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Understanding part of derivation of Chebychev's Theorem
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I cannot understand this result from pages 17–18 of Tenenbaum and Mendes's The Prime Numbers and Their Distribution on how the summation of $fracxlog(2)2^j+O(log(x))$ results in $2x log(2) + O(log(x)^2).$
What are the steps in the summation that give the result? Did they use a telescoping series?
It is given that $B_2(x)=log[2x]! - 2log[x]! = B(x)-2B(x/2)$, and from the text,
The lower estimate for $B_2 (x)$ is used inductively: we have
beginalign psi(x) & le B_2(x) + psi(x/2) le B_2(x) + B_2(x/2) + psi(x/4) \
& le cdots le sum_0 le j le k B_2(x/2^j) + psi(x/2^k+1). endalign
Here, $k$ is an arbitrary integer. Let us choose $$k = K(x) := [(log x)/ log 2],$$
so that $psi(x/2^k+1) = 0$. It follows that
beginalign psi(x) & le sum_0 le j le K(x) leftlbrace fracx log 22^j + O(log x)rightrbrace \
& le 2x log 2 + O((log x)^2).
endalign
number-theory elementary-number-theory prime-numbers analytic-number-theory prime-factorization
asked Aug 5 at 19:18
onepound
181115
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I cannot understand this result from pages 17–18 of Tenenbaum and Mendes's The Prime Numbers and Their Distribution on how the summation of $fracxlog(2)2^j+O(log(x))$ results in $2x log(2) + O(log(x)^2).$
What are the steps in the summation that give the result? Did they use a telescoping series?
It is given that $B_2(x)=log[2x]! - 2log[x]! = B(x)-2B(x/2)$, and from the text,
The lower estimate for $B_2 (x)$ is used inductively: we have
beginalign psi(x) & le B_2(x) + psi(x/2) le B_2(x) + B_2(x/2) + psi(x/4) \
& le cdots le sum_0 le j le k B_2(x/2^j) + psi(x/2^k+1). endalign
Here, $k$ is an arbitrary integer. Let us choose $$k = K(x) := [(log x)/ log 2],$$
so that $psi(x/2^k+1) = 0$. It follows that
beginalign psi(x) & le sum_0 le j le K(x) leftlbrace fracx log 22^j + O(log x)rightrbrace \
& le 2x log 2 + O((log x)^2).
endalign
number-theory elementary-number-theory prime-numbers analytic-number-theory prime-factorization
asked Aug 5 at 19:18
onepound
181115
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I cannot understand this result from pages 17–18 of Tenenbaum and Mendes's The Prime Numbers and Their Distribution on how the summation of $fracxlog(2)2^j+O(log(x))$ results in $2x log(2) + O(log(x)^2).$
What are the steps in the summation that give the result? Did they use a telescoping series?
It is given that $B_2(x)=log[2x]! - 2log[x]! = B(x)-2B(x/2)$, and from the text,
The lower estimate for $B_2 (x)$ is used inductively: we have
beginalign psi(x) & le B_2(x) + psi(x/2) le B_2(x) + B_2(x/2) + psi(x/4) \
& le cdots le sum_0 le j le k B_2(x/2^j) + psi(x/2^k+1). endalign
Here, $k$ is an arbitrary integer. Let us choose $$k = K(x) := [(log x)/ log 2],$$
so that $psi(x/2^k+1) = 0$. It follows that
beginalign psi(x) & le sum_0 le j le K(x) leftlbrace fracx log 22^j + O(log x)rightrbrace \
& le 2x log 2 + O((log x)^2).
endalign
number-theory elementary-number-theory prime-numbers analytic-number-theory prime-factorization
asked Aug 5 at 19:18
onepound
181115
I cannot understand this result from pages 17–18 of Tenenbaum and Mendes's The Prime Numbers and Their Distribution on how the summation of $fracxlog(2)2^j+O(log(x))$ results in $2x log(2) + O(log(x)^2).$
What are the steps in the summation that give the result? Did they use a telescoping series?
It is given that $B_2(x)=log[2x]! - 2log[x]! = B(x)-2B(x/2)$, and from the text,
The lower estimate for $B_2 (x)$ is used inductively: we have
beginalign psi(x) & le B_2(x) + psi(x/2) le B_2(x) + B_2(x/2) + psi(x/4) \
& le cdots le sum_0 le j le k B_2(x/2^j) + psi(x/2^k+1). endalign
Here, $k$ is an arbitrary integer. Let us choose $$k = K(x) := [(log x)/ log 2],$$
so that $psi(x/2^k+1) = 0$. It follows that
beginalign psi(x) & le sum_0 le j le K(x) leftlbrace fracx log 22^j + O(log x)rightrbrace \
& le 2x log 2 + O((log x)^2).
endalign
number-theory elementary-number-theory prime-numbers analytic-number-theory prime-factorization
asked Aug 5 at 19:18
onepound
181115
edited Aug 6 at 9:50
asked Aug 5 at 19:18
onepound
181115
asked Aug 5 at 19:18
onepound
181115
asked Aug 5 at 19:18
onepound
181115
181115
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add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
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accepted
$$sum_jleft(xlog2over2^j+O(log x)right)=xlog2sum_j(1/2^j)+sum_jO(log x)le2xlog2+O(K(x)(log x))$$
$$=2xlog2+O((log x)^2)$$
answered Aug 6 at 10:04
Gerry Myerson
143k7145294
1
I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
– onepound
Aug 7 at 9:27
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$sum_jleft(xlog2over2^j+O(log x)right)=xlog2sum_j(1/2^j)+sum_jO(log x)le2xlog2+O(K(x)(log x))$$
$$=2xlog2+O((log x)^2)$$
answered Aug 6 at 10:04
Gerry Myerson
143k7145294
1
I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
– onepound
Aug 7 at 9:27
add a comment |Â
up vote
1
down vote
accepted
$$sum_jleft(xlog2over2^j+O(log x)right)=xlog2sum_j(1/2^j)+sum_jO(log x)le2xlog2+O(K(x)(log x))$$
$$=2xlog2+O((log x)^2)$$
answered Aug 6 at 10:04
Gerry Myerson
143k7145294
1
I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
– onepound
Aug 7 at 9:27
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$sum_jleft(xlog2over2^j+O(log x)right)=xlog2sum_j(1/2^j)+sum_jO(log x)le2xlog2+O(K(x)(log x))$$
$$=2xlog2+O((log x)^2)$$
answered Aug 6 at 10:04
Gerry Myerson
143k7145294
$$sum_jleft(xlog2over2^j+O(log x)right)=xlog2sum_j(1/2^j)+sum_jO(log x)le2xlog2+O(K(x)(log x))$$
$$=2xlog2+O((log x)^2)$$
answered Aug 6 at 10:04
Gerry Myerson
143k7145294
answered Aug 6 at 10:04
Gerry Myerson
143k7145294
answered Aug 6 at 10:04
Gerry Myerson
143k7145294
answered Aug 6 at 10:04
Gerry Myerson
143k7145294
143k7145294
1
I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
– onepound
Aug 7 at 9:27
add a comment |Â
1
I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
– onepound
Aug 7 at 9:27
1
1
I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
– onepound
Aug 7 at 9:27
I see because $sum_j(1/2^j)=2$ and $log 2$ in asymptotic $((log x)/ log 2) log x$ can be ignored. Thanks.
– onepound
Aug 7 at 9:27
add a comment |Â
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