Mixing
Show that...
$begingroup$
I have a hard time to understand a certain part of the proof below
Let $e_n:=(1+frac{1}{n})^n$ and $s_n:=sum_{v=0}^{n}frac{x^v}{v!}$
According to the binomial Theorem we have
$$e_n=1+ncdotfrac{1}{n}+frac{n(n-1)}{2!}cdotfrac{1}{n^2}+...+frac{n(n-1)(n-2)...1}{n^n}$$
$$=1+1+frac{1}{2!}(1-frac{1}{n})+...+frac{1}{n!}(1-frac{1}{n})(1-frac{2}{n})...(1-frac{n-1}{n})$$
$$<s_n(1)$$
this gives us
$$e=lim_{nrightarrowinfty}e_nleqlim_{nrightarrowinfty}s_n(1)=exp(1)$$
What is the justification for the $leq$ symbol?
I know the limits exists, the left Limit was proved by constructing nested intervals $I_n=[(1+frac{1}{n})^n,(1+frac{1}{n})^{n+1}]$, the right Limit exists because of the quotient criterion.
The proof continues with:
Let $k>n$
Then
$$e_k>1+1+frac{1}{2!}(1-frac{1}{k})+...+frac{1}{n!}(1-frac{1}{k})(1-frac{2}{k})...(1-frac{n-1}{k})(*)$$
If we fixate a $n$ and let $k$ approach infinity we get $egeq s_n(1)$ (Why?) which gives us
$$egeq lim_{nrightarrowinfty}s_n(1)$$
What was the logic of fixating a $k$ and why is it still the same Limit after the Fixation and why are we allowed to do that ?
Please shed some light on those equations.
exponential-function proof-explanation
asked Jan 23 at 19:16
New2MathNew2Math
11812
$endgroup$
add a comment |
$begingroup$
I have a hard time to understand a certain part of the proof below
Let $e_n:=(1+frac{1}{n})^n$ and $s_n:=sum_{v=0}^{n}frac{x^v}{v!}$
According to the binomial Theorem we have
$$e_n=1+ncdotfrac{1}{n}+frac{n(n-1)}{2!}cdotfrac{1}{n^2}+...+frac{n(n-1)(n-2)...1}{n^n}$$
$$=1+1+frac{1}{2!}(1-frac{1}{n})+...+frac{1}{n!}(1-frac{1}{n})(1-frac{2}{n})...(1-frac{n-1}{n})$$
$$<s_n(1)$$
this gives us
$$e=lim_{nrightarrowinfty}e_nleqlim_{nrightarrowinfty}s_n(1)=exp(1)$$
What is the justification for the $leq$ symbol?
I know the limits exists, the left Limit was proved by constructing nested intervals $I_n=[(1+frac{1}{n})^n,(1+frac{1}{n})^{n+1}]$, the right Limit exists because of the quotient criterion.
The proof continues with:
Let $k>n$
Then
$$e_k>1+1+frac{1}{2!}(1-frac{1}{k})+...+frac{1}{n!}(1-frac{1}{k})(1-frac{2}{k})...(1-frac{n-1}{k})(*)$$
If we fixate a $n$ and let $k$ approach infinity we get $egeq s_n(1)$ (Why?) which gives us
$$egeq lim_{nrightarrowinfty}s_n(1)$$
What was the logic of fixating a $k$ and why is it still the same Limit after the Fixation and why are we allowed to do that ?
Please shed some light on those equations.
exponential-function proof-explanation
asked Jan 23 at 19:16
New2MathNew2Math
11812
$endgroup$
$begingroup$
It's because $e_n<s_n(1)$.
$endgroup$
– Lord Shark the Unknown
Jan 23 at 19:18
$begingroup$
For the second question, the lower bound is obtained by taking only the first $(n+1)$ terms of the expansion of $e_k$ and dropping the rest. Since all the terms appearing in the binomial expansion of $e_k$ is positive, dropping some terms will only lower the value.
