Proving a sequence of functions converges uniformly
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Let ${a_{n}}$ be a bounded sequence of numbers, and for each $n in
mathbb{N}$ and each number $x$, define
$$f_{n}(x) = a_{0} + a_{1}x + frac{a_{2}x^{2}}{2!} + cdots +
frac{a_{n}x^{n}}{n!}.$$
Prove that for each $r > 0,$ the sequence of functions ${f_{n} : [-r,
r] rightarrow mathbb{R}}$ is uniformly convergent.
My attempt:
Let $M$ be a bound for the sequence ${a_{n}}$. By definition, we have $|a_{n}| leq M$ for every natural number $n$. To show $sum_{k=0}^{infty} frac{a_{k}x^{k}}{k!}$ converges uniformly, we can just show $lim_{ntoinfty} sum_{i=n+1}^{infty}frac{a_{k}x^{k}}{k!} = 0$, which means that the convergence to $0$ is independent of $x$.
We have
$$lim_{ntoinfty} sum_{k=n+1}^{infty} frac{a_{k}x^{k}}{k!} leq lim_{ntoinfty} sum_{k=n+1}^{infty} left|frac{a_{k}x^{k}}{k!}right| $$
$$= lim_{ntoinfty} sum_{k=n+1}^{infty} frac{|a_{k}||x|^{k}}{k!} leq lim_{ntoinfty} sum_{k=n+1}^{infty} frac{Mr^{k}}{k!} $$
$$= M cdot lim_{ntoinfty} sum_{k=n+1}^{infty} frac{r^{k}}{k!}.$$
But since $sum_{k=0}^{infty} frac{r^{k}}{k!} = e^{r}$, we have
$$M cdot lim_{ntoinfty} sum_{k=n+1}^{infty} frac{r^{k}}{k!} leq Me^{r}, $$
which proves that the convergence is uniform.
Is my proof right? What can I improve?
EDIT: It is wrong because we need the limit to equal $0$.
sequences-and-series functions proof-verification convergence uniform-convergence
asked Dec 3 at 18:36
joseph
3139
add a comment |
up vote
0
down vote
favorite
Let ${a_{n}}$ be a bounded sequence of numbers, and for each $n in
mathbb{N}$ and each number $x$, define
$$f_{n}(x) = a_{0} + a_{1}x + frac{a_{2}x^{2}}{2!} + cdots +
frac{a_{n}x^{n}}{n!}.$$
Prove that for each $r > 0,$ the sequence of functions ${f_{n} : [-r,
r] rightarrow mathbb{R}}$ is uniformly convergent.
My attempt:
Let $M$ be a bound for the sequence ${a_{n}}$. By definition, we have $|a_{n}| leq M$ for every natural number $n$. To show $sum_{k=0}^{infty} frac{a_{k}x^{k}}{k!}$ converges uniformly, we can just show $lim_{ntoinfty} sum_{i=n+1}^{infty}frac{a_{k}x^{k}}{k!} = 0$, which means that the convergence to $0$ is independent of $x$.
We have
$$lim_{ntoinfty} sum_{k=n+1}^{infty} frac{a_{k}x^{k}}{k!} leq lim_{ntoinfty} sum_{k=n+1}^{infty} left|frac{a_{k}x^{k}}{k!}right| $$
$$= lim_{ntoinfty} sum_{k=n+1}^{infty} frac{|a_{k}||x|^{k}}{k!} leq lim_{ntoinfty} sum_{k=n+1}^{infty} frac{Mr^{k}}{k!} $$
$$= M cdot lim_{ntoinfty} sum_{k=n+1}^{infty} frac{r^{k}}{k!}.$$
But since $sum_{k=0}^{infty} frac{r^{k}}{k!} = e^{r}$, we have
$$M cdot lim_{ntoinfty} sum_{k=n+1}^{infty} frac{r^{k}}{k!} leq Me^{r}, $$
which proves that the convergence is uniform.
Is my proof right? What can I improve?
EDIT: It is wrong because we need the limit to equal $0$.
sequences-and-series functions proof-verification convergence uniform-convergence
asked Dec 3 at 18:36
joseph
3139
"Since $sum_0^infty r^k/k! = mathrm e^r$, we have" $color{red}{lim_n sum_{n+1}^infty r^k/k! = 0}$. Also as an alternative, Weierstrass--$M$ test would work.
– xbh
Dec 3 at 19:02
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let ${a_{n}}$ be a bounded sequence of numbers, and for each $n in
mathbb{N}$ and each number $x$, define
$$f_{n}(x) = a_{0} + a_{1}x + frac{a_{2}x^{2}}{2!} + cdots +
frac{a_{n}x^{n}}{n!}.$$
Prove that for each $r > 0,$ the sequence of functions ${f_{n} : [-r,
r] rightarrow mathbb{R}}$ is uniformly convergent.
