The last digit of pi (in terms of Banach limits)
Let $phi : l^infty to mathbb C$ be a Banach limit, and define the sequence ${x_k}_{kgeq 0}$ to be the digits in the 10-base decimal expansion of $pi$. Note that
$${x_k}_{kgeq 0} in l^infty$$
and so we can talk about $phi({x_k}_{kgeq 0})$.
What it is? Note that Banach limits don't have to be unique.
Now consider the real number $sqrt 2$. What is its last number? Finally, consider any element $xin mathbb R$. Can we say about the last digit of $x$, in the sense of Banach limits as considered above?
sequences-and-series functional-analysis limits banach-spaces
asked Nov 27 '18 at 18:15
Markus Klyver
390314
add a comment |
Let $phi : l^infty to mathbb C$ be a Banach limit, and define the sequence ${x_k}_{kgeq 0}$ to be the digits in the 10-base decimal expansion of $pi$. Note that
$${x_k}_{kgeq 0} in l^infty$$
and so we can talk about $phi({x_k}_{kgeq 0})$.
What it is? Note that Banach limits don't have to be unique.
Now consider the real number $sqrt 2$. What is its last number? Finally, consider any element $xin mathbb R$. Can we say about the last digit of $x$, in the sense of Banach limits as considered above?
sequences-and-series functional-analysis limits banach-spaces
asked Nov 27 '18 at 18:15
Markus Klyver
390314
2
Perhaps this is naïve, but the shift invariance property might tell you that $phi$ is just a long-term average of the digits (and so it's $4.5$ for almost any real number).
– T. Bongers
Nov 27 '18 at 18:18
1
@T.Bongers looking at en.m.wikipedia.org/wiki/Almost_convergent_sequence I'd say the opposite is true: for almost all real nmbers there is no unique Banach limit
– Bananach
Nov 27 '18 at 19:20
1
Chuck Norris knows the last digit of pi.
– gerw
Nov 28 '18 at 7:58
1
@gerw Well, Gauß knew it before him!
– Dirk
Nov 29 '18 at 8:33
add a comment |
Let $phi : l^infty to mathbb C$ be a Banach limit, and define the sequence ${x_k}_{kgeq 0}$ to be the digits in the 10-base decimal expansion of $pi$. Note that
$${x_k}_{kgeq 0} in l^infty$$
and so we can talk about $phi({x_k}_{kgeq 0})$.
What it is? Note that Banach limits don't have to be unique.
Now consider the real number $sqrt 2$. What is its last number? Finally, consider any element $xin mathbb R$. Can we say about the last digit of $x$, in the sense of Banach limits as considered above?
sequences-and-series functional-analysis limits banach-spaces
asked Nov 27 '18 at 18:15
Markus Klyver
390314
Let $phi : l^infty to mathbb C$ be a Banach limit, and define the sequence ${x_k}_{kgeq 0}$ to be the digits in the 10-base decimal expansion of $pi$. Note that
$${x_k}_{kgeq 0} in l^infty$$
and so we can talk about $phi({x_k}_{kgeq 0})$.
What it is? Note that Banach limits don't have to be unique.
Now consider the real number $sqrt 2$. What is its last number? Finally, consider any element $xin mathbb R$. Can we say about the last digit of $x$, in the sense of Banach limits as considered above?
sequences-and-series functional-analysis limits banach-spaces
sequences-and-series functional-analysis limits banach-spaces
asked Nov 27 '18 at 18:15
Markus Klyver
390314
asked Nov 27 '18 at 18:15
Markus Klyver
390314
edited Nov 29 '18 at 9:12
asked Nov 27 '18 at 18:15
Markus Klyver
390314
asked Nov 27 '18 at 18:15
Markus Klyver
390314
asked Nov 27 '18 at 18:15
Markus Klyver
390314
390314
2
Perhaps this is naïve, but the shift invariance property might tell you that $phi$ is just a long-term average of the digits (and so it's $4.5$ for almost any real number).
– T. Bongers
Nov 27 '18 at 18:18
1
@T.Bongers looking at en.m.wikipedia.org/wiki/Almost_convergent_sequence I'd say the opposite is true: for almost all real nmbers there is no unique Banach limit
– Bananach
Nov 27 '18 at 19:20
1
Chuck Norris knows the last digit of pi.
– gerw
Nov 28 '18 at 7:58
1
@gerw Well, Gauß knew it before him!
– Dirk
Nov 29 '18 at 8:33
add a comment |
2
Perhaps this is naïve, but the shift invariance property might tell you that $phi$ is just a long-term average of the digits (and so it's $4.5$ for almost any real number).