$endgroup$
– Sangchul Lee
Jan 23 at 19:21
$begingroup$
@SangchulLee I think I have understood a little bit now correct me if I am wrong but the answer to my first Question is actually a rule for limits. And to my second question the Supremum of the Right side of (*) is $s_n(1)$ therefor the smallest upperbound, and because $e_k$ is also an upper Bound we have $egeq e_kgeq s_n(1)$ but I still dont know why $egeq lim_{nrightarrowinfty}s_n(1)$ what is the explaination for that?
$endgroup$
– New2Math
Jan 23 at 19:44
$begingroup$
Okay I think I got it now because $forall nin mathbb{N}:e>s_n(1)$. We know that $e$ is an upper bound. Because $s_n(1)$ is monotonously rising the Limit of $s_n(1)$ also must be the Supremum ,i.e. the smallest upperbound that's why $egeq lim_{nrightarrowinfty}s_n(1)$
$endgroup$
– New2Math
Jan 23 at 20:01
1
$begingroup$
For a more transparent way of explaining the second step is that, taking limit as $ktoinfty$ to the second step, begin{align*}e=lim_{ktoinfty} e_k&geq lim_{ktoinfty} sum_{p=0}^{n} frac{1}{p!}left(1-frac{1}{k}right)cdotsleft(1-frac{p-1}{k}right)\&=sum_{p=0}^{n} lim_{ktoinfty} frac{1}{p!}left(1-frac{1}{k}right)cdotsleft(1-frac{p-1}{k}right)=sum_{p=0}^{n}frac{1}{p!}=s_n(1).end{align*} Now this inequality gives an upper bound of the sequence $(s_n(1))_{n=1}^{infty}$, and so, taking $ntoinfty$ shows $$egeqlim_m s_n(1).$$
$endgroup$
– Sangchul Lee
Jan 23 at 20:08
add a comment |
$begingroup$
I have a hard time to understand a certain part of the proof below
Let $e_n:=(1+frac{1}{n})^n$ and $s_n:=sum_{v=0}^{n}frac{x^v}{v!}$
According to the binomial Theorem we have
$$e_n=1+ncdotfrac{1}{n}+frac{n(n-1)}{2!}cdotfrac{1}{n^2}+...+frac{n(n-1)(n-2)...1}{n^n}$$
$$=1+1+frac{1}{2!}(1-frac{1}{n})+...+frac{1}{n!}(1-frac{1}{n})(1-frac{2}{n})...(1-frac{n-1}{n})$$
$$<s_n(1)$$
this gives us
$$e=lim_{nrightarrowinfty}e_nleqlim_{nrightarrowinfty}s_n(1)=exp(1)$$
What is the justification for the $leq$ symbol?
I know the limits exists, the left Limit was proved by constructing nested intervals $I_n=[(1+frac{1}{n})^n,(1+frac{1}{n})^{n+1}]$, the right Limit exists because of the quotient criterion.
The proof continues with:
Let $k>n$
Then
$$e_k>1+1+frac{1}{2!}(1-frac{1}{k})+...+frac{1}{n!}(1-frac{1}{k})(1-frac{2}{k})...(1-frac{n-1}{k})(*)$$
If we fixate a $n$ and let $k$ approach infinity we get $egeq s_n(1)$ (Why?) which gives us
$$egeq lim_{nrightarrowinfty}s_n(1)$$
What was the logic of fixating a $k$ and why is it still the same Limit after the Fixation and why are we allowed to do that ?
Please shed some light on those equations.
exponential-function proof-explanation
asked Jan 23 at 19:16
New2MathNew2Math
11812
$endgroup$
I have a hard time to understand a certain part of the proof below
Let $e_n:=(1+frac{1}{n})^n$ and $s_n:=sum_{v=0}^{n}frac{x^v}{v!}$
According to the binomial Theorem we have
$$e_n=1+ncdotfrac{1}{n}+frac{n(n-1)}{2!}cdotfrac{1}{n^2}+...+frac{n(n-1)(n-2)...1}{n^n}$$
$$=1+1+frac{1}{2!}(1-frac{1}{n})+...+frac{1}{n!}(1-frac{1}{n})(1-frac{2}{n})...(1-frac{n-1}{n})$$
$$<s_n(1)$$
this gives us
$$e=lim_{nrightarrowinfty}e_nleqlim_{nrightarrowinfty}s_n(1)=exp(1)$$
What is the justification for the $leq$ symbol?
