Mixing
Practice qualifying question Analysis
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So I am looking at old qualifying exams for an entrance program in august. I am struggling with a few of the problems and the following one in particular.
Let $f:[0,infty) rightarrow mathbb{R} $ be continuous, and assume that $lim_{x to infty} f(x)$ exists and is a finite number $L$. What can be said about
$$lim_{n to infty}int_{0}^{1} f(nx)dx?$$
Prove your answer.
I am confused as to what the goal of this exercise is and I am not sure where to begin. Any guidance woud be appreciated! Thanks!
real-analysis
asked Jan 22 at 22:27
geoplantedgeoplanted
617
$endgroup$
|
show 1 more comment
$begingroup$
So I am looking at old qualifying exams for an entrance program in august. I am struggling with a few of the problems and the following one in particular.
Let $f:[0,infty) rightarrow mathbb{R} $ be continuous, and assume that $lim_{x to infty} f(x)$ exists and is a finite number $L$. What can be said about
$$lim_{n to infty}int_{0}^{1} f(nx)dx?$$
Prove your answer.
I am confused as to what the goal of this exercise is and I am not sure where to begin. Any guidance woud be appreciated! Thanks!
real-analysis
asked Jan 22 at 22:27
geoplantedgeoplanted
617
$endgroup$
$begingroup$
Have you tried making a picture? Maybe it helps to visualize with a good example such as $f(x)=frac{1+x}{2+x}$.
$endgroup$
– SmileyCraft
Jan 22 at 22:36
$begingroup$
Making the substitution $u=nx$, the integral is $int_{0}^{n} f(u),du$. So if $L ne 0$, the limit definitely doesn't exist. If $L=0$, it's more complicated?
$endgroup$
– preferred_anon
Jan 22 at 22:54
$begingroup$
It is a good problem in analysis, related to convergence results for passing limits through integrals. It is also one where you might "want" to use some type of Lebesgue dominated convergence theorem, but you can also prove it directly without knowing such dominated convergence theorems (the assumptions will help out for some pesky particulars). The first step is to get some intuition, what do you think the answer should be? Can you take steps to prove it?
$endgroup$
– Michael
Jan 22 at 22:55
1
$begingroup$
@preferred_anon : Minor fix it would be $frac{1}{n}int_0^n f(u)du$.
$endgroup$
– Michael
Jan 22 at 22:56
$begingroup$
@Michael oops, thank you!
$endgroup$
– preferred_anon
Jan 22 at 23:02
|
show 1 more comment
$begingroup$
So I am looking at old qualifying exams for an entrance program in august. I am struggling with a few of the problems and the following one in particular.
Let $f:[0,infty) rightarrow mathbb{R} $ be continuous, and assume that $lim_{x to infty} f(x)$ exists and is a finite number $L$. What can be said about
$$lim_{n to infty}int_{0}^{1} f(nx)dx?$$
Prove your answer.
I am confused as to what the goal of this exercise is and I am not sure where to begin. Any guidance woud be appreciated! Thanks!
real-analysis
asked Jan 22 at 22:27
geoplantedgeoplanted
617
$endgroup$
So I am looking at old qualifying exams for an entrance program in august. I am struggling with a few of the problems and the following one in particular.
Let $f:[0,infty) rightarrow mathbb{R} $ be continuous, and assume that $lim_{x to infty} f(x)$ exists and is a finite number $L$. What can be said about
$$lim_{n to infty}int_{0}^{1} f(nx)dx?$$
Prove your answer.
I am confused as to what the goal of this exercise is and I am not sure where to begin. Any guidance woud be appreciated! Thanks!
real-analysis
real-analysis
asked Jan 22 at 22:27
geoplantedgeoplanted
617
asked Jan 22 at 22:27
geoplantedgeoplanted
617
asked Jan 22 at 22:27
geoplantedgeoplanted
617
asked Jan 22 at 22:27
geoplantedgeoplanted
617
asked Jan 22 at 22:27
geoplantedgeoplanted
617
617
$begingroup$
Have you tried making a picture? Maybe it helps to visualize with a good example such as $f(x)=frac{1+x}{2+x}$.
