We have to show that $f_n(x)=x^ne^{-nx}$.
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So, we have to show that $f_n(x)=x^ne^{-nx}$ converges uniformly on $[0,infty)$ ?
Consider pointwise, in case $xin[0,infty),f_nto0,(x^n<e^{nx})$.
Let $epsilon>0,$ choose $N=frac{ln(epsilon)}{lnx -x},$ if $nle N$ then $|f_n(x)-f(x)|=|frac{x^n}{e^{nx}}-0|<epsilon$.
My problem, when i choose $N=frac{ln(epsilon)}{lnx -x}$ if $x=0,N$ dose not exists ?
real-analysis
asked Nov 14 at 13:20
kim wenasa
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up vote
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So, we have to show that $f_n(x)=x^ne^{-nx}$ converges uniformly on $[0,infty)$ ?
Consider pointwise, in case $xin[0,infty),f_nto0,(x^n<e^{nx})$.
Let $epsilon>0,$ choose $N=frac{ln(epsilon)}{lnx -x},$ if $nle N$ then $|f_n(x)-f(x)|=|frac{x^n}{e^{nx}}-0|<epsilon$.
My problem, when i choose $N=frac{ln(epsilon)}{lnx -x}$ if $x=0,N$ dose not exists ?
real-analysis
asked Nov 14 at 13:20
kim wenasa
1
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So, we have to show that $f_n(x)=x^ne^{-nx}$ converges uniformly on $[0,infty)$ ?
Consider pointwise, in case $xin[0,infty),f_nto0,(x^n<e^{nx})$.
Let $epsilon>0,$ choose $N=frac{ln(epsilon)}{lnx -x},$ if $nle N$ then $|f_n(x)-f(x)|=|frac{x^n}{e^{nx}}-0|<epsilon$.
My problem, when i choose $N=frac{ln(epsilon)}{lnx -x}$ if $x=0,N$ dose not exists ?
real-analysis
asked Nov 14 at 13:20
kim wenasa
1
So, we have to show that $f_n(x)=x^ne^{-nx}$ converges uniformly on $[0,infty)$ ?
Consider pointwise, in case $xin[0,infty),f_nto0,(x^n<e^{nx})$.
Let $epsilon>0,$ choose $N=frac{ln(epsilon)}{lnx -x},$ if $nle N$ then $|f_n(x)-f(x)|=|frac{x^n}{e^{nx}}-0|<epsilon$.
My problem, when i choose $N=frac{ln(epsilon)}{lnx -x}$ if $x=0,N$ dose not exists ?
real-analysis
real-analysis
asked Nov 14 at 13:20
kim wenasa
1
asked Nov 14 at 13:20
kim wenasa
1
asked Nov 14 at 13:20
kim wenasa
1
asked Nov 14 at 13:20
kim wenasa
1
asked Nov 14 at 13:20
kim wenasa
1
1
add a comment |
add a comment |
1 Answer
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Note that $f_n'(x)=nx^{n-1}e^{-nx}-nx^ne^{-nx}=nx^{n-1}(1-x)e^{-nx}$. So, $f_n(x)$ is increasing in $[0,1)$ and decreasing in $(1,infty)$. Therefore, $max f_n=f_n(1)=e^{-n}$. Since $lim_{ntoinfty}e^{-n}=0$, $(f_n)_{ninmathbb N}$ converges uniformly to the null function.
answered Nov 14 at 13:28
José Carlos Santos
140k18111203
Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
– kim wenasa
Nov 14 at 14:10
That's another question. I suggest that you post it as such.
– José Carlos Santos
Nov 14 at 14:13
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Note that $f_n'(x)=nx^{n-1}e^{-nx}-nx^ne^{-nx}=nx^{n-1}(1-x)e^{-nx}$. So, $f_n(x)$ is increasing in $[0,1)$ and decreasing in $(1,infty)$. Therefore, $max f_n=f_n(1)=e^{-n}$. Since $lim_{ntoinfty}e^{-n}=0$, $(f_n)_{ninmathbb N}$ converges uniformly to the null function.
answered Nov 14 at 13:28
José Carlos Santos
140k18111203
Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
– kim wenasa
Nov 14 at 14:10
That's another question. I suggest that you post it as such.
– José Carlos Santos
Nov 14 at 14:13
add a comment |
up vote
1
down vote
Note that $f_n'(x)=nx^{n-1}e^{-nx}-nx^ne^{-nx}=nx^{n-1}(1-x)e^{-nx}$. So, $f_n(x)$ is increasing in $[0,1)$ and decreasing in $(1,infty)$. Therefore, $max f_n=f_n(1)=e^{-n}$. Since $lim_{ntoinfty}e^{-n}=0$, $(f_n)_{ninmathbb N}$ converges uniformly to the null function.
answered Nov 14 at 13:28
José Carlos Santos
140k18111203
Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
– kim wenasa
Nov 14 at 14:10
That's another question. I suggest that you post it as such.
– José Carlos Santos
Nov 14 at 14:13
add a comment |
up vote
1
down vote
up vote
1
down vote
Note that $f_n'(x)=nx^{n-1}e^{-nx}-nx^ne^{-nx}=nx^{n-1}(1-x)e^{-nx}$. So, $f_n(x)$ is increasing in $[0,1)$ and decreasing in $(1,infty)$. Therefore, $max f_n=f_n(1)=e^{-n}$. Since $lim_{ntoinfty}e^{-n}=0$, $(f_n)_{ninmathbb N}$ converges uniformly to the null function.
answered Nov 14 at 13:28
José Carlos Santos
140k18111203
Note that $f_n'(x)=nx^{n-1}e^{-nx}-nx^ne^{-nx}=nx^{n-1}(1-x)e^{-nx}$. So, $f_n(x)$ is increasing in $[0,1)$ and decreasing in $(1,infty)$. Therefore, $max f_n=f_n(1)=e^{-n}$. Since $lim_{ntoinfty}e^{-n}=0$, $(f_n)_{ninmathbb N}$ converges uniformly to the null function.
answered Nov 14 at 13:28
José Carlos Santos
140k18111203
answered Nov 14 at 13:28
José Carlos Santos
140k18111203
answered Nov 14 at 13:28
José Carlos Santos
140k18111203
answered Nov 14 at 13:28
José Carlos Santos
140k18111203
140k18111203
Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
– kim wenasa
Nov 14 at 14:10
That's another question. I suggest that you post it as such.
– José Carlos Santos
Nov 14 at 14:13
add a comment |
Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
– kim wenasa
Nov 14 at 14:10
That's another question. I suggest that you post it as such.
– José Carlos Santos
Nov 14 at 14:13
Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
– kim wenasa
Nov 14 at 14:10
Can i ask some question this $f_n(x)=x^n$ is converges uniformly on $(-1,1)$?, i think it true, check $f_nto 0 $ on $(-1,1)$ @José Carlos Santos.
– kim wenasa
Nov 14 at 14:10
That's another question. I suggest that you post it as such.
– José Carlos Santos
Nov 14 at 14:13
That's another question. I suggest that you post it as such.
– José Carlos Santos
Nov 14 at 14:13
add a comment |
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