Prove uniform convergence of $f_{n} (x) = sqrt{frac{1}{n^2} +x^2}$ to $f(x) = |x|$
$begingroup$
For each $n ∈ Bbb N$, let $f_{n} (x) = sqrt{frac{1}{n^2} +x^2}. $ Show $(f_{n})^{infty}_{n=0}$ converges uniformly on $Bbb R$ to $f(x) = |x|$
I'm trying to find the $N$ such that for all $x in Bbb R$ and $n > N$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg| < epsilon$$ but I'm struggling with proving convergence for all real $x$. I'v seen that for $x∈ [0,1]$, for $n$ greater than or equal to $1$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg|leq frac{1}{sqrt{n}} $$ but I'm unsure of how to generalize. Any help is appreciated!
real-analysis sequences-and-series uniform-convergence
edited Dec 5 '18 at 9:36
Chinnapparaj R
5,4331928
asked Dec 5 '18 at 9:28
user613048user613048
322
$endgroup$
add a comment |
$begingroup$
For each $n ∈ Bbb N$, let $f_{n} (x) = sqrt{frac{1}{n^2} +x^2}. $ Show $(f_{n})^{infty}_{n=0}$ converges uniformly on $Bbb R$ to $f(x) = |x|$
I'm trying to find the $N$ such that for all $x in Bbb R$ and $n > N$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg| < epsilon$$ but I'm struggling with proving convergence for all real $x$. I'v seen that for $x∈ [0,1]$, for $n$ greater than or equal to $1$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg|leq frac{1}{sqrt{n}} $$ but I'm unsure of how to generalize. Any help is appreciated!
real-analysis sequences-and-series uniform-convergence
edited Dec 5 '18 at 9:36
Chinnapparaj R
5,4331928
asked Dec 5 '18 at 9:28
user613048user613048
322
$endgroup$
add a comment |
$begingroup$
For each $n ∈ Bbb N$, let $f_{n} (x) = sqrt{frac{1}{n^2} +x^2}. $ Show $(f_{n})^{infty}_{n=0}$ converges uniformly on $Bbb R$ to $f(x) = |x|$
I'm trying to find the $N$ such that for all $x in Bbb R$ and $n > N$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg| < epsilon$$ but I'm struggling with proving convergence for all real $x$. I'v seen that for $x∈ [0,1]$, for $n$ greater than or equal to $1$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg|leq frac{1}{sqrt{n}} $$ but I'm unsure of how to generalize. Any help is appreciated!
real-analysis sequences-and-series uniform-convergence
edited Dec 5 '18 at 9:36
Chinnapparaj R
5,4331928
asked Dec 5 '18 at 9:28
user613048user613048
322
$endgroup$
For each $n ∈ Bbb N$, let $f_{n} (x) = sqrt{frac{1}{n^2} +x^2}. $ Show $(f_{n})^{infty}_{n=0}$ converges uniformly on $Bbb R$ to $f(x) = |x|$
I'm trying to find the $N$ such that for all $x in Bbb R$ and $n > N$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg| < epsilon$$ but I'm struggling with proving convergence for all real $x$. I'v seen that for $x∈ [0,1]$, for $n$ greater than or equal to $1$, $$Bigg|sqrt{frac{1}{n^2} +x^2} - |x|Bigg|leq frac{1}{sqrt{n}} $$ but I'm unsure of how to generalize. Any help is appreciated!
real-analysis sequences-and-series uniform-convergence
real-analysis sequences-and-series uniform-convergence
edited Dec 5 '18 at 9:36
Chinnapparaj R
5,4331928
asked Dec 5 '18 at 9:28
user613048user613048
322
edited Dec 5 '18 at 9:36
Chinnapparaj R
5,4331928
asked Dec 5 '18 at 9:28
user613048user613048
322
edited Dec 5 '18 at 9:36
Chinnapparaj R
5,4331928
edited Dec 5 '18 at 9:36
Chinnapparaj R
5,4331928
edited Dec 5 '18 at 9:36
Chinnapparaj R
5,4331928
5,4331928
asked Dec 5 '18 at 9:28
user613048user613048
322
asked Dec 5 '18 at 9:28
user613048user613048
322
asked Dec 5 '18 at 9:28
user613048user613048
322
322
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Note that$$sqrt{x^2+frac1{n^2}}-lvert xrvert=frac{frac1{n^2}}{sqrt{x^2+frac1{n^2}}+lvert xrvert}.$$Since the numerator is constant and the denominator is positive and increases to $infty$ when $lvert xrverttoinfty$, the maximum of $sqrt{x^2+frac1{n^2}}-lvert xrvert$ is attained when $x=0$. And that maximum is $frac1n$. So$$(forall xinmathbb{R}):leftlvertsqrt{x^2+frac1{n^2}}-lvert xrvertrightrvertleqslantfrac1n,$$from which the uniform convergence follows.