– T. Bongers
Nov 27 '18 at 18:18
1
@T.Bongers looking at en.m.wikipedia.org/wiki/Almost_convergent_sequence I'd say the opposite is true: for almost all real nmbers there is no unique Banach limit
– Bananach
Nov 27 '18 at 19:20
1
Chuck Norris knows the last digit of pi.
– gerw
Nov 28 '18 at 7:58
1
@gerw Well, Gauß knew it before him!
– Dirk
Nov 29 '18 at 8:33
2
2
Perhaps this is naïve, but the shift invariance property might tell you that $phi$ is just a long-term average of the digits (and so it's $4.5$ for almost any real number).
– T. Bongers
Nov 27 '18 at 18:18
Perhaps this is naïve, but the shift invariance property might tell you that $phi$ is just a long-term average of the digits (and so it's $4.5$ for almost any real number).
– T. Bongers
Nov 27 '18 at 18:18
1
1
@T.Bongers looking at en.m.wikipedia.org/wiki/Almost_convergent_sequence I'd say the opposite is true: for almost all real nmbers there is no unique Banach limit
– Bananach
Nov 27 '18 at 19:20
@T.Bongers looking at en.m.wikipedia.org/wiki/Almost_convergent_sequence I'd say the opposite is true: for almost all real nmbers there is no unique Banach limit
– Bananach
Nov 27 '18 at 19:20
1
1
Chuck Norris knows the last digit of pi.
– gerw
Nov 28 '18 at 7:58
Chuck Norris knows the last digit of pi.
– gerw
Nov 28 '18 at 7:58
1
1
@gerw Well, Gauß knew it before him!
– Dirk
Nov 29 '18 at 8:33
@gerw Well, Gauß knew it before him!
– Dirk
Nov 29 '18 at 8:33
add a comment |
1 Answer
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Concerning last question about general $xinmathbb{R}$ and concerning the twice upvoted comment (under the original question) that claims the limit is 4.5 for almost all $x$:
I think the opposite is true (see proof below): almost all (in the sense of the Lebesgue measure) real numbers do not have a unique Banach limit. Unless $pi$ is an exception to this (and I don't know whether it is), your first question does not have a well-defined answer: The 'last digit' of $pi$ is different for different choices of the Banach limit $phi$.
Proof of claim: According to a theorem of Lorentz, a sequence $(x_k)_{k=0}^{infty}$ has a unique Banach limit $Linmathbb{R}$ if and only if its averages $overline{x}_{n,p}:=frac{x_n+dots+x_{n+p-1}}{p}$ converge to $L$ as $ptoinfty$, uniformly in $n$. In formulas $forall epsilon>0: exists p_0inmathbb{N}: forall pgeq p_0: forall ninmathbb{N}:|overline{x}_{n,p}-L|<epsilon$. In particular, just by rewriting this definition in set form and choosing $epsilon:=1$, this implies that $xin bigcap_{minmathbb{N}}A_{p_0,m}$ for some $p_0inmathbb{N}$, where $A_{p_0,m}:={x:|overline{x}_{mp_0,p_0}-L|<1}$.
If we assume for simplicity that $(x_k)_{k=0}^{infty}$ are the digits of $xin[0,1]$, the events $(A_{p_0,m})_{minmathbb{N}}$ are independent under the Lebesgue probability measure on $[0,1]$ (since the digits themselves are independent random variables) and have probability less than $1$. This shows $P(bigcap_{minmathbb{N}}A_{p_0,m})=0$ for all $p_0inmathbb{N}$, or, if we denote by $B_L$ the reals with unique Banach limit $L$, that $P(B_L)=0$ for all $Linmathbb{R}$. Since $mathbb{R}$ is uncountable, it could happen that $P(bigcup_{L} B_L)>0$. However, this is not the case since $P(bigcup_{Lnot= 4.5}B_L)=0$ by the law of large numbers.
answered Nov 29 '18 at 8:02
Bananach
3,74111229
add a comment |
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Concerning last question about general $xinmathbb{R}$ and concerning the twice upvoted comment (under the original question) that claims the limit is 4.5 for almost all $x$:
I think the opposite is true (see proof below): almost all (in the sense of the Lebesgue measure) real numbers do not have a unique Banach limit. Unless $pi$ is an exception to this (and I don't know whether it is), your first question does not have a well-defined answer: The 'last digit' of $pi$ is different for different choices of the Banach limit $phi$.
