Proof of infinite limit












1












$begingroup$


My question is really as to whether I can consider the following results as proof of one another, since $(2a)$&$(2b)$ cannot be true unless $(0)$ is true, and vice versa.



So if for below I prove $(0)$ using the elementary means in Real Analysis, can I also then consider this to be proof for $(2a)$&$(2b)$?



I am looking to prove with the standard epsilon approach that:



$$x in mathbb R Rightarrowlim _{nrightarrow infty }Bigl({frac { lfloor nx rfloor }{n}
}Bigr)=x
$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(0)}$$



Which has become necessary in the following context:



It's well known that in the p-adic coefficient of a natural number:
$$d_{n,p,j}=
Bigllfloor{n,{p}^{,j- {lfloor
ln_p(n)rfloor} -1}} Bigrrfloor-pBigllfloor{n,{p}^{,j- {lfloor
ln_p(n)rfloor} -2}} Bigrrfloor$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(1)}$$
as seen in the p-adic expansion of that number:
$$n=sum_{j=1}^{lfloor
ln_p(n)rfloor+1}d_{n,p,j},p^{,lfloor
ln_p(n)rfloor+1-j}quadforall n in mathbb N$$



can likewise be used to separate any positive real number into it's integer and fractional parts:



$$lfloor xrfloor=sum_{j=1}^{lfloor
ln_p(x)rfloor+1}d_{x,p,j},p^{,lfloor
ln_p(x)rfloor+1-j}quadforall x in mathbb R^{+}$$



$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(2a)}$$
$${{x}}=sum_{j=lfloor
ln_p(x)rfloor+2}^{infty}d_{x,p,j},p^{,lfloor
ln_p(x)rfloor+1-j}quadforall x in mathbb R^{+}$$



$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(2b)}$$



And because the corresponding finite sum for the above always computes a value much like the one I am trying to prove:
$$sum_{j=1}^{n}d_{x,p,j},p^{,lfloor
ln_p(x)rfloor+1-j}=frac{lfloor Nxrfloor}{N}$$

Then if it is true that taking the limit $n rightarrow infty$ results in convergence to $x$ in the manner described above, this means that proving $(0)$ to be true also concurrently proves $(2a)$ & $(2b)$ to be true.










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The simplest proof of (0) starts by noting that for all real $r$, $r-1<lfloor rrfloorle r$.
    $endgroup$
    – Gerry Myerson
    Dec 30 '18 at 2:45










  • $begingroup$
    Sure but more to the point of what I am asking, can the final remark I've made being that the sum of $(2a)$ and $(2b)$ in the infinite limit for the upper bound being the same form as $(0)$ allow me to state QED for those having proven $(0)$ in an elementary manner as you stated, or do I need to prove them separately?
    $endgroup$
    – Adam
    Dec 30 '18 at 2:58


















1












$begingroup$


My question is really as to whether I can consider the following results as proof of one another, since $(2a)$&$(2b)$ cannot be true unless $(0)$ is true, and vice versa.



So if for below I prove $(0)$ using the elementary means in Real Analysis, can I also then consider this to be proof for $(2a)$&$(2b)$?



I am looking to prove with the standard epsilon approach that:



$$x in mathbb R Rightarrowlim _{nrightarrow infty }Bigl({frac { lfloor nx rfloor }{n}
}Bigr)=x
$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(0)}$$



Which has become necessary in the following context:



It's well known that in the p-adic coefficient of a natural number:
$$d_{n,p,j}=
Bigllfloor{n,{p}^{,j- {lfloor
ln_p(n)rfloor} -1}} Bigrrfloor-pBigllfloor{n,{p}^{,j- {lfloor
ln_p(n)rfloor} -2}} Bigrrfloor$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(1)}$$
as seen in the p-adic expansion of that number:
$$n=sum_{j=1}^{lfloor
ln_p(n)rfloor+1}d_{n,p,j},p^{,lfloor
ln_p(n)rfloor+1-j}quadforall n in mathbb N$$



can likewise be used to separate any positive real number into it's integer and fractional parts:



$$lfloor xrfloor=sum_{j=1}^{lfloor
ln_p(x)rfloor+1}d_{x,p,j},p^{,lfloor
ln_p(x)rfloor+1-j}quadforall x in mathbb R^{+}$$