I know the limits exists, the left Limit was proved by constructing nested intervals $I_n=[(1+frac{1}{n})^n,(1+frac{1}{n})^{n+1}]$, the right Limit exists because of the quotient criterion.
The proof continues with:
Let $k>n$
Then
$$e_k>1+1+frac{1}{2!}(1-frac{1}{k})+...+frac{1}{n!}(1-frac{1}{k})(1-frac{2}{k})...(1-frac{n-1}{k})(*)$$
If we fixate a $n$ and let $k$ approach infinity we get $egeq s_n(1)$ (Why?) which gives us
$$egeq lim_{nrightarrowinfty}s_n(1)$$
What was the logic of fixating a $k$ and why is it still the same Limit after the Fixation and why are we allowed to do that ?
Please shed some light on those equations.
exponential-function proof-explanation
exponential-function proof-explanation
asked Jan 23 at 19:16
New2MathNew2Math
11812
asked Jan 23 at 19:16
New2MathNew2Math
11812
edited Jan 23 at 19:35
New2Math
asked Jan 23 at 19:16
New2MathNew2Math
11812
asked Jan 23 at 19:16
New2MathNew2Math
11812
asked Jan 23 at 19:16
New2MathNew2Math
11812
11812
$begingroup$
It's because $e_n<s_n(1)$.
$endgroup$
– Lord Shark the Unknown
Jan 23 at 19:18
$begingroup$
For the second question, the lower bound is obtained by taking only the first $(n+1)$ terms of the expansion of $e_k$ and dropping the rest. Since all the terms appearing in the binomial expansion of $e_k$ is positive, dropping some terms will only lower the value.
$endgroup$
– Sangchul Lee
Jan 23 at 19:21
$begingroup$
@SangchulLee I think I have understood a little bit now correct me if I am wrong but the answer to my first Question is actually a rule for limits. And to my second question the Supremum of the Right side of (*) is $s_n(1)$ therefor the smallest upperbound, and because $e_k$ is also an upper Bound we have $egeq e_kgeq s_n(1)$ but I still dont know why $egeq lim_{nrightarrowinfty}s_n(1)$ what is the explaination for that?
$endgroup$
– New2Math
Jan 23 at 19:44
$begingroup$
Okay I think I got it now because $forall nin mathbb{N}:e>s_n(1)$. We know that $e$ is an upper bound. Because $s_n(1)$ is monotonously rising the Limit of $s_n(1)$ also must be the Supremum ,i.e. the smallest upperbound that's why $egeq lim_{nrightarrowinfty}s_n(1)$
$endgroup$
– New2Math
Jan 23 at 20:01
1
$begingroup$
For a more transparent way of explaining the second step is that, taking limit as $ktoinfty$ to the second step, begin{align*}e=lim_{ktoinfty} e_k&geq lim_{ktoinfty} sum_{p=0}^{n} frac{1}{p!}left(1-frac{1}{k}right)cdotsleft(1-frac{p-1}{k}right)\&=sum_{p=0}^{n} lim_{ktoinfty} frac{1}{p!}left(1-frac{1}{k}right)cdotsleft(1-frac{p-1}{k}right)=sum_{p=0}^{n}frac{1}{p!}=s_n(1).end{align*} Now this inequality gives an upper bound of the sequence $(s_n(1))_{n=1}^{infty}$, and so, taking $ntoinfty$ shows $$egeqlim_m s_n(1).$$
$endgroup$
– Sangchul Lee
Jan 23 at 20:08
add a comment |
$begingroup$
It's because $e_n<s_n(1)$.
$endgroup$
– Lord Shark the Unknown
Jan 23 at 19:18
$begingroup$
For the second question, the lower bound is obtained by taking only the first $(n+1)$ terms of the expansion of $e_k$ and dropping the rest. Since all the terms appearing in the binomial expansion of $e_k$ is positive, dropping some terms will only lower the value.
