Why we need to have $S(x)=O(x)$?
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Currently, I'm working on problem 4.21 in Apostol's Analytic number theory and I solved that in different ways. The original problem is as follows:
Given two real-valued functions $S(x)$ and $T(x)$ such that
$$T(x)=sum_{nle x}S(frac xn),xge1.$$
If $S(x)=O(x)$ and if $c$ is a positive costant, prove that the relation
$$S(x)sim cx~text{as}~xtoinfty$$ implies
$$T(x)sim cxlog x~text{as}~x rightarrow infty.$$
This is one of my answers:
$$lim_{xtoinfty}frac{T(x)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}S(frac xn)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}(frac{cx}n+o(x))}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}frac 1n}{log x}=lim_{xtoinfty}frac{log x+O(1)}{log x}=1$$
As you see, I didn't use the relation $S(x)=O(x)$. But the problem has said that "If $S(x)=O(x)$". Could anyone help me to know what is wrong in my answer? Thanks!
number-theory analytic-number-theory
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add a comment |
$begingroup$
Currently, I'm working on problem 4.21 in Apostol's Analytic number theory and I solved that in different ways. The original problem is as follows:
Given two real-valued functions $S(x)$ and $T(x)$ such that
$$T(x)=sum_{nle x}S(frac xn),xge1.$$
If $S(x)=O(x)$ and if $c$ is a positive costant, prove that the relation
$$S(x)sim cx~text{as}~xtoinfty$$ implies
$$T(x)sim cxlog x~text{as}~x rightarrow infty.$$
This is one of my answers:
$$lim_{xtoinfty}frac{T(x)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}S(frac xn)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}(frac{cx}n+o(x))}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}frac 1n}{log x}=lim_{xtoinfty}frac{log x+O(1)}{log x}=1$$
As you see, I didn't use the relation $S(x)=O(x)$. But the problem has said that "If $S(x)=O(x)$". Could anyone help me to know what is wrong in my answer? Thanks!
number-theory analytic-number-theory
$endgroup$
add a comment |
$begingroup$
Currently, I'm working on problem 4.21 in Apostol's Analytic number theory and I solved that in different ways. The original problem is as follows:
Given two real-valued functions $S(x)$ and $T(x)$ such that
$$T(x)=sum_{nle x}S(frac xn),xge1.$$
If $S(x)=O(x)$ and if $c$ is a positive costant, prove that the relation
$$S(x)sim cx~text{as}~xtoinfty$$ implies
$$T(x)sim cxlog x~text{as}~x rightarrow infty.$$
This is one of my answers:
$$lim_{xtoinfty}frac{T(x)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}S(frac xn)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}(frac{cx}n+o(x))}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}frac 1n}{log x}=lim_{xtoinfty}frac{log x+O(1)}{log x}=1$$
As you see, I didn't use the relation $S(x)=O(x)$. But the problem has said that "If $S(x)=O(x)$". Could anyone help me to know what is wrong in my answer? Thanks!
number-theory analytic-number-theory
$endgroup$
Currently, I'm working on problem 4.21 in Apostol's Analytic number theory and I solved that in different ways. The original problem is as follows:
Given two real-valued functions $S(x)$ and $T(x)$ such that
$$T(x)=sum_{nle x}S(frac xn),xge1.$$
If $S(x)=O(x)$ and if $c$ is a positive costant, prove that the relation
$$S(x)sim cx~text{as}~xtoinfty$$ implies
$$T(x)sim cxlog x~text{as}~x rightarrow infty.$$
This is one of my answers:
$$lim_{xtoinfty}frac{T(x)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}S(frac xn)}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}(frac{cx}n+o(x))}{cxlog(x)}=lim_{xtoinfty}frac{sum_{nle x}frac 1n}{log x}=lim_{xtoinfty}frac{log x+O(1)}{log x}=1$$
As you see, I didn't use the relation $S(x)=O(x)$. But the problem has said that "If $S(x)=O(x)$". Could anyone help me to know what is wrong in my answer? Thanks!
number-theory analytic-number-theory
number-theory analytic-number-theory
edited Jan 8 at 12:33
asked Jan 8 at 11:16
user604594
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The assumption $S(x)=O(x)$ follows from $S(x)sim cx$ and therefore is totally redundant.
On the other hand, there is an inaccuracy in your argument: your $o(x)$ term actually depends not only on $x$, but also on $n$ and, more importantly, you cannot dispose of this term the way you did. You should instead define $phi(x)$ by $S(x)=cx(1+phi(x))$, so that $phi(x)to 0$ and argue more accurately in terms of this function.
answered Jan 8 at 15:47
W-t-PW-t-P
1,749612
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Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
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– Peter Humphries
Jan 9 at 14:50
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@Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
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– W-t-P
Jan 9 at 15:55
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sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
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– Peter Humphries
Jan 9 at 15:57
add a comment |
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$begingroup$
The assumption $S(x)=O(x)$ follows from $S(x)sim cx$ and therefore is totally redundant.