answered Dec 5 '18 at 9:35
José Carlos SantosJosé Carlos Santos
159k22126229
$endgroup$
add a comment |
$begingroup$
If you use $sqrt a -sqrt b=frac {a-b} {sqrt a +sqrt b}$ you will get the result easily.
answered Dec 5 '18 at 9:31
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
add a comment |
$begingroup$
Hint: We have
$$|f_n(x) - |x|| = frac{1}{n^2} frac{1}{|x|+sqrt{1/n^2 +x^2}} le frac{1}{n}.$$
answered Dec 5 '18 at 9:32
p4schp4sch
5,245217
$endgroup$
add a comment |
$begingroup$
Hint: $$left(vert x vert +frac{1}{n}right)^2=x^2+frac{1}{n^2}+2vert x vert frac{1}{n}=f_n^2+2vert x vert frac{1}{n} geq f_n^2 geq 0$$ so $$0 leq f_n le vert x vert+frac{1}{n}$$ and so $$0 leq f_n - vert x vert leq frac{1}{n} rightarrow 0$$
Moral:
Uniform limit of sequence of continuously differentiable function need not be continuously differentiable!
answered Dec 5 '18 at 9:33
Chinnapparaj RChinnapparaj R
5,4331928
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that$$sqrt{x^2+frac1{n^2}}-lvert xrvert=frac{frac1{n^2}}{sqrt{x^2+frac1{n^2}}+lvert xrvert}.$$Since the numerator is constant and the denominator is positive and increases to $infty$ when $lvert xrverttoinfty$, the maximum of $sqrt{x^2+frac1{n^2}}-lvert xrvert$ is attained when $x=0$. And that maximum is $frac1n$. So$$(forall xinmathbb{R}):leftlvertsqrt{x^2+frac1{n^2}}-lvert xrvertrightrvertleqslantfrac1n,$$from which the uniform convergence follows.
answered Dec 5 '18 at 9:35
José Carlos SantosJosé Carlos Santos
159k22126229
$endgroup$
add a comment |
$begingroup$
Note that$$sqrt{x^2+frac1{n^2}}-lvert xrvert=frac{frac1{n^2}}{sqrt{x^2+frac1{n^2}}+lvert xrvert}.$$Since the numerator is constant and the denominator is positive and increases to $infty$ when $lvert xrverttoinfty$, the maximum of $sqrt{x^2+frac1{n^2}}-lvert xrvert$ is attained when $x=0$. And that maximum is $frac1n$. So$$(forall xinmathbb{R}):leftlvertsqrt{x^2+frac1{n^2}}-lvert xrvertrightrvertleqslantfrac1n,$$from which the uniform convergence follows.
answered Dec 5 '18 at 9:35
José Carlos SantosJosé Carlos Santos
159k22126229
$endgroup$
add a comment |
$begingroup$
Note that$$sqrt{x^2+frac1{n^2}}-lvert xrvert=frac{frac1{n^2}}{sqrt{x^2+frac1{n^2}}+lvert xrvert}.$$Since the numerator is constant and the denominator is positive and increases to $infty$ when $lvert xrverttoinfty$, the maximum of $sqrt{x^2+frac1{n^2}}-lvert xrvert$ is attained when $x=0$. And that maximum is $frac1n$. So$$(forall xinmathbb{R}):leftlvertsqrt{x^2+frac1{n^2}}-lvert xrvertrightrvertleqslantfrac1n,$$from which the uniform convergence follows.
answered Dec 5 '18 at 9:35
José Carlos SantosJosé Carlos Santos
159k22126229
$endgroup$
Note that$$sqrt{x^2+frac1{n^2}}-lvert xrvert=frac{frac1{n^2}}{sqrt{x^2+frac1{n^2}}+lvert xrvert}.$$Since the numerator is constant and the denominator is positive and increases to $infty$ when $lvert xrverttoinfty$, the maximum of $sqrt{x^2+frac1{n^2}}-lvert xrvert$ is attained when $x=0$. And that maximum is $frac1n$. So$$(forall xinmathbb{R}):leftlvertsqrt{x^2+frac1{n^2}}-lvert xrvertrightrvertleqslantfrac1n,$$from which the uniform convergence follows.
answered Dec 5 '18 at 9:35
José Carlos SantosJosé Carlos Santos
159k22126229
answered Dec 5 '18 at 9:35
José Carlos SantosJosé Carlos Santos
159k22126229
answered Dec 5 '18 at 9:35
José Carlos SantosJosé Carlos Santos
159k22126229
answered Dec 5 '18 at 9:35
José Carlos SantosJosé Carlos Santos
159k22126229
159k22126229
add a comment |
add a comment |
$begingroup$
If you use $sqrt a -sqrt b=frac {a-b} {sqrt a +sqrt b}$ you will get the result easily.
answered Dec 5 '18 at 9:31
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
add a comment |
$begingroup$
If you use $sqrt a -sqrt b=frac {a-b} {sqrt a +sqrt b}$ you will get the result easily.