$endgroup$
– Sangchul Lee
Jan 23 at 19:21
$begingroup$
@SangchulLee I think I have understood a little bit now correct me if I am wrong but the answer to my first Question is actually a rule for limits. And to my second question the Supremum of the Right side of (*) is $s_n(1)$ therefor the smallest upperbound, and because $e_k$ is also an upper Bound we have $egeq e_kgeq s_n(1)$ but I still dont know why $egeq lim_{nrightarrowinfty}s_n(1)$ what is the explaination for that?
$endgroup$
– New2Math
Jan 23 at 19:44
$begingroup$
Okay I think I got it now because $forall nin mathbb{N}:e>s_n(1)$. We know that $e$ is an upper bound. Because $s_n(1)$ is monotonously rising the Limit of $s_n(1)$ also must be the Supremum ,i.e. the smallest upperbound that's why $egeq lim_{nrightarrowinfty}s_n(1)$
$endgroup$
– New2Math
Jan 23 at 20:01
1
$begingroup$
For a more transparent way of explaining the second step is that, taking limit as $ktoinfty$ to the second step, begin{align*}e=lim_{ktoinfty} e_k&geq lim_{ktoinfty} sum_{p=0}^{n} frac{1}{p!}left(1-frac{1}{k}right)cdotsleft(1-frac{p-1}{k}right)\&=sum_{p=0}^{n} lim_{ktoinfty} frac{1}{p!}left(1-frac{1}{k}right)cdotsleft(1-frac{p-1}{k}right)=sum_{p=0}^{n}frac{1}{p!}=s_n(1).end{align*} Now this inequality gives an upper bound of the sequence $(s_n(1))_{n=1}^{infty}$, and so, taking $ntoinfty$ shows $$egeqlim_m s_n(1).$$
$endgroup$
– Sangchul Lee
Jan 23 at 20:08
$begingroup$
It's because $e_n<s_n(1)$.
$endgroup$
– Lord Shark the Unknown
Jan 23 at 19:18
$begingroup$
It's because $e_n<s_n(1)$.
$endgroup$
– Lord Shark the Unknown
Jan 23 at 19:18
$begingroup$
For the second question, the lower bound is obtained by taking only the first $(n+1)$ terms of the expansion of $e_k$ and dropping the rest. Since all the terms appearing in the binomial expansion of $e_k$ is positive, dropping some terms will only lower the value.
$endgroup$
– Sangchul Lee
Jan 23 at 19:21
$begingroup$
For the second question, the lower bound is obtained by taking only the first $(n+1)$ terms of the expansion of $e_k$ and dropping the rest. Since all the terms appearing in the binomial expansion of $e_k$ is positive, dropping some terms will only lower the value.
$endgroup$
– Sangchul Lee
Jan 23 at 19:21
$begingroup$
@SangchulLee I think I have understood a little bit now correct me if I am wrong but the answer to my first Question is actually a rule for limits. And to my second question the Supremum of the Right side of (*) is $s_n(1)$ therefor the smallest upperbound, and because $e_k$ is also an upper Bound we have $egeq e_kgeq s_n(1)$ but I still dont know why $egeq lim_{nrightarrowinfty}s_n(1)$ what is the explaination for that?
$endgroup$
– New2Math
Jan 23 at 19:44
$begingroup$
@SangchulLee I think I have understood a little bit now correct me if I am wrong but the answer to my first Question is actually a rule for limits. And to my second question the Supremum of the Right side of (*) is $s_n(1)$ therefor the smallest upperbound, and because $e_k$ is also an upper Bound we have $egeq e_kgeq s_n(1)$ but I still dont know why $egeq lim_{nrightarrowinfty}s_n(1)$ what is the explaination for that?