My attempt:
Let $M$ be a bound for the sequence ${a_{n}}$. By definition, we have $|a_{n}| leq M$ for every natural number $n$. To show $sum_{k=0}^{infty} frac{a_{k}x^{k}}{k!}$ converges uniformly, we can just show $lim_{ntoinfty} sum_{i=n+1}^{infty}frac{a_{k}x^{k}}{k!} = 0$, which means that the convergence to $0$ is independent of $x$.
We have
$$lim_{ntoinfty} sum_{k=n+1}^{infty} frac{a_{k}x^{k}}{k!} leq lim_{ntoinfty} sum_{k=n+1}^{infty} left|frac{a_{k}x^{k}}{k!}right| $$
$$= lim_{ntoinfty} sum_{k=n+1}^{infty} frac{|a_{k}||x|^{k}}{k!} leq lim_{ntoinfty} sum_{k=n+1}^{infty} frac{Mr^{k}}{k!} $$
$$= M cdot lim_{ntoinfty} sum_{k=n+1}^{infty} frac{r^{k}}{k!}.$$
But since $sum_{k=0}^{infty} frac{r^{k}}{k!} = e^{r}$, we have
$$M cdot lim_{ntoinfty} sum_{k=n+1}^{infty} frac{r^{k}}{k!} leq Me^{r}, $$
which proves that the convergence is uniform.
Is my proof right? What can I improve?
EDIT: It is wrong because we need the limit to equal $0$.
sequences-and-series functions proof-verification convergence uniform-convergence
asked Dec 3 at 18:36
joseph
3139
Let ${a_{n}}$ be a bounded sequence of numbers, and for each $n in
mathbb{N}$ and each number $x$, define
$$f_{n}(x) = a_{0} + a_{1}x + frac{a_{2}x^{2}}{2!} + cdots +
frac{a_{n}x^{n}}{n!}.$$
Prove that for each $r > 0,$ the sequence of functions ${f_{n} : [-r,
r] rightarrow mathbb{R}}$ is uniformly convergent.
My attempt:
Let $M$ be a bound for the sequence ${a_{n}}$. By definition, we have $|a_{n}| leq M$ for every natural number $n$. To show $sum_{k=0}^{infty} frac{a_{k}x^{k}}{k!}$ converges uniformly, we can just show $lim_{ntoinfty} sum_{i=n+1}^{infty}frac{a_{k}x^{k}}{k!} = 0$, which means that the convergence to $0$ is independent of $x$.
We have
$$lim_{ntoinfty} sum_{k=n+1}^{infty} frac{a_{k}x^{k}}{k!} leq lim_{ntoinfty} sum_{k=n+1}^{infty} left|frac{a_{k}x^{k}}{k!}right| $$
$$= lim_{ntoinfty} sum_{k=n+1}^{infty} frac{|a_{k}||x|^{k}}{k!} leq lim_{ntoinfty} sum_{k=n+1}^{infty} frac{Mr^{k}}{k!} $$
$$= M cdot lim_{ntoinfty} sum_{k=n+1}^{infty} frac{r^{k}}{k!}.$$
But since $sum_{k=0}^{infty} frac{r^{k}}{k!} = e^{r}$, we have
$$M cdot lim_{ntoinfty} sum_{k=n+1}^{infty} frac{r^{k}}{k!} leq Me^{r}, $$
which proves that the convergence is uniform.
Is my proof right? What can I improve?
EDIT: It is wrong because we need the limit to equal $0$.
sequences-and-series functions proof-verification convergence uniform-convergence
sequences-and-series functions proof-verification convergence uniform-convergence
asked Dec 3 at 18:36
joseph
3139
asked Dec 3 at 18:36
joseph
3139
edited Dec 3 at 18:49
asked Dec 3 at 18:36
joseph
3139
asked Dec 3 at 18:36
joseph
3139
asked Dec 3 at 18:36
joseph
3139
3139
"Since $sum_0^infty r^k/k! = mathrm e^r$, we have" $color{red}{lim_n sum_{n+1}^infty r^k/k! = 0}$. Also as an alternative, Weierstrass--$M$ test would work.
– xbh
Dec 3 at 19:02
add a comment |
"Since $sum_0^infty r^k/k! = mathrm e^r$, we have" $color{red}{lim_n sum_{n+1}^infty r^k/k! = 0}$. Also as an alternative, Weierstrass--$M$ test would work.
– xbh
Dec 3 at 19:02
"Since $sum_0^infty r^k/k! = mathrm e^r$, we have" $color{red}{lim_n sum_{n+1}^infty r^k/k! = 0}$. Also as an alternative, Weierstrass--$M$ test would work.
– xbh
Dec 3 at 19:02
"Since $sum_0^infty r^k/k! = mathrm e^r$, we have" $color{red}{lim_n sum_{n+1}^infty r^k/k! = 0}$. Also as an alternative, Weierstrass--$M$ test would work.
– xbh
Dec 3 at 19:02
add a comment |
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"Since $sum_0^infty r^k/k! = mathrm e^r$, we have" $color{red}{lim_n sum_{n+1}^infty r^k/k! = 0}$. Also as an alternative, Weierstrass--$M$ test would work.
– xbh
Dec 3 at 19:02