Since $zeta(1) neq 1$, does this mean that $zeta$ is not multiplicative?












0












$begingroup$


Let $zeta(s)$ be the Riemann zeta function, that is,
$$zeta(s) = sum_{n=1}^{infty}{frac{1}{n^s}}.$$



A function $g$ is said to be multiplicative if, whenever $gcd(x,y)=1$, we have
$$g(xy) = g(x)g(y).$$
Examples of multiplicative (arithmetic) functions include the divisor-sum $sigma(z)=sigma_{1}(z)$ and the abundancy index $I(z)=sigma(z)/z$.



It is known that, if a function $f$ is multiplicative (and not identically zero), then $f(1)=1$.




Proof: Suppose that $f$ is multiplicative (and not identically zero). Since $gcd(m,1)=1$ and $f neq 0$ is multiplicative, then
$$f(m)=f(mcdot{1})=f(m)f(1)$$
which implies that $f(1)=1$.




Since we know that $zeta(1) neq 1$ (in fact, the sum
$$zeta(1) = sum_{n=1}^{infty}{frac{1}{n}}$$
is known to diverge), then can we already conclude that the Riemann zeta function $zeta$ is not multiplicative?






number-theory elementary-number-theory riemann-zeta multiplicative-function







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The term "multiplicative" is for functions defined on the positive integers. Zeta isn't defined at $1.$
    $endgroup$
    – coffeemath
    Jan 8 at 12:02








  • 1




    $begingroup$
    @coffeemath, ohh yeah right! Can you write your last comment as an actual answer, so that I would be able to accept it? Thanks!
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 8 at 12:06
















0












$begingroup$


Let $zeta(s)$ be the Riemann zeta function, that is,
$$zeta(s) = sum_{n=1}^{infty}{frac{1}{n^s}}.$$



A function $g$ is said to be multiplicative if, whenever $gcd(x,y)=1$, we have
$$g(xy) = g(x)g(y).$$
Examples of multiplicative (arithmetic) functions include the divisor-sum $sigma(z)=sigma_{1}(z)$ and the abundancy index $I(z)=sigma(z)/z$.



It is known that, if a function $f$ is multiplicative (and not identically zero), then $f(1)=1$.




Proof: Suppose that $f$ is multiplicative (and not identically zero). Since $gcd(m,1)=1$ and $f neq 0$ is multiplicative, then
$$f(m)=f(mcdot{1})=f(m)f(1)$$
which implies that $f(1)=1$.




Since we know that $zeta(1) neq 1$ (in fact, the sum
$$zeta(1) = sum_{n=1}^{infty}{frac{1}{n}}$$
is known to diverge), then can we already conclude that the Riemann zeta function $zeta$ is not multiplicative?






number-theory elementary-number-theory riemann-zeta multiplicative-function







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The term "multiplicative" is for functions defined on the positive integers. Zeta isn't defined at $1.$
    $endgroup$
    – coffeemath
    Jan 8 at 12:02








  • 1




    $begingroup$
    @coffeemath, ohh yeah right! Can you write your last comment as an actual answer, so that I would be able to accept it? Thanks!
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 8 at 12:06














0












0








0


2



$begingroup$


Let $zeta(s)$ be the Riemann zeta function, that is,
$$zeta(s) = sum_{n=1}^{infty}{frac{1}{n^s}}.$$



A function $g$ is said to be multiplicative if, whenever $gcd(x,y)=1$, we have
$$g(xy) = g(x)g(y).$$
Examples of multiplicative (arithmetic) functions include the divisor-sum $sigma(z)=sigma_{1}(z)$ and the abundancy index $I(z)=sigma(z)/z$.



It is known that, if a function $f$ is multiplicative (and not identically zero), then $f(1)=1$.




Proof: Suppose that $f$ is multiplicative (and not identically zero). Since $gcd(m,1)=1$ and $f neq 0$ is multiplicative, then
$$f(m)=f(mcdot{1})=f(m)f(1)$$
which implies that $f(1)=1$.




