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Proving a sequence with infinitely many zeros is not zero heavy?
Clash Royale CLAN TAG#URR8PPP
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Is my counterexample correct?
If a sequence contains an infinite number of zeroes, is it necessarily
zero-heavy? If not, provide a counterexample.
Solution. Consider the sequence $(a_n)$ defined such that $a_n = 0$ whenever $n = 2^k$ for some $kinmathbfN$ and is $1$ otherwise, then given any $MinmathbfN$ we may choose a $kinmathbfN$ such that $2^k+1-(2^k+1)>M$ then given the construction of the sequence if $nin2^k+1,2^k+1+2,dots,(2^k+1)+M$ we have $a_nneq 0$.The sequence in question then cannot possibly be zero heavy.
NOTE: We define a sequence to be zero heavy if $$exists MinmathbfNforall NinmathbfNexists ninN,N+1,dots,N+M(x_n=0)$$
real-analysis sequences-and-series proof-verification
asked Jul 19 at 11:56
Atif Farooq
2,7092824
add a comment |Â
up vote
3
down vote
favorite
Is my counterexample correct?
If a sequence contains an infinite number of zeroes, is it necessarily
zero-heavy? If not, provide a counterexample.
Solution. Consider the sequence $(a_n)$ defined such that $a_n = 0$ whenever $n = 2^k$ for some $kinmathbfN$ and is $1$ otherwise, then given any $MinmathbfN$ we may choose a $kinmathbfN$ such that $2^k+1-(2^k+1)>M$ then given the construction of the sequence if $nin2^k+1,2^k+1+2,dots,(2^k+1)+M$ we have $a_nneq 0$.The sequence in question then cannot possibly be zero heavy.
NOTE: We define a sequence to be zero heavy if $$exists MinmathbfNforall NinmathbfNexists ninN,N+1,dots,N+M(x_n=0)$$
real-analysis sequences-and-series proof-verification
asked Jul 19 at 11:56
Atif Farooq
2,7092824
2
Your construction works.
– Suzet
Jul 19 at 11:59
It's correct. Another example: Let $a_n=0$ if $n$ is prime and $a_n=1$ otherwise. If $2leq K>M$ then $K!+j:2leq jleq K$ contains no primes because $K!+j >j>1$ and $ j$ divides $K!+j$ (... when $Kgeq j >1).$
– DanielWainfleet
Jul 20 at 6:57
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Is my counterexample correct?
If a sequence contains an infinite number of zeroes, is it necessarily
zero-heavy? If not, provide a counterexample.
Solution. Consider the sequence $(a_n)$ defined such that $a_n = 0$ whenever $n = 2^k$ for some $kinmathbfN$ and is $1$ otherwise, then given any $MinmathbfN$ we may choose a $kinmathbfN$ such that $2^k+1-(2^k+1)>M$ then given the construction of the sequence if $nin2^k+1,2^k+1+2,dots,(2^k+1)+M$ we have $a_nneq 0$.The sequence in question then cannot possibly be zero heavy.
NOTE: We define a sequence to be zero heavy if $$exists MinmathbfNforall NinmathbfNexists ninN,N+1,dots,N+M(x_n=0)$$
real-analysis sequences-and-series proof-verification
asked Jul 19 at 11:56
Atif Farooq
2,7092824
Is my counterexample correct?
If a sequence contains an infinite number of zeroes, is it necessarily
zero-heavy? If not, provide a counterexample.
Solution. Consider the sequence $(a_n)$ defined such that $a_n = 0$ whenever $n = 2^k$ for some $kinmathbfN$ and is $1$ otherwise, then given any $MinmathbfN$ we may choose a $kinmathbfN$ such that $2^k+1-(2^k+1)>M$ then given the construction of the sequence if $nin2^k+1,2^k+1+2,dots,(2^k+1)+M$ we have $a_nneq 0$.The sequence in question then cannot possibly be zero heavy.
