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February 22nd, 2019, 08:44 AM   #1
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Lightbulb How to approach this integral?

How approach this integral? I have assumed z=nx and hence
Then derivative gives #dz=ndx#. Now it becomes this
Integral from #\frac{n} {n+1}# to 1 ##\frac{arctan(z)dz} {n arcsin(z)}##. After this I am not able to proceed further.
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February 22nd, 2019, 10:33 AM   #2
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Where did you get the question from? The fourth answer seems to be correct.
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February 22nd, 2019, 01:09 PM   #3
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To get the answer, note that for large n the integrand $\approx \frac{tan^{-1}(1)}{sin^{-1}(1)}=\frac{\pi/4}{\pi/2}=\frac{1}{2}$, so $C_n\approx \frac{1}{2n^2}$
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February 22nd, 2019, 04:12 PM   #4
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Quote:
Originally Posted by mathman View Post
To get the answer, note that for large n the integrand $\approx \frac{tan^{-1}(1)}{sin^{-1}(1)}=\frac{\pi/4}{\pi/2}=\frac{1}{2}$, so $C_n\approx \frac{1}{2n^2}$
Interchanging limits like that is not always allowed. For example, your analysis applied to the following sequence would give a wrong answer.
\[ C_n = \int_0^{n} 1/n \ dx \]
In general I would say its a bad idea to think about it this way even if just for "intuition" since the intuition gained is wrong.
Justifying when you can interchange limits or not typically requires the dominated convergence theorem or monotone convergence theorem. Both of which are a bit advanced for a calculus student.
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February 23rd, 2019, 01:58 PM   #5
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Quote:
Originally Posted by SDK View Post
Interchanging limits like that is not always allowed. For example, your analysis applied to the following sequence would give a wrong answer.
\[ C_n = \int_0^{n} 1/n \ dx \]
In general I would say its a bad idea to think about it this way even if just for "intuition" since the intuition gained is wrong.
Justifying when you can interchange limits or not typically requires the dominated convergence theorem or monotone convergence theorem. Both of which are a bit advanced for a calculus student.
In your example $C_n=1$ for all $n$. What is the problem?

What is wrong with my answer. It could have been expressed with usual $\epsilon, n$ notation, but the result would be the same.
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February 23rd, 2019, 04:04 PM   #6
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In your example $C_n=1$ for all $n$. What is the problem?

What is wrong with my answer. It could have been expressed with usual $\epsilon, n$ notation, but the result would be the same.
Yes $C_n = 1$ for all $n$. The problem is that
$\int_0^n \lim_{n \to \infty} 1/n \ dx = 0 $ for all $n$ so obviously the limit exchange,
\[ \lim_{n \to \infty} \int_0^n 1/n \ dx = \int_0^n \lim_{n \to \infty} 1/n \ dx \]
fails in this example. But this limit exchange is exactly what you implied in your example without justifying why it is allowed in that case.
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February 24th, 2019, 01:29 PM   #7
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Quote:
Originally Posted by SDK View Post
Yes $C_n = 1$ for all $n$. The problem is that
$\int_0^n \lim_{n \to \infty} 1/n \ dx = 0 $ for all $n$ so obviously the limit exchange,
\[ \lim_{n \to \infty} \int_0^n 1/n \ dx = \int_0^n \lim_{n \to \infty} 1/n \ dx \]
fails in this example. But this limit exchange is exactly what you implied in your example without justifying why it is allowed in that case.
I should have given more detail for the first step. In the range of integration $tan^{-1}(1-\frac{1}{n+1})\le tan^{-1}(nx) \le tan^{-1}(1)$, similarly for $sin^{-1}(nx)$, so that the integrand can be approximated by a constant $k=\frac{tan^{-1}(1)}{sin^{-1}(1)}$ and the integral $\int_{1/(n+1)}^{1/n}dx$ becomes $\approx \frac{k}{n(n+1)}\approx \frac{k}{n^2}$, where $k=1/2$.
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