Since we know that $zeta(1) neq 1$ (in fact, the sum
$$zeta(1) = sum_{n=1}^{infty}{frac{1}{n}}$$
is known to diverge), then can we already conclude that the Riemann zeta function $zeta$ is not multiplicative?






number-theory elementary-number-theory riemann-zeta multiplicative-function







share|cite|improve this question









$endgroup$




Let $zeta(s)$ be the Riemann zeta function, that is,
$$zeta(s) = sum_{n=1}^{infty}{frac{1}{n^s}}.$$



A function $g$ is said to be multiplicative if, whenever $gcd(x,y)=1$, we have
$$g(xy) = g(x)g(y).$$
Examples of multiplicative (arithmetic) functions include the divisor-sum $sigma(z)=sigma_{1}(z)$ and the abundancy index $I(z)=sigma(z)/z$.



It is known that, if a function $f$ is multiplicative (and not identically zero), then $f(1)=1$.




Proof: Suppose that $f$ is multiplicative (and not identically zero). Since $gcd(m,1)=1$ and $f neq 0$ is multiplicative, then
$$f(m)=f(mcdot{1})=f(m)f(1)$$
which implies that $f(1)=1$.




Since we know that $zeta(1) neq 1$ (in fact, the sum
$$zeta(1) = sum_{n=1}^{infty}{frac{1}{n}}$$
is known to diverge), then can we already conclude that the Riemann zeta function $zeta$ is not multiplicative?






number-theory elementary-number-theory riemann-zeta multiplicative-function




number-theory elementary-number-theory riemann-zeta multiplicative-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 11:55









Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris

5,38141944




5,38141944








  • 2




    $begingroup$
    The term "multiplicative" is for functions defined on the positive integers. Zeta isn't defined at $1.$
    $endgroup$
    – coffeemath
    Jan 8 at 12:02








  • 1




    $begingroup$
    @coffeemath, ohh yeah right! Can you write your last comment as an actual answer, so that I would be able to accept it? Thanks!
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 8 at 12:06














  • 2




    $begingroup$
    The term "multiplicative" is for functions defined on the positive integers. Zeta isn't defined at $1.$
    $endgroup$
    – coffeemath
    Jan 8 at 12:02








  • 1




    $begingroup$
    @coffeemath, ohh yeah right! Can you write your last comment as an actual answer, so that I would be able to accept it? Thanks!
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 8 at 12:06








2




2




$begingroup$
The term "multiplicative" is for functions defined on the positive integers. Zeta isn't defined at $1.$
$endgroup$
– coffeemath
Jan 8 at 12:02






$begingroup$
The term "multiplicative" is for functions defined on the positive integers. Zeta isn't defined at $1.$
$endgroup$
– coffeemath
Jan 8 at 12:02






1




1




$begingroup$
@coffeemath, ohh yeah right! Can you write your last comment as an actual answer, so that I would be able to accept it? Thanks!
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 8 at 12:06




$begingroup$
@coffeemath, ohh yeah right! Can you write your last comment as an actual answer, so that I would be able to accept it? Thanks!
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 8 at 12:06










1 Answer
1






active

oldest

votes


















8












$begingroup$

The term "multiplicative" is used for a function $f$ defined on the positive integers, for which $f(mn)=f(m)f(n)$ whenever $gcd(m,n)=1.$ The Riemann zeta function $zeta$ isn't even defined at $1$ so is not a candidate to be called multiplicative. Even ignoring the problem at $1$ it could be checked numerically that $zeta(2)zeta(3) neq zeta(6)$.



Added: No need for a numerical check. Exact values for $zeta(2k)$ are known, and $zeta(6)/zeta(2)<0.7.$ So it can't equal $zeta(3),$ which is greater than $1.$



Another note: $zeta(n)$is strictly monotone decreasing as $n>1$ increases. So if for $a,b$ with $1 <a,b$ we had $zeta(ab)=zeta(a)zeta(b),$ then since $ab>a$ it would follow that $zeta(ab)/zeta(a)<1,$ which cannot be $zeta(b)>1.$ So $zeta(n)$ is not multiplicative for any values of $a,b>1,$ coprime or not.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you for your answer. I hope you don't mind my minor edit.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 8 at 12:18











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$begingroup$

The term "multiplicative" is used for a function $f$ defined on the positive integers, for which $f(mn)=f(m)f(n)$ whenever $gcd(m,n)=1.$ The Riemann zeta function $zeta$ isn't even defined at $1$ so is not a candidate to be called multiplicative. Even ignoring the problem at $1$ it could be checked numerically that $zeta(2)zeta(3) neq zeta(6)$.