answered Dec 5 '18 at 9:31
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
add a comment |
$begingroup$
If you use $sqrt a -sqrt b=frac {a-b} {sqrt a +sqrt b}$ you will get the result easily.
answered Dec 5 '18 at 9:31
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
$endgroup$
If you use $sqrt a -sqrt b=frac {a-b} {sqrt a +sqrt b}$ you will get the result easily.
answered Dec 5 '18 at 9:31
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Dec 5 '18 at 9:31
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Dec 5 '18 at 9:31
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
answered Dec 5 '18 at 9:31
Kavi Rama MurthyKavi Rama Murthy
57.7k42160
57.7k42160
add a comment |
add a comment |
$begingroup$
Hint: We have
$$|f_n(x) - |x|| = frac{1}{n^2} frac{1}{|x|+sqrt{1/n^2 +x^2}} le frac{1}{n}.$$
answered Dec 5 '18 at 9:32
p4schp4sch
5,245217
$endgroup$
add a comment |
$begingroup$
Hint: We have
$$|f_n(x) - |x|| = frac{1}{n^2} frac{1}{|x|+sqrt{1/n^2 +x^2}} le frac{1}{n}.$$
answered Dec 5 '18 at 9:32
p4schp4sch
5,245217
$endgroup$
add a comment |
$begingroup$
Hint: We have
$$|f_n(x) - |x|| = frac{1}{n^2} frac{1}{|x|+sqrt{1/n^2 +x^2}} le frac{1}{n}.$$
answered Dec 5 '18 at 9:32
p4schp4sch
5,245217
$endgroup$
Hint: We have
$$|f_n(x) - |x|| = frac{1}{n^2} frac{1}{|x|+sqrt{1/n^2 +x^2}} le frac{1}{n}.$$
answered Dec 5 '18 at 9:32
p4schp4sch
5,245217
answered Dec 5 '18 at 9:32
p4schp4sch
5,245217
answered Dec 5 '18 at 9:32
p4schp4sch
5,245217
answered Dec 5 '18 at 9:32
p4schp4sch
5,245217
5,245217
add a comment |
add a comment |
$begingroup$
Hint: $$left(vert x vert +frac{1}{n}right)^2=x^2+frac{1}{n^2}+2vert x vert frac{1}{n}=f_n^2+2vert x vert frac{1}{n} geq f_n^2 geq 0$$ so $$0 leq f_n le vert x vert+frac{1}{n}$$ and so $$0 leq f_n - vert x vert leq frac{1}{n} rightarrow 0$$
Moral:
Uniform limit of sequence of continuously differentiable function need not be continuously differentiable!
answered Dec 5 '18 at 9:33
Chinnapparaj RChinnapparaj R
5,4331928
$endgroup$
add a comment |
$begingroup$
Hint: $$left(vert x vert +frac{1}{n}right)^2=x^2+frac{1}{n^2}+2vert x vert frac{1}{n}=f_n^2+2vert x vert frac{1}{n} geq f_n^2 geq 0$$ so $$0 leq f_n le vert x vert+frac{1}{n}$$ and so $$0 leq f_n - vert x vert leq frac{1}{n} rightarrow 0$$
Moral:
Uniform limit of sequence of continuously differentiable function need not be continuously differentiable!
answered Dec 5 '18 at 9:33
Chinnapparaj RChinnapparaj R
5,4331928
$endgroup$
add a comment |
$begingroup$
Hint: $$left(vert x vert +frac{1}{n}right)^2=x^2+frac{1}{n^2}+2vert x vert frac{1}{n}=f_n^2+2vert x vert frac{1}{n} geq f_n^2 geq 0$$ so $$0 leq f_n le vert x vert+frac{1}{n}$$ and so $$0 leq f_n - vert x vert leq frac{1}{n} rightarrow 0$$
Moral:
Uniform limit of sequence of continuously differentiable function need not be continuously differentiable!
answered Dec 5 '18 at 9:33
Chinnapparaj RChinnapparaj R
5,4331928
$endgroup$
Hint: $$left(vert x vert +frac{1}{n}right)^2=x^2+frac{1}{n^2}+2vert x vert frac{1}{n}=f_n^2+2vert x vert frac{1}{n} geq f_n^2 geq 0$$ so $$0 leq f_n le vert x vert+frac{1}{n}$$ and so $$0 leq f_n - vert x vert leq frac{1}{n} rightarrow 0$$
Moral:
Uniform limit of sequence of continuously differentiable function need not be continuously differentiable!
answered Dec 5 '18 at 9:33
Chinnapparaj RChinnapparaj R
5,4331928
edited Dec 5 '18 at 9:42
answered Dec 5 '18 at 9:33
Chinnapparaj RChinnapparaj R
5,4331928
answered Dec 5 '18 at 9:33
Chinnapparaj RChinnapparaj R
5,4331928
answered Dec 5 '18 at 9:33
Chinnapparaj RChinnapparaj R
5,4331928
5,4331928
add a comment |
add a comment |
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