$endgroup$
– SmileyCraft
Jan 22 at 22:36
$begingroup$
Making the substitution $u=nx$, the integral is $int_{0}^{n} f(u),du$. So if $L ne 0$, the limit definitely doesn't exist. If $L=0$, it's more complicated?
$endgroup$
– preferred_anon
Jan 22 at 22:54
$begingroup$
It is a good problem in analysis, related to convergence results for passing limits through integrals. It is also one where you might "want" to use some type of Lebesgue dominated convergence theorem, but you can also prove it directly without knowing such dominated convergence theorems (the assumptions will help out for some pesky particulars). The first step is to get some intuition, what do you think the answer should be? Can you take steps to prove it?
$endgroup$
– Michael
Jan 22 at 22:55
1
$begingroup$
@preferred_anon : Minor fix it would be $frac{1}{n}int_0^n f(u)du$.
$endgroup$
– Michael
Jan 22 at 22:56
$begingroup$
@Michael oops, thank you!
$endgroup$
– preferred_anon
Jan 22 at 23:02
|
show 1 more comment
$begingroup$
Have you tried making a picture? Maybe it helps to visualize with a good example such as $f(x)=frac{1+x}{2+x}$.
$endgroup$
– SmileyCraft
Jan 22 at 22:36
$begingroup$
Making the substitution $u=nx$, the integral is $int_{0}^{n} f(u),du$. So if $L ne 0$, the limit definitely doesn't exist. If $L=0$, it's more complicated?
$endgroup$
– preferred_anon
Jan 22 at 22:54
$begingroup$
It is a good problem in analysis, related to convergence results for passing limits through integrals. It is also one where you might "want" to use some type of Lebesgue dominated convergence theorem, but you can also prove it directly without knowing such dominated convergence theorems (the assumptions will help out for some pesky particulars). The first step is to get some intuition, what do you think the answer should be? Can you take steps to prove it?
$endgroup$
– Michael
Jan 22 at 22:55
1
$begingroup$
@preferred_anon : Minor fix it would be $frac{1}{n}int_0^n f(u)du$.
$endgroup$
– Michael
Jan 22 at 22:56
$begingroup$
@Michael oops, thank you!
$endgroup$
– preferred_anon
Jan 22 at 23:02
$begingroup$
Have you tried making a picture? Maybe it helps to visualize with a good example such as $f(x)=frac{1+x}{2+x}$.
$endgroup$
– SmileyCraft
Jan 22 at 22:36
$begingroup$
Have you tried making a picture? Maybe it helps to visualize with a good example such as $f(x)=frac{1+x}{2+x}$.
$endgroup$
– SmileyCraft
Jan 22 at 22:36
$begingroup$
Making the substitution $u=nx$, the integral is $int_{0}^{n} f(u),du$. So if $L ne 0$, the limit definitely doesn't exist. If $L=0$, it's more complicated?
$endgroup$
– preferred_anon
Jan 22 at 22:54
$begingroup$
Making the substitution $u=nx$, the integral is $int_{0}^{n} f(u),du$. So if $L ne 0$, the limit definitely doesn't exist. If $L=0$, it's more complicated?
$endgroup$
– preferred_anon
Jan 22 at 22:54
$begingroup$
It is a good problem in analysis, related to convergence results for passing limits through integrals. It is also one where you might "want" to use some type of Lebesgue dominated convergence theorem, but you can also prove it directly without knowing such dominated convergence theorems (the assumptions will help out for some pesky particulars). The first step is to get some intuition, what do you think the answer should be? Can you take steps to prove it?
$endgroup$
– Michael
Jan 22 at 22:55
$begingroup$
It is a good problem in analysis, related to convergence results for passing limits through integrals. It is also one where you might "want" to use some type of Lebesgue dominated convergence theorem, but you can also prove it directly without knowing such dominated convergence theorems (the assumptions will help out for some pesky particulars). The first step is to get some intuition, what do you think the answer should be? Can you take steps to prove it?