$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(2a)}$$
$${{x}}=sum_{j=lfloor
ln_p(x)rfloor+2}^{infty}d_{x,p,j},p^{,lfloor
ln_p(x)rfloor+1-j}quadforall x in mathbb R^{+}$$



$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(2b)}$$



And because the corresponding finite sum for the above always computes a value much like the one I am trying to prove:
$$sum_{j=1}^{n}d_{x,p,j},p^{,lfloor
ln_p(x)rfloor+1-j}=frac{lfloor Nxrfloor}{N}$$

Then if it is true that taking the limit $n rightarrow infty$ results in convergence to $x$ in the manner described above, this means that proving $(0)$ to be true also concurrently proves $(2a)$ & $(2b)$ to be true.










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The simplest proof of (0) starts by noting that for all real $r$, $r-1<lfloor rrfloorle r$.
    $endgroup$
    – Gerry Myerson
    Dec 30 '18 at 2:45










  • $begingroup$
    Sure but more to the point of what I am asking, can the final remark I've made being that the sum of $(2a)$ and $(2b)$ in the infinite limit for the upper bound being the same form as $(0)$ allow me to state QED for those having proven $(0)$ in an elementary manner as you stated, or do I need to prove them separately?
    $endgroup$
    – Adam
    Dec 30 '18 at 2:58
















1












1








1





$begingroup$


My question is really as to whether I can consider the following results as proof of one another, since $(2a)$&$(2b)$ cannot be true unless $(0)$ is true, and vice versa.



So if for below I prove $(0)$ using the elementary means in Real Analysis, can I also then consider this to be proof for $(2a)$&$(2b)$?



I am looking to prove with the standard epsilon approach that:



$$x in mathbb R Rightarrowlim _{nrightarrow infty }Bigl({frac { lfloor nx rfloor }{n}
}Bigr)=x
$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(0)}$$



Which has become necessary in the following context:



It's well known that in the p-adic coefficient of a natural number:
$$d_{n,p,j}=
Bigllfloor{n,{p}^{,j- {lfloor
ln_p(n)rfloor} -1}} Bigrrfloor-pBigllfloor{n,{p}^{,j- {lfloor
ln_p(n)rfloor} -2}} Bigrrfloor$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(1)}$$
as seen in the p-adic expansion of that number:
$$n=sum_{j=1}^{lfloor
ln_p(n)rfloor+1}d_{n,p,j},p^{,lfloor
ln_p(n)rfloor+1-j}quadforall n in mathbb N$$



can likewise be used to separate any positive real number into it's integer and fractional parts:



$$lfloor xrfloor=sum_{j=1}^{lfloor
ln_p(x)rfloor+1}d_{x,p,j},p^{,lfloor
ln_p(x)rfloor+1-j}quadforall x in mathbb R^{+}$$



$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(2a)}$$
$${{x}}=sum_{j=lfloor
ln_p(x)rfloor+2}^{infty}d_{x,p,j},p^{,lfloor
ln_p(x)rfloor+1-j}quadforall x in mathbb R^{+}$$



$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(2b)}$$



And because the corresponding finite sum for the above always computes a value much like the one I am trying to prove:
$$sum_{j=1}^{n}d_{x,p,j},p^{,lfloor
ln_p(x)rfloor+1-j}=frac{lfloor Nxrfloor}{N}$$

Then if it is true that taking the limit $n rightarrow infty$ results in convergence to $x$ in the manner described above, this means that proving $(0)$ to be true also concurrently proves $(2a)$ & $(2b)$ to be true.










share|cite|improve this question











$endgroup$




My question is really as to whether I can consider the following results as proof of one another, since $(2a)$&$(2b)$ cannot be true unless $(0)$ is true, and vice versa.



So if for below I prove $(0)$ using the elementary means in Real Analysis, can I also then consider this to be proof for $(2a)$&$(2b)$?



