why is the sequence not uniformly convergent on the whole C? Can one show, that sup(f_n-f) is unequal zero? Thanks
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What does it mean for a sequence to be uniformly continuous? – copper.hat Nov 24 '16 at 01:03
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I am sorry. I have corrected the question – Solidbridges mathematics Nov 24 '16 at 01:06
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$|e^z-(1+z/n)^n|$ is unbounded on $\Bbb C$ for all $n$. In fact it's unbounded on $\Bbb R$: exponentials dominate polynomials. – anon Nov 24 '16 at 01:09
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thank you artic tern! why is it unbounded? – Solidbridges mathematics Nov 24 '16 at 01:17
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We have $\exp(x)>x^n/n!$ for all naturals $n$ and positive reals $x$. (Just look at the power series defining $\exp$ after all.) You can use this to prove $\exp(x)-f(x)$ is unbounded on $\mathbb{R}$ for any polynomial $f(x)$. – anon Nov 24 '16 at 01:25
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but isn't the limit of(1+z/n)^n exactly e(z)? so the distance between them is getting less and less? – Solidbridges mathematics Nov 24 '16 at 01:29
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It is enough to show that the convergence is not uniform on $\mathbb{R}$. We need to show that $(1+z/n)^n$ does not converge too fast to $e^z$. We should also know that $(1+z/n)^n$ is increasing for $z\geq 0$ (see I have to show $(1+\frac1n)^n$ is monotonically increasing sequence and planetmath), so we have to show that $e^z-(1+z/n)^n$ is not too small, for a given $n$, when $z$ is very large.
Take $n\in\mathbb{N}$. Take $k\geq 0$ and $z=kn$. Then $$\left(1+\frac{z}{n}\right)^n=(1+k)^n=\sum_{j=1}^n n(n-1)\cdots (n-j+1)\frac{k^j}{j!}\geq n^j\frac{k^j}{j!}=\sum_{j=1}^n\frac{z^j}{j!}$$ Also, $e^z=\sum_{j=1}^\frac{z^j}{j!}$, so $$e^z-(1+z/n)^n\geq\sum_{j=n+1}^\infty \frac{z^j}{j!}\geq\frac{z^{n+1}}{(n+1)!}$$ If $k$ is very large, then $\frac{z^{n+1}}{(n+1)!}$ is also very large. This shows that for all $\epsilon>0$ (and in particular there exists some $\epsilon>0$) such that for all $n$, there exists $z$ with $|e^z-(1+z/n)^n|>\epsilon$. This is the negation of uniform convergence.
Luiz Cordeiro
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I will show, using only $\ln(1+z) =\int_0^z \dfrac{dt}{1+t} $, that, if $0 < z < n$ then $z-\ln (1+z/n)^n =c\dfrac{z^{2}}{2n} $ where $\frac12 < c < 1$.
Therefore if $z = n/2$, $z-\ln (1+z/n)^n =c\dfrac{(n/2)^{2}}{2n} =cn/8 $ so that the convergence is not uniform.
Here we go.
Since, for positive integer $m$, $\dfrac1{1+v} =\sum_{k=0}^{2m-1} (-1)^k v^k+\dfrac{v^{2m}}{1+v} $, integrating,
$\begin{array}\\ \ln(1+v) &=\int_0^v \dfrac{dw}{1+w}\\ &=\int_0^v(\sum_{k=0}^{2m-1} (-1)^k w^k+\dfrac{w^{2m}}{1+w})dw\\ &=\sum_{k=0}^{2m-1} \dfrac{(-1)^k v^{k+1}}{k+1}+\int_0^v\dfrac{w^{2m}}{1+w}dw \end{array} $
Therefore
$\begin{array}\\ z-\ln (1+z/n)^n &=z-n\ln (1+z/n)\\ &=z-n(\sum_{k=0}^{2m-1} \dfrac{(-1)^k (z/n)^{k+1}}{k+1}+\int_0^{z/n}\dfrac{w^{2m}}{1+w}dw)\\ &=z-\sum_{k=0}^{2m-1} \dfrac{(-1)^k z^{k+1}}{n^k(k+1)} -n\int_0^{z/n}\dfrac{w^{2m}}{1+w}dw)\\ &=\sum_{k=1}^{2m-1} \dfrac{(-1)^{k+1} z^{k+1}}{n^k(k+1)} -\int_0^{z}\dfrac{(w/n)^{2m}}{1+w/n}dw\\ \end{array} $
If $0 < z < 1$, $1-z \lt \frac1{1+z} \lt 1-z/2 $, so $\frac1{1+z} =1-cz $ where $\frac12 < c < 1$.
Therefore, if $n > z$,
$\begin{array}\\ \int_0^{z}\dfrac{(w/n)^{2m}}{1+w/n}dw &=\int_0^{z}(1-cw/n)(w/n)^{2m}dw\\ &=\int_0^{z}((w/n)^{2m}-c(w/n)^{2m+1})dw\\ &=\dfrac{z^{2m+1}}{(2m+1)n^{2m}}-c\dfrac{z^{2m+2}}{(2m+2)n^{2m+1}}\\ \end{array} $
so that
$\begin{array}\\ z-\ln (1+z/n)^n &=\sum_{k=1}^{2m-1} \dfrac{(-1)^{k+1} z^{k+1}}{n^k(k+1)} -\int_0^{z}\dfrac{(w/n)^{2m}}{1+w/n}dw\\ &=\sum_{k=1}^{2m-1} \dfrac{(-1)^{k+1} z^{k+1}}{n^k(k+1)} -(\dfrac{z^{2m+1}}{(2m+1)n^{2m}}-c\dfrac{z^{2m+2}}{(2m+2)n^{2m+1}})\\ &=\sum_{k=1}^{2m} \dfrac{(-1)^{k+1} z^{k+1}}{n^k(k+1)} +c\dfrac{z^{2m+2}}{(2m+2)n^{2m+1}}\\ \end{array} $
If $m=0$, this is $z-\ln (1+z/n)^n =c\dfrac{z^{2}}{2n} $.
If $m=1$, this is $z-\ln (1+z/n)^n =\dfrac{ z^{2}}{2n} -\dfrac{ z^{3}}{3n^2} +c\dfrac{z^{4}}{4n^3} $.
In particular, when $m=0$, if $z = n/2$, $z-\ln (1+z/n)^n =c\dfrac{(n/2)^{2}}{2n} =cn/8 $.
marty cohen
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