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Math Help - DCT

  1. #1
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    DCT

    Here is the question:

    Evaluate:

    <br />
\lim_{n\to\infty} \int_0^{\infty} (1 + {\frac{x}{n}})^{-n} sin{\frac{x}{n}} dx<br />

    <br /> <br /> <br />
\lim_{n\to\infty} \int_0^{1} (1+nx^{2})(1+x^{2})^{-n}dx<br /> <br />

    For the first integral I use the DCT, noting that (1 + {\frac{x}{n}})^{-n} converges to e^{-x} and that sin is bounded by 1. Then taking the limit inside, the integrand converges to 0 and therefore the upper integral is equal to 0. Is this correct? Any hints on the second one would be appreciated.
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  2. #2
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    Quote Originally Posted by davidmccormick View Post
    Here is the question:

    Evaluate:

    <br />
\lim_{n\to\infty} \int_0^{\infty} (1 + {\frac{x}{n}})^{-n} sin{\frac{x}{n}} dx<br />

    <br /> <br /> <br />
\lim_{n\to\infty} \int_0^{1} (1+nx^{2})(1+x^{2})^{-n}dx<br /> <br />

    For the first integral I use the DCT, noting that (1 + {\frac{x}{n}})^{-n} converges to e^{-x} and that sin is bounded by 1. Then taking the limit inside, the integrand converges to 0 and therefore the upper integral is equal to 0. Is this correct? Any hints on the second one would be appreciated.
    DCT looks correct for the first one. The answer to the second one also appears to be 0. Notice that the integrand (1+nx^{2})(1+x^{2})^{-n} is a decreasing function on the interval [0,1], so the upper Riemann sum for any partition is obtained by evaluating the function at the left endpoint of each subinterval. Form that sum for the partition [0,1] = [0,\varepsilon]\cup [\varepsilon,1], and show that it is less than 2\varepsilon for all sufficiently large n.
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  3. #3
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    Quote Originally Posted by davidmccormick View Post
    Evaluate:

    <br />
\lim_{n\to\infty} \int_0^{\infty} (1 + {\frac{x}{n}})^{-n} sin{\frac{x}{n}} dx<br />

    For the first integral I use the DCT, noting that (1 + {\frac{x}{n}})^{-n} converges to e^{-x} and that sin is bounded by 1. Then taking the limit inside, the integrand converges to 0 and therefore the upper integral is equal to 0. Is this correct?
    So what "dominating" function are you using for the first one? Sure (1 + {\frac{x}{n}})^{-n} converges to e^{-x} but you need a bound that wouldn't depend on n and would be integrable. To that aim, you can use (if proven) the fact that (1+\frac{x}{n})^n increases with n. Alternatively, you can use (1+\frac{x}{n})^n\geq 1+x+\frac{(n-1)}{2n}x^2\geq 1+x+\frac{1}{4}x^2 for n\geq 2 from the binomial expansion.


    <br /> <br /> <br />
\lim_{n\to\infty} \int_0^{1} (1+nx^{2})(1+x^{2})^{-n}dx<br /> <br />
    With DCT, the easiest here (this is a one-liner) is just to use the inequality 1+nx^2\leq (1+x^2)^n, that follows from the binomial formula. And notice that the integrand converges to 0 as n\to\infty (polynomial vs. exponential).
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  4. #4
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    Thanks, you've got brilliant insight. How do you do it, just by doing lots of problems and practice or you'd be able to do the same with an unseen problem?
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  5. #5
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    Quote Originally Posted by davidmccormick View Post
    Thanks, you've got brilliant insight. How do you do it, just by doing lots of problems and practice or you'd be able to do the same with an unseen problem?
    Sure there's a lot of practice, and after some time, you also will develop a kind of familiarity with functions that helps visualize their main features without graphing them. I didn't know this specific problem, and certainly didn't Opalg either, but there are numerous similar ones circulating. Keep working; after a while, this becomes simpler! Look at what you did last year, or the year before. A good thing about maths is that we keep dealing with the same objects but with a gradually increasing precision and generality, so that the same problem becomes easier as the years pass...
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