Added: No need for a numerical check. Exact values for $zeta(2k)$ are known, and $zeta(6)/zeta(2)<0.7.$ So it can't equal $zeta(3),$ which is greater than $1.$



Another note: $zeta(n)$is strictly monotone decreasing as $n>1$ increases. So if for $a,b$ with $1 <a,b$ we had $zeta(ab)=zeta(a)zeta(b),$ then since $ab>a$ it would follow that $zeta(ab)/zeta(a)<1,$ which cannot be $zeta(b)>1.$ So $zeta(n)$ is not multiplicative for any values of $a,b>1,$ coprime or not.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you for your answer. I hope you don't mind my minor edit.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 8 at 12:18
















8












$begingroup$

The term "multiplicative" is used for a function $f$ defined on the positive integers, for which $f(mn)=f(m)f(n)$ whenever $gcd(m,n)=1.$ The Riemann zeta function $zeta$ isn't even defined at $1$ so is not a candidate to be called multiplicative. Even ignoring the problem at $1$ it could be checked numerically that $zeta(2)zeta(3) neq zeta(6)$.



Added: No need for a numerical check. Exact values for $zeta(2k)$ are known, and $zeta(6)/zeta(2)<0.7.$ So it can't equal $zeta(3),$ which is greater than $1.$



Another note: $zeta(n)$is strictly monotone decreasing as $n>1$ increases. So if for $a,b$ with $1 <a,b$ we had $zeta(ab)=zeta(a)zeta(b),$ then since $ab>a$ it would follow that $zeta(ab)/zeta(a)<1,$ which cannot be $zeta(b)>1.$ So $zeta(n)$ is not multiplicative for any values of $a,b>1,$ coprime or not.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Thank you for your answer. I hope you don't mind my minor edit.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 8 at 12:18














8












8








8





$begingroup$

The term "multiplicative" is used for a function $f$ defined on the positive integers, for which $f(mn)=f(m)f(n)$ whenever $gcd(m,n)=1.$ The Riemann zeta function $zeta$ isn't even defined at $1$ so is not a candidate to be called multiplicative. Even ignoring the problem at $1$ it could be checked numerically that $zeta(2)zeta(3) neq zeta(6)$.



Added: No need for a numerical check. Exact values for $zeta(2k)$ are known, and $zeta(6)/zeta(2)<0.7.$ So it can't equal $zeta(3),$ which is greater than $1.$



Another note: $zeta(n)$is strictly monotone decreasing as $n>1$ increases. So if for $a,b$ with $1 <a,b$ we had $zeta(ab)=zeta(a)zeta(b),$ then since $ab>a$ it would follow that $zeta(ab)/zeta(a)<1,$ which cannot be $zeta(b)>1.$ So $zeta(n)$ is not multiplicative for any values of $a,b>1,$ coprime or not.






share|cite|improve this answer











$endgroup$



The term "multiplicative" is used for a function $f$ defined on the positive integers, for which $f(mn)=f(m)f(n)$ whenever $gcd(m,n)=1.$ The Riemann zeta function $zeta$ isn't even defined at $1$ so is not a candidate to be called multiplicative. Even ignoring the problem at $1$ it could be checked numerically that $zeta(2)zeta(3) neq zeta(6)$.



Added: No need for a numerical check. Exact values for $zeta(2k)$ are known, and $zeta(6)/zeta(2)<0.7.$ So it can't equal $zeta(3),$ which is greater than $1.$



Another note: $zeta(n)$is strictly monotone decreasing as $n>1$ increases. So if for $a,b$ with $1 <a,b$ we had $zeta(ab)=zeta(a)zeta(b),$ then since $ab>a$ it would follow that $zeta(ab)/zeta(a)<1,$ which cannot be $zeta(b)>1.$ So $zeta(n)$ is not multiplicative for any values of $a,b>1,$ coprime or not.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 8 at 21:16

























answered Jan 8 at 12:12









coffeemathcoffeemath

2,7361415




2,7361415








  • 1




    $begingroup$
    Thank you for your answer. I hope you don't mind my minor edit.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 8 at 12:18














  • 1




    $begingroup$
    Thank you for your answer. I hope you don't mind my minor edit.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 8 at 12:18








1




1




$begingroup$
Thank you for your answer. I hope you don't mind my minor edit.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 8 at 12:18




$begingroup$
Thank you for your answer. I hope you don't mind my minor edit.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 8 at 12:18


















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