NOTE: We define a sequence to be zero heavy if $$exists MinmathbfNforall NinmathbfNexists ninN,N+1,dots,N+M(x_n=0)$$
real-analysis sequences-and-series proof-verification
asked Jul 19 at 11:56
Atif Farooq
2,7092824
asked Jul 19 at 11:56
Atif Farooq
2,7092824
asked Jul 19 at 11:56
Atif Farooq
2,7092824
asked Jul 19 at 11:56
Atif Farooq
2,7092824
2,7092824
2
Your construction works.
– Suzet
Jul 19 at 11:59
It's correct. Another example: Let $a_n=0$ if $n$ is prime and $a_n=1$ otherwise. If $2leq K>M$ then $K!+j:2leq jleq K$ contains no primes because $K!+j >j>1$ and $ j$ divides $K!+j$ (... when $Kgeq j >1).$
– DanielWainfleet
Jul 20 at 6:57
add a comment |Â
2
Your construction works.
– Suzet
Jul 19 at 11:59
It's correct. Another example: Let $a_n=0$ if $n$ is prime and $a_n=1$ otherwise. If $2leq K>M$ then $K!+j:2leq jleq K$ contains no primes because $K!+j >j>1$ and $ j$ divides $K!+j$ (... when $Kgeq j >1).$
– DanielWainfleet
Jul 20 at 6:57
2
2
Your construction works.
– Suzet
Jul 19 at 11:59
Your construction works.
– Suzet
Jul 19 at 11:59
It's correct. Another example: Let $a_n=0$ if $n$ is prime and $a_n=1$ otherwise. If $2leq K>M$ then $K!+j:2leq jleq K$ contains no primes because $K!+j >j>1$ and $ j$ divides $K!+j$ (... when $Kgeq j >1).$
– DanielWainfleet
Jul 20 at 6:57
It's correct. Another example: Let $a_n=0$ if $n$ is prime and $a_n=1$ otherwise. If $2leq K>M$ then $K!+j:2leq jleq K$ contains no primes because $K!+j >j>1$ and $ j$ divides $K!+j$ (... when $Kgeq j >1).$
– DanielWainfleet
Jul 20 at 6:57
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Looks good! The only part I might do/phrase at all differently is:
...we may choose $kinmathbfN$ such that $k>log_2(M+1),$ so that $2^k+1-(2^k+1)>M,$ and so, given the construction of the sequence, if....
answered Jul 19 at 12:05
Cameron Buie
83.5k771153
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Looks good! The only part I might do/phrase at all differently is:
...we may choose $kinmathbfN$ such that $k>log_2(M+1),$ so that $2^k+1-(2^k+1)>M,$ and so, given the construction of the sequence, if....
answered Jul 19 at 12:05
Cameron Buie
83.5k771153
add a comment |Â
up vote
3
down vote
accepted
Looks good! The only part I might do/phrase at all differently is:
...we may choose $kinmathbfN$ such that $k>log_2(M+1),$ so that $2^k+1-(2^k+1)>M,$ and so, given the construction of the sequence, if....
answered Jul 19 at 12:05
Cameron Buie
83.5k771153
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Looks good! The only part I might do/phrase at all differently is:
...we may choose $kinmathbfN$ such that $k>log_2(M+1),$ so that $2^k+1-(2^k+1)>M,$ and so, given the construction of the sequence, if....
answered Jul 19 at 12:05
Cameron Buie
83.5k771153
Looks good! The only part I might do/phrase at all differently is:
...we may choose $kinmathbfN$ such that $k>log_2(M+1),$ so that $2^k+1-(2^k+1)>M,$ and so, given the construction of the sequence, if....
answered Jul 19 at 12:05
Cameron Buie
83.5k771153
answered Jul 19 at 12:05
Cameron Buie
83.5k771153
answered Jul 19 at 12:05
Cameron Buie
83.5k771153
answered Jul 19 at 12:05
Cameron Buie
83.5k771153
83.5k771153
add a comment |Â
add a comment |Â
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2
Your construction works.
– Suzet
Jul 19 at 11:59
It's correct. Another example: Let $a_n=0$ if $n$ is prime and $a_n=1$ otherwise. If $2leq K>M$ then $K!+j:2leq jleq K$ contains no primes because $K!+j >j>1$ and $ j$ divides $K!+j$ (... when $Kgeq j >1).$
– DanielWainfleet
Jul 20 at 6:57