$endgroup$
– Michael
Jan 22 at 22:55
1
1
$begingroup$
@preferred_anon : Minor fix it would be $frac{1}{n}int_0^n f(u)du$.
$endgroup$
– Michael
Jan 22 at 22:56
$begingroup$
@preferred_anon : Minor fix it would be $frac{1}{n}int_0^n f(u)du$.
$endgroup$
– Michael
Jan 22 at 22:56
$begingroup$
@Michael oops, thank you!
$endgroup$
– preferred_anon
Jan 22 at 23:02
$begingroup$
@Michael oops, thank you!
$endgroup$
– preferred_anon
Jan 22 at 23:02
|
show 1 more comment
1 Answer
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$|frac 1 n int_0 ^{n} (f(x)-L), dx leq |frac 1 n int_0 ^{c} (f(x)-L), dx+|frac 1 n int_c ^{n} (f(x)-L), dx$. First choose $c$ such that $|f(x)-L| <epsilon$ for $x >c$. Then the second term is less than $epsilon$. Now let $n to infty$ in the first term. Thus $frac 1 n int_0 ^{n} (f(x)-L), dx to 0$. This shows the given limit is $L$ (because $int_0^{1} f(nx), dx =frac 1 n int_0 ^{n} f(x), dx$).
answered Jan 23 at 0:01
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
$endgroup$
$begingroup$
This neglects an important continuity assumption (on the domain of $f$) that is given in the problem and that deals with a "pesky particular" as I mentioned in my comment above. So, part of this problem is to prove something about $f$ based on the given assumptions (after which an argument such as the above, or other similar arguments, or Lebesgue Dominated Convergence, will work).
$endgroup$
– Michael
Jan 25 at 22:08
add a comment |
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$begingroup$
$|frac 1 n int_0 ^{n} (f(x)-L), dx leq |frac 1 n int_0 ^{c} (f(x)-L), dx+|frac 1 n int_c ^{n} (f(x)-L), dx$. First choose $c$ such that $|f(x)-L| <epsilon$ for $x >c$. Then the second term is less than $epsilon$. Now let $n to infty$ in the first term. Thus $frac 1 n int_0 ^{n} (f(x)-L), dx to 0$. This shows the given limit is $L$ (because $int_0^{1} f(nx), dx =frac 1 n int_0 ^{n} f(x), dx$).
answered Jan 23 at 0:01
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
$endgroup$
$begingroup$
This neglects an important continuity assumption (on the domain of $f$) that is given in the problem and that deals with a "pesky particular" as I mentioned in my comment above. So, part of this problem is to prove something about $f$ based on the given assumptions (after which an argument such as the above, or other similar arguments, or Lebesgue Dominated Convergence, will work).
$endgroup$
– Michael
Jan 25 at 22:08
add a comment |
$begingroup$
$|frac 1 n int_0 ^{n} (f(x)-L), dx leq |frac 1 n int_0 ^{c} (f(x)-L), dx+|frac 1 n int_c ^{n} (f(x)-L), dx$. First choose $c$ such that $|f(x)-L| <epsilon$ for $x >c$. Then the second term is less than $epsilon$. Now let $n to infty$ in the first term. Thus $frac 1 n int_0 ^{n} (f(x)-L), dx to 0$. This shows the given limit is $L$ (because $int_0^{1} f(nx), dx =frac 1 n int_0 ^{n} f(x), dx$).
answered Jan 23 at 0:01
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
$endgroup$
$begingroup$
This neglects an important continuity assumption (on the domain of $f$) that is given in the problem and that deals with a "pesky particular" as I mentioned in my comment above. So, part of this problem is to prove something about $f$ based on the given assumptions (after which an argument such as the above, or other similar arguments, or Lebesgue Dominated Convergence, will work).
