the series $sum_{n=1}^{infty}f_n(x)$ converges uniformly Choose the correct option
let $f :mathbb{R} rightarrow mathbb{R}$ be non zero function such that $|f(x)| le frac{1}{1+2x^2}$ for all $x in mathbb{R}$ .Define real value function $f_n$ on $mathbb{R}$ for all $n in mathbb{N}$ by $f_n(x) = f(x +n)$.
Then ,the series $sum_{n=1}^{infty}f_n(x)$ converges uniformly
Choose the correct option
$a)$ on $[0,1]$
$b)$ on $[-1,0]$
$c)$ on $[-1,1]$
$d)$ None of these
I thinks option $a) ,b)$ and $c)$ will correct because derivative of $f$ is bounded
Is its True??
real-analysis
add a comment |
let $f :mathbb{R} rightarrow mathbb{R}$ be non zero function such that $|f(x)| le frac{1}{1+2x^2}$ for all $x in mathbb{R}$ .Define real value function $f_n$ on $mathbb{R}$ for all $n in mathbb{N}$ by $f_n(x) = f(x +n)$.
Then ,the series $sum_{n=1}^{infty}f_n(x)$ converges uniformly
Choose the correct option
$a)$ on $[0,1]$
$b)$ on $[-1,0]$
$c)$ on $[-1,1]$
$d)$ None of these
I thinks option $a) ,b)$ and $c)$ will correct because derivative of $f$ is bounded
Is its True??
real-analysis
See $displaystyle f_n<dfrac{1}{1+2x^2} :forall:n$. So what can you say about the sum?
– Yadati Kiran
Nov 21 '18 at 1:04
add a comment |
let $f :mathbb{R} rightarrow mathbb{R}$ be non zero function such that $|f(x)| le frac{1}{1+2x^2}$ for all $x in mathbb{R}$ .Define real value function $f_n$ on $mathbb{R}$ for all $n in mathbb{N}$ by $f_n(x) = f(x +n)$.
Then ,the series $sum_{n=1}^{infty}f_n(x)$ converges uniformly
Choose the correct option
$a)$ on $[0,1]$
$b)$ on $[-1,0]$
$c)$ on $[-1,1]$
$d)$ None of these
I thinks option $a) ,b)$ and $c)$ will correct because derivative of $f$ is bounded
Is its True??
real-analysis
let $f :mathbb{R} rightarrow mathbb{R}$ be non zero function such that $|f(x)| le frac{1}{1+2x^2}$ for all $x in mathbb{R}$ .Define real value function $f_n$ on $mathbb{R}$ for all $n in mathbb{N}$ by $f_n(x) = f(x +n)$.
Then ,the series $sum_{n=1}^{infty}f_n(x)$ converges uniformly
Choose the correct option
$a)$ on $[0,1]$
$b)$ on $[-1,0]$
$c)$ on $[-1,1]$
$d)$ None of these
I thinks option $a) ,b)$ and $c)$ will correct because derivative of $f$ is bounded
Is its True??
real-analysis
real-analysis
asked Nov 21 '18 at 0:51
Messi fifa
51611
51611
See $displaystyle f_n<dfrac{1}{1+2x^2} :forall:n$. So what can you say about the sum?
– Yadati Kiran
Nov 21 '18 at 1:04
add a comment |
See $displaystyle f_n<dfrac{1}{1+2x^2} :forall:n$. So what can you say about the sum?
– Yadati Kiran
Nov 21 '18 at 1:04
See $displaystyle f_n<dfrac{1}{1+2x^2} :forall:n$. So what can you say about the sum?
– Yadati Kiran
Nov 21 '18 at 1:04
See $displaystyle f_n<dfrac{1}{1+2x^2} :forall:n$. So what can you say about the sum?
– Yadati Kiran
Nov 21 '18 at 1:04
add a comment |
1 Answer
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HINT:
We are not given that $f$ is differentiable.
So, note that for $xin [-1,1]$
$$|f_n(x)|=|f(x+n)|le frac{1}{1+2(x+n)^2}le frac1{1+2(n-1)^2}$$
Appeal to the Weierstrass M-Test. Can you finish?
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
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HINT:
We are not given that $f$ is differentiable.
So, note that for $xin [-1,1]$
$$|f_n(x)|=|f(x+n)|le frac{1}{1+2(x+n)^2}le frac1{1+2(n-1)^2}$$
Appeal to the Weierstrass M-Test. Can you finish?
add a comment |
HINT:
We are not given that $f$ is differentiable.
So, note that for $xin [-1,1]$
$$|f_n(x)|=|f(x+n)|le frac{1}{1+2(x+n)^2}le frac1{1+2(n-1)^2}$$
Appeal to the Weierstrass M-Test. Can you finish?
add a comment |
HINT:
We are not given that $f$ is differentiable.
So, note that for $xin [-1,1]$
$$|f_n(x)|=|f(x+n)|le frac{1}{1+2(x+n)^2}le frac1{1+2(n-1)^2}$$
Appeal to the Weierstrass M-Test. Can you finish?
HINT:
We are not given that $f$ is differentiable.
So, note that for $xin [-1,1]$
$$|f_n(x)|=|f(x+n)|le frac{1}{1+2(x+n)^2}le frac1{1+2(n-1)^2}$$
Appeal to the Weierstrass M-Test. Can you finish?
answered Nov 21 '18 at 2:15
Mark Viola
130k1274170
130k1274170
add a comment |
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See $displaystyle f_n<dfrac{1}{1+2x^2} :forall:n$. So what can you say about the sum?
– Yadati Kiran
Nov 21 '18 at 1:04