kP_k(n)$.
– packetpacket
Aug 6 at 14:17
add a comment |Â
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I don't know if this is helpful, but it's possible to put all into a power of $-1$ (and remove the trig function): It can be shown $f(m) = Im[sum_n=1^m (-1)^g(n)]$, where $g(n) = n+ sum_k=1^infty fraca_kb_kP_k(n)$.
– packetpacket
Aug 6 at 14:17