Proof of claim: According to a theorem of Lorentz, a sequence $(x_k)_{k=0}^{infty}$ has a unique Banach limit $Linmathbb{R}$ if and only if its averages $overline{x}_{n,p}:=frac{x_n+dots+x_{n+p-1}}{p}$ converge to $L$ as $ptoinfty$, uniformly in $n$. In formulas $forall epsilon>0: exists p_0inmathbb{N}: forall pgeq p_0: forall ninmathbb{N}:|overline{x}_{n,p}-L|<epsilon$. In particular, just by rewriting this definition in set form and choosing $epsilon:=1$, this implies that $xin bigcap_{minmathbb{N}}A_{p_0,m}$ for some $p_0inmathbb{N}$, where $A_{p_0,m}:={x:|overline{x}_{mp_0,p_0}-L|<1}$.
If we assume for simplicity that $(x_k)_{k=0}^{infty}$ are the digits of $xin[0,1]$, the events $(A_{p_0,m})_{minmathbb{N}}$ are independent under the Lebesgue probability measure on $[0,1]$ (since the digits themselves are independent random variables) and have probability less than $1$. This shows $P(bigcap_{minmathbb{N}}A_{p_0,m})=0$ for all $p_0inmathbb{N}$, or, if we denote by $B_L$ the reals with unique Banach limit $L$, that $P(B_L)=0$ for all $Linmathbb{R}$. Since $mathbb{R}$ is uncountable, it could happen that $P(bigcup_{L} B_L)>0$. However, this is not the case since $P(bigcup_{Lnot= 4.5}B_L)=0$ by the law of large numbers.
answered Nov 29 '18 at 8:02
Bananach
3,74111229
add a comment |
Concerning last question about general $xinmathbb{R}$ and concerning the twice upvoted comment (under the original question) that claims the limit is 4.5 for almost all $x$:
I think the opposite is true (see proof below): almost all (in the sense of the Lebesgue measure) real numbers do not have a unique Banach limit. Unless $pi$ is an exception to this (and I don't know whether it is), your first question does not have a well-defined answer: The 'last digit' of $pi$ is different for different choices of the Banach limit $phi$.
Proof of claim: According to a theorem of Lorentz, a sequence $(x_k)_{k=0}^{infty}$ has a unique Banach limit $Linmathbb{R}$ if and only if its averages $overline{x}_{n,p}:=frac{x_n+dots+x_{n+p-1}}{p}$ converge to $L$ as $ptoinfty$, uniformly in $n$. In formulas $forall epsilon>0: exists p_0inmathbb{N}: forall pgeq p_0: forall ninmathbb{N}:|overline{x}_{n,p}-L|<epsilon$. In particular, just by rewriting this definition in set form and choosing $epsilon:=1$, this implies that $xin bigcap_{minmathbb{N}}A_{p_0,m}$ for some $p_0inmathbb{N}$, where $A_{p_0,m}:={x:|overline{x}_{mp_0,p_0}-L|<1}$.
If we assume for simplicity that $(x_k)_{k=0}^{infty}$ are the digits of $xin[0,1]$, the events $(A_{p_0,m})_{minmathbb{N}}$ are independent under the Lebesgue probability measure on $[0,1]$ (since the digits themselves are independent random variables) and have probability less than $1$. This shows $P(bigcap_{minmathbb{N}}A_{p_0,m})=0$ for all $p_0inmathbb{N}$, or, if we denote by $B_L$ the reals with unique Banach limit $L$, that $P(B_L)=0$ for all $Linmathbb{R}$. Since $mathbb{R}$ is uncountable, it could happen that $P(bigcup_{L} B_L)>0$. However, this is not the case since $P(bigcup_{Lnot= 4.5}B_L)=0$ by the law of large numbers.
answered Nov 29 '18 at 8:02
Bananach
3,74111229
add a comment |
Concerning last question about general $xinmathbb{R}$ and concerning the twice upvoted comment (under the original question) that claims the limit is 4.5 for almost all $x$:
I think the opposite is true (see proof below): almost all (in the sense of the Lebesgue measure) real numbers do not have a unique Banach limit. Unless $pi$ is an exception to this (and I don't know whether it is), your first question does not have a well-defined answer: The 'last digit' of $pi$ is different for different choices of the Banach limit $phi$.