$endgroup$
– Michael
Jan 25 at 22:08
add a comment |
$begingroup$
$|frac 1 n int_0 ^{n} (f(x)-L), dx leq |frac 1 n int_0 ^{c} (f(x)-L), dx+|frac 1 n int_c ^{n} (f(x)-L), dx$. First choose $c$ such that $|f(x)-L| <epsilon$ for $x >c$. Then the second term is less than $epsilon$. Now let $n to infty$ in the first term. Thus $frac 1 n int_0 ^{n} (f(x)-L), dx to 0$. This shows the given limit is $L$ (because $int_0^{1} f(nx), dx =frac 1 n int_0 ^{n} f(x), dx$).
answered Jan 23 at 0:01
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
$endgroup$
$|frac 1 n int_0 ^{n} (f(x)-L), dx leq |frac 1 n int_0 ^{c} (f(x)-L), dx+|frac 1 n int_c ^{n} (f(x)-L), dx$. First choose $c$ such that $|f(x)-L| <epsilon$ for $x >c$. Then the second term is less than $epsilon$. Now let $n to infty$ in the first term. Thus $frac 1 n int_0 ^{n} (f(x)-L), dx to 0$. This shows the given limit is $L$ (because $int_0^{1} f(nx), dx =frac 1 n int_0 ^{n} f(x), dx$).
answered Jan 23 at 0:01
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
answered Jan 23 at 0:01
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
answered Jan 23 at 0:01
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
answered Jan 23 at 0:01
Kavi Rama MurthyKavi Rama Murthy
66.6k52867
66.6k52867
$begingroup$
This neglects an important continuity assumption (on the domain of $f$) that is given in the problem and that deals with a "pesky particular" as I mentioned in my comment above. So, part of this problem is to prove something about $f$ based on the given assumptions (after which an argument such as the above, or other similar arguments, or Lebesgue Dominated Convergence, will work).
$endgroup$
– Michael
Jan 25 at 22:08
add a comment |
$begingroup$
This neglects an important continuity assumption (on the domain of $f$) that is given in the problem and that deals with a "pesky particular" as I mentioned in my comment above. So, part of this problem is to prove something about $f$ based on the given assumptions (after which an argument such as the above, or other similar arguments, or Lebesgue Dominated Convergence, will work).
$endgroup$
– Michael
Jan 25 at 22:08
$begingroup$
This neglects an important continuity assumption (on the domain of $f$) that is given in the problem and that deals with a "pesky particular" as I mentioned in my comment above. So, part of this problem is to prove something about $f$ based on the given assumptions (after which an argument such as the above, or other similar arguments, or Lebesgue Dominated Convergence, will work).
$endgroup$
– Michael
Jan 25 at 22:08
$begingroup$
This neglects an important continuity assumption (on the domain of $f$) that is given in the problem and that deals with a "pesky particular" as I mentioned in my comment above. So, part of this problem is to prove something about $f$ based on the given assumptions (after which an argument such as the above, or other similar arguments, or Lebesgue Dominated Convergence, will work).
$endgroup$
– Michael
Jan 25 at 22:08
add a comment |
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$begingroup$
Have you tried making a picture? Maybe it helps to visualize with a good example such as $f(x)=frac{1+x}{2+x}$.
$endgroup$
– SmileyCraft
Jan 22 at 22:36
$begingroup$
Making the substitution $u=nx$, the integral is $int_{0}^{n} f(u),du$. So if $L ne 0$, the limit definitely doesn't exist. If $L=0$, it's more complicated?
$endgroup$
– preferred_anon
Jan 22 at 22:54
$begingroup$
It is a good problem in analysis, related to convergence results for passing limits through integrals. It is also one where you might "want" to use some type of Lebesgue dominated convergence theorem, but you can also prove it directly without knowing such dominated convergence theorems (the assumptions will help out for some pesky particulars). The first step is to get some intuition, what do you think the answer should be? Can you take steps to prove it?
$endgroup$
– Michael
Jan 22 at 22:55
1
$begingroup$
@preferred_anon : Minor fix it would be $frac{1}{n}int_0^n f(u)du$.
$endgroup$
– Michael
Jan 22 at 22:56
$begingroup$
@Michael oops, thank you!
$endgroup$
– preferred_anon
Jan 22 at 23:02