I am looking to prove with the standard epsilon approach that:



$$x in mathbb R Rightarrowlim _{nrightarrow infty }Bigl({frac { lfloor nx rfloor }{n}
}Bigr)=x
$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(0)}$$



Which has become necessary in the following context:



It's well known that in the p-adic coefficient of a natural number:
$$d_{n,p,j}=
Bigllfloor{n,{p}^{,j- {lfloor
ln_p(n)rfloor} -1}} Bigrrfloor-pBigllfloor{n,{p}^{,j- {lfloor
ln_p(n)rfloor} -2}} Bigrrfloor$$

$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(1)}$$
as seen in the p-adic expansion of that number:
$$n=sum_{j=1}^{lfloor
ln_p(n)rfloor+1}d_{n,p,j},p^{,lfloor
ln_p(n)rfloor+1-j}quadforall n in mathbb N$$



can likewise be used to separate any positive real number into it's integer and fractional parts:



$$lfloor xrfloor=sum_{j=1}^{lfloor
ln_p(x)rfloor+1}d_{x,p,j},p^{,lfloor
ln_p(x)rfloor+1-j}quadforall x in mathbb R^{+}$$



$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(2a)}$$
$${{x}}=sum_{j=lfloor
ln_p(x)rfloor+2}^{infty}d_{x,p,j},p^{,lfloor
ln_p(x)rfloor+1-j}quadforall x in mathbb R^{+}$$



$$quadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadquadoperatorname{(2b)}$$



And because the corresponding finite sum for the above always computes a value much like the one I am trying to prove:
$$sum_{j=1}^{n}d_{x,p,j},p^{,lfloor
ln_p(x)rfloor+1-j}=frac{lfloor Nxrfloor}{N}$$

Then if it is true that taking the limit $n rightarrow infty$ results in convergence to $x$ in the manner described above, this means that proving $(0)$ to be true also concurrently proves $(2a)$ & $(2b)$ to be true.







number-theory limits p-adic-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 11:48







Adam

















asked Dec 30 '18 at 2:32









AdamAdam

560114




560114








  • 2




    $begingroup$
    The simplest proof of (0) starts by noting that for all real $r$, $r-1<lfloor rrfloorle r$.
    $endgroup$
    – Gerry Myerson
    Dec 30 '18 at 2:45










  • $begingroup$
    Sure but more to the point of what I am asking, can the final remark I've made being that the sum of $(2a)$ and $(2b)$ in the infinite limit for the upper bound being the same form as $(0)$ allow me to state QED for those having proven $(0)$ in an elementary manner as you stated, or do I need to prove them separately?
    $endgroup$
    – Adam
    Dec 30 '18 at 2:58
















  • 2




    $begingroup$
    The simplest proof of (0) starts by noting that for all real $r$, $r-1<lfloor rrfloorle r$.
    $endgroup$
    – Gerry Myerson
    Dec 30 '18 at 2:45










  • $begingroup$
    Sure but more to the point of what I am asking, can the final remark I've made being that the sum of $(2a)$ and $(2b)$ in the infinite limit for the upper bound being the same form as $(0)$ allow me to state QED for those having proven $(0)$ in an elementary manner as you stated, or do I need to prove them separately?
    $endgroup$
    – Adam
    Dec 30 '18 at 2:58










2




2




$begingroup$
The simplest proof of (0) starts by noting that for all real $r$, $r-1<lfloor rrfloorle r$.
$endgroup$
– Gerry Myerson
Dec 30 '18 at 2:45




$begingroup$
The simplest proof of (0) starts by noting that for all real $r$, $r-1<lfloor rrfloorle r$.
$endgroup$
– Gerry Myerson
Dec 30 '18 at 2:45












$begingroup$
Sure but more to the point of what I am asking, can the final remark I've made being that the sum of $(2a)$ and $(2b)$ in the infinite limit for the upper bound being the same form as $(0)$ allow me to state QED for those having proven $(0)$ in an elementary manner as you stated, or do I need to prove them separately?
$endgroup$
– Adam
Dec 30 '18 at 2:58






$begingroup$
Sure but more to the point of what I am asking, can the final remark I've made being that the sum of $(2a)$ and $(2b)$ in the infinite limit for the upper bound being the same form as $(0)$ allow me to state QED for those having proven $(0)$ in an elementary manner as you stated, or do I need to prove them separately?
$endgroup$
– Adam
Dec 30 '18 at 2:58












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