$endgroup$
– New2Math
Jan 23 at 19:44
$begingroup$
Okay I think I got it now because $forall nin mathbb{N}:e>s_n(1)$. We know that $e$ is an upper bound. Because $s_n(1)$ is monotonously rising the Limit of $s_n(1)$ also must be the Supremum ,i.e. the smallest upperbound that's why $egeq lim_{nrightarrowinfty}s_n(1)$
$endgroup$
– New2Math
Jan 23 at 20:01
$begingroup$
Okay I think I got it now because $forall nin mathbb{N}:e>s_n(1)$. We know that $e$ is an upper bound. Because $s_n(1)$ is monotonously rising the Limit of $s_n(1)$ also must be the Supremum ,i.e. the smallest upperbound that's why $egeq lim_{nrightarrowinfty}s_n(1)$
$endgroup$
– New2Math
Jan 23 at 20:01
1
1
$begingroup$
For a more transparent way of explaining the second step is that, taking limit as $ktoinfty$ to the second step, begin{align*}e=lim_{ktoinfty} e_k&geq lim_{ktoinfty} sum_{p=0}^{n} frac{1}{p!}left(1-frac{1}{k}right)cdotsleft(1-frac{p-1}{k}right)\&=sum_{p=0}^{n} lim_{ktoinfty} frac{1}{p!}left(1-frac{1}{k}right)cdotsleft(1-frac{p-1}{k}right)=sum_{p=0}^{n}frac{1}{p!}=s_n(1).end{align*} Now this inequality gives an upper bound of the sequence $(s_n(1))_{n=1}^{infty}$, and so, taking $ntoinfty$ shows $$egeqlim_m s_n(1).$$
$endgroup$
– Sangchul Lee
Jan 23 at 20:08
$begingroup$
For a more transparent way of explaining the second step is that, taking limit as $ktoinfty$ to the second step, begin{align*}e=lim_{ktoinfty} e_k&geq lim_{ktoinfty} sum_{p=0}^{n} frac{1}{p!}left(1-frac{1}{k}right)cdotsleft(1-frac{p-1}{k}right)\&=sum_{p=0}^{n} lim_{ktoinfty} frac{1}{p!}left(1-frac{1}{k}right)cdotsleft(1-frac{p-1}{k}right)=sum_{p=0}^{n}frac{1}{p!}=s_n(1).end{align*} Now this inequality gives an upper bound of the sequence $(s_n(1))_{n=1}^{infty}$, and so, taking $ntoinfty$ shows $$egeqlim_m s_n(1).$$
$endgroup$
– Sangchul Lee
Jan 23 at 20:08
add a comment |
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$begingroup$
It's because $e_n<s_n(1)$.
$endgroup$
– Lord Shark the Unknown
Jan 23 at 19:18
$begingroup$
For the second question, the lower bound is obtained by taking only the first $(n+1)$ terms of the expansion of $e_k$ and dropping the rest. Since all the terms appearing in the binomial expansion of $e_k$ is positive, dropping some terms will only lower the value.
$endgroup$
– Sangchul Lee
Jan 23 at 19:21
$begingroup$
@SangchulLee I think I have understood a little bit now correct me if I am wrong but the answer to my first Question is actually a rule for limits. And to my second question the Supremum of the Right side of (*) is $s_n(1)$ therefor the smallest upperbound, and because $e_k$ is also an upper Bound we have $egeq e_kgeq s_n(1)$ but I still dont know why $egeq lim_{nrightarrowinfty}s_n(1)$ what is the explaination for that?
$endgroup$
– New2Math
Jan 23 at 19:44
$begingroup$
Okay I think I got it now because $forall nin mathbb{N}:e>s_n(1)$. We know that $e$ is an upper bound. Because $s_n(1)$ is monotonously rising the Limit of $s_n(1)$ also must be the Supremum ,i.e. the smallest upperbound that's why $egeq lim_{nrightarrowinfty}s_n(1)$
$endgroup$
– New2Math
Jan 23 at 20:01
1
$begingroup$
For a more transparent way of explaining the second step is that, taking limit as $ktoinfty$ to the second step, begin{align*}e=lim_{ktoinfty} e_k&geq lim_{ktoinfty} sum_{p=0}^{n} frac{1}{p!}left(1-frac{1}{k}right)cdotsleft(1-frac{p-1}{k}right)\&=sum_{p=0}^{n} lim_{ktoinfty} frac{1}{p!}left(1-frac{1}{k}right)cdotsleft(1-frac{p-1}{k}right)=sum_{p=0}^{n}frac{1}{p!}=s_n(1).end{align*} Now this inequality gives an upper bound of the sequence $(s_n(1))_{n=1}^{infty}$, and so, taking $ntoinfty$ shows $$egeqlim_m s_n(1).$$
$endgroup$
– Sangchul Lee
Jan 23 at 20:08