Proof of claim: According to a theorem of Lorentz, a sequence $(x_k)_{k=0}^{infty}$ has a unique Banach limit $Linmathbb{R}$ if and only if its averages $overline{x}_{n,p}:=frac{x_n+dots+x_{n+p-1}}{p}$ converge to $L$ as $ptoinfty$, uniformly in $n$. In formulas $forall epsilon>0: exists p_0inmathbb{N}: forall pgeq p_0: forall ninmathbb{N}:|overline{x}_{n,p}-L|<epsilon$. In particular, just by rewriting this definition in set form and choosing $epsilon:=1$, this implies that $xin bigcap_{minmathbb{N}}A_{p_0,m}$ for some $p_0inmathbb{N}$, where $A_{p_0,m}:={x:|overline{x}_{mp_0,p_0}-L|<1}$.
If we assume for simplicity that $(x_k)_{k=0}^{infty}$ are the digits of $xin[0,1]$, the events $(A_{p_0,m})_{minmathbb{N}}$ are independent under the Lebesgue probability measure on $[0,1]$ (since the digits themselves are independent random variables) and have probability less than $1$. This shows $P(bigcap_{minmathbb{N}}A_{p_0,m})=0$ for all $p_0inmathbb{N}$, or, if we denote by $B_L$ the reals with unique Banach limit $L$, that $P(B_L)=0$ for all $Linmathbb{R}$. Since $mathbb{R}$ is uncountable, it could happen that $P(bigcup_{L} B_L)>0$. However, this is not the case since $P(bigcup_{Lnot= 4.5}B_L)=0$ by the law of large numbers.
answered Nov 29 '18 at 8:02
Bananach
3,74111229
Concerning last question about general $xinmathbb{R}$ and concerning the twice upvoted comment (under the original question) that claims the limit is 4.5 for almost all $x$:
I think the opposite is true (see proof below): almost all (in the sense of the Lebesgue measure) real numbers do not have a unique Banach limit. Unless $pi$ is an exception to this (and I don't know whether it is), your first question does not have a well-defined answer: The 'last digit' of $pi$ is different for different choices of the Banach limit $phi$.
Proof of claim: According to a theorem of Lorentz, a sequence $(x_k)_{k=0}^{infty}$ has a unique Banach limit $Linmathbb{R}$ if and only if its averages $overline{x}_{n,p}:=frac{x_n+dots+x_{n+p-1}}{p}$ converge to $L$ as $ptoinfty$, uniformly in $n$. In formulas $forall epsilon>0: exists p_0inmathbb{N}: forall pgeq p_0: forall ninmathbb{N}:|overline{x}_{n,p}-L|<epsilon$. In particular, just by rewriting this definition in set form and choosing $epsilon:=1$, this implies that $xin bigcap_{minmathbb{N}}A_{p_0,m}$ for some $p_0inmathbb{N}$, where $A_{p_0,m}:={x:|overline{x}_{mp_0,p_0}-L|<1}$.
If we assume for simplicity that $(x_k)_{k=0}^{infty}$ are the digits of $xin[0,1]$, the events $(A_{p_0,m})_{minmathbb{N}}$ are independent under the Lebesgue probability measure on $[0,1]$ (since the digits themselves are independent random variables) and have probability less than $1$. This shows $P(bigcap_{minmathbb{N}}A_{p_0,m})=0$ for all $p_0inmathbb{N}$, or, if we denote by $B_L$ the reals with unique Banach limit $L$, that $P(B_L)=0$ for all $Linmathbb{R}$. Since $mathbb{R}$ is uncountable, it could happen that $P(bigcup_{L} B_L)>0$. However, this is not the case since $P(bigcup_{Lnot= 4.5}B_L)=0$ by the law of large numbers.
answered Nov 29 '18 at 8:02
Bananach
3,74111229
answered Nov 29 '18 at 8:02
Bananach
3,74111229
answered Nov 29 '18 at 8:02
Bananach
3,74111229
answered Nov 29 '18 at 8:02
Bananach
3,74111229
3,74111229
add a comment |
add a comment |
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2
Perhaps this is naïve, but the shift invariance property might tell you that $phi$ is just a long-term average of the digits (and so it's $4.5$ for almost any real number).
– T. Bongers
Nov 27 '18 at 18:18
1
@T.Bongers looking at en.m.wikipedia.org/wiki/Almost_convergent_sequence I'd say the opposite is true: for almost all real nmbers there is no unique Banach limit
– Bananach
Nov 27 '18 at 19:20
1
Chuck Norris knows the last digit of pi.
– gerw
Nov 28 '18 at 7:58
1
@gerw Well, Gauß knew it before him!
– Dirk
Nov 29 '18 at 8:33