On the other hand, there is an inaccuracy in your argument: your $o(x)$ term actually depends not only on $x$, but also on $n$ and, more importantly, you cannot dispose of this term the way you did. You should instead define $phi(x)$ by $S(x)=cx(1+phi(x))$, so that $phi(x)to 0$ and argue more accurately in terms of this function.
answered Jan 8 at 15:47
W-t-PW-t-P
1,749612
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$begingroup$
Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
$endgroup$
– Peter Humphries
Jan 9 at 14:50
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@Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
$endgroup$
– W-t-P
Jan 9 at 15:55
$begingroup$
sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
$endgroup$
– Peter Humphries
Jan 9 at 15:57
add a comment |
$begingroup$
The assumption $S(x)=O(x)$ follows from $S(x)sim cx$ and therefore is totally redundant.
On the other hand, there is an inaccuracy in your argument: your $o(x)$ term actually depends not only on $x$, but also on $n$ and, more importantly, you cannot dispose of this term the way you did. You should instead define $phi(x)$ by $S(x)=cx(1+phi(x))$, so that $phi(x)to 0$ and argue more accurately in terms of this function.
answered Jan 8 at 15:47
W-t-PW-t-P
1,749612
$endgroup$
$begingroup$
Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
$endgroup$
– Peter Humphries
Jan 9 at 14:50
$begingroup$
@Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
$endgroup$
– W-t-P
Jan 9 at 15:55
$begingroup$
sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
$endgroup$
– Peter Humphries
Jan 9 at 15:57
add a comment |
$begingroup$
The assumption $S(x)=O(x)$ follows from $S(x)sim cx$ and therefore is totally redundant.
On the other hand, there is an inaccuracy in your argument: your $o(x)$ term actually depends not only on $x$, but also on $n$ and, more importantly, you cannot dispose of this term the way you did. You should instead define $phi(x)$ by $S(x)=cx(1+phi(x))$, so that $phi(x)to 0$ and argue more accurately in terms of this function.
answered Jan 8 at 15:47
W-t-PW-t-P
1,749612
$endgroup$
The assumption $S(x)=O(x)$ follows from $S(x)sim cx$ and therefore is totally redundant.
On the other hand, there is an inaccuracy in your argument: your $o(x)$ term actually depends not only on $x$, but also on $n$ and, more importantly, you cannot dispose of this term the way you did. You should instead define $phi(x)$ by $S(x)=cx(1+phi(x))$, so that $phi(x)to 0$ and argue more accurately in terms of this function.
answered Jan 8 at 15:47
W-t-PW-t-P
1,749612
answered Jan 8 at 15:47
W-t-PW-t-P
1,749612
answered Jan 8 at 15:47
W-t-PW-t-P
1,749612
answered Jan 8 at 15:47
W-t-PW-t-P
1,749612
1,749612
$begingroup$
Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
$endgroup$
– Peter Humphries
Jan 9 at 14:50
$begingroup$
@Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
$endgroup$
– W-t-P
Jan 9 at 15:55
$begingroup$
sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
$endgroup$
– Peter Humphries
Jan 9 at 15:57
add a comment |
$begingroup$
Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
$endgroup$
– Peter Humphries
Jan 9 at 14:50
$begingroup$
@Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
$endgroup$
– W-t-P
Jan 9 at 15:55
$begingroup$
sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
$endgroup$
– Peter Humphries
Jan 9 at 15:57
$begingroup$
Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
$endgroup$
– Peter Humphries
Jan 9 at 14:50
$begingroup$
Nah, just write $S(frac{x}{n}) = frac{cx}{n} + o(frac{x}{n})$ and use the fact that $sum_{n leq x} o(frac{x}{n}) = o(x sum_{n leq x} frac{1}{n}) = o(x log x)$.
$endgroup$
– Peter Humphries
Jan 9 at 14:50
$begingroup$
@Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
$endgroup$
– W-t-P
Jan 9 at 15:55
$begingroup$
@Peter Huphries: The equality $sum_{nle x} o(x/n)=o(xsum_{nle x} frac1n)$ is correct but not immediate!
$endgroup$
– W-t-P
Jan 9 at 15:55
$begingroup$
sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
$endgroup$
– Peter Humphries
Jan 9 at 15:57
$begingroup$
sure, but it's a completely standard calculation in analytic number theory (which I expect is justified in detail somewhere in Apostol's book).
$endgroup$
– Peter Humphries
Jan 9 at 15:57
add a comment |
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