How to use Lebesgue Dominated Convergence theorem in this example?
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I have to use the DCT in order to get the value of this limit:
$$ lim_{nto infty} int_0^n biggl(frac{sin x}{x} biggr)^n dx$$
If i take $f_n = bigl(frac{sin x}{x} bigr)^n, lim_{nto infty} f_n =0 $ so it converges, but i can't tell what function to use for the domination.
My thought is that as $|sin x| le |x| $, and $space bigl|left(frac{sin x}{x} right)^n bigr| le space left|frac{sin x}{x} right|^n $ then $ space left|frac{sin x}{x} right|^n le 1$ but i feel like I'm messing up hard somewhere.
How would you proof that function is dominated?
Thanks in advance.
real-analysis measure-theory convergence
asked Jan 6 at 11:21
MontyroMontyro
32
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add a comment |
$begingroup$
I have to use the DCT in order to get the value of this limit:
$$ lim_{nto infty} int_0^n biggl(frac{sin x}{x} biggr)^n dx$$
If i take $f_n = bigl(frac{sin x}{x} bigr)^n, lim_{nto infty} f_n =0 $ so it converges, but i can't tell what function to use for the domination.
My thought is that as $|sin x| le |x| $, and $space bigl|left(frac{sin x}{x} right)^n bigr| le space left|frac{sin x}{x} right|^n $ then $ space left|frac{sin x}{x} right|^n le 1$ but i feel like I'm messing up hard somewhere.
How would you proof that function is dominated?
Thanks in advance.
real-analysis measure-theory convergence
asked Jan 6 at 11:21
MontyroMontyro
32
$endgroup$
add a comment |
$begingroup$
I have to use the DCT in order to get the value of this limit:
$$ lim_{nto infty} int_0^n biggl(frac{sin x}{x} biggr)^n dx$$
If i take $f_n = bigl(frac{sin x}{x} bigr)^n, lim_{nto infty} f_n =0 $ so it converges, but i can't tell what function to use for the domination.
My thought is that as $|sin x| le |x| $, and $space bigl|left(frac{sin x}{x} right)^n bigr| le space left|frac{sin x}{x} right|^n $ then $ space left|frac{sin x}{x} right|^n le 1$ but i feel like I'm messing up hard somewhere.
How would you proof that function is dominated?
Thanks in advance.
real-analysis measure-theory convergence
asked Jan 6 at 11:21
MontyroMontyro
32
$endgroup$
I have to use the DCT in order to get the value of this limit:
$$ lim_{nto infty} int_0^n biggl(frac{sin x}{x} biggr)^n dx$$
If i take $f_n = bigl(frac{sin x}{x} bigr)^n, lim_{nto infty} f_n =0 $ so it converges, but i can't tell what function to use for the domination.
My thought is that as $|sin x| le |x| $, and $space bigl|left(frac{sin x}{x} right)^n bigr| le space left|frac{sin x}{x} right|^n $ then $ space left|frac{sin x}{x} right|^n le 1$ but i feel like I'm messing up hard somewhere.
How would you proof that function is dominated?
Thanks in advance.
real-analysis measure-theory convergence
real-analysis measure-theory convergence
asked Jan 6 at 11:21
MontyroMontyro
32
asked Jan 6 at 11:21
MontyroMontyro
32
asked Jan 6 at 11:21
MontyroMontyro
32
asked Jan 6 at 11:21
MontyroMontyro
32
asked Jan 6 at 11:21
MontyroMontyro
32
32
add a comment |
add a comment |
1 Answer
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oldest
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To use DCT, you have to be on the same measure space. So you should view the original integral as $int_mathbb{R} (frac{sin x}{x})^n 1_{0 le x le n} dx$. Let $f_n(x) = (frac{sin x}{x})^n 1_{0 le x le n}$ and $f(x) = 1$ for $0 le x le 1$ and $f(x) = frac{1}{x^2}$ for $x > 1$. Then $|f_n(x)| le f(x)$ for each $n ge 2$ and $x ge 0$. Since $f in L^1$, we can apply DCT.
answered Jan 6 at 11:26
mathworker21mathworker21
9,4721929
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$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To use DCT, you have to be on the same measure space. So you should view the original integral as $int_mathbb{R} (frac{sin x}{x})^n 1_{0 le x le n} dx$. Let $f_n(x) = (frac{sin x}{x})^n 1_{0 le x le n}$ and $f(x) = 1$ for $0 le x le 1$ and $f(x) = frac{1}{x^2}$ for $x > 1$. Then $|f_n(x)| le f(x)$ for each $n ge 2$ and $x ge 0$. Since $f in L^1$, we can apply DCT.
answered Jan 6 at 11:26
mathworker21mathworker21
9,4721929
$endgroup$
$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
add a comment |
$begingroup$
To use DCT, you have to be on the same measure space. So you should view the original integral as $int_mathbb{R} (frac{sin x}{x})^n 1_{0 le x le n} dx$. Let $f_n(x) = (frac{sin x}{x})^n 1_{0 le x le n}$ and $f(x) = 1$ for $0 le x le 1$ and $f(x) = frac{1}{x^2}$ for $x > 1$. Then $|f_n(x)| le f(x)$ for each $n ge 2$ and $x ge 0$. Since $f in L^1$, we can apply DCT.
answered Jan 6 at 11:26
mathworker21mathworker21
9,4721929
$endgroup$
$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
add a comment |
$begingroup$
To use DCT, you have to be on the same measure space. So you should view the original integral as $int_mathbb{R} (frac{sin x}{x})^n 1_{0 le x le n} dx$. Let $f_n(x) = (frac{sin x}{x})^n 1_{0 le x le n}$ and $f(x) = 1$ for $0 le x le 1$ and $f(x) = frac{1}{x^2}$ for $x > 1$. Then $|f_n(x)| le f(x)$ for each $n ge 2$ and $x ge 0$. Since $f in L^1$, we can apply DCT.
answered Jan 6 at 11:26
mathworker21mathworker21
9,4721929
$endgroup$
To use DCT, you have to be on the same measure space. So you should view the original integral as $int_mathbb{R} (frac{sin x}{x})^n 1_{0 le x le n} dx$. Let $f_n(x) = (frac{sin x}{x})^n 1_{0 le x le n}$ and $f(x) = 1$ for $0 le x le 1$ and $f(x) = frac{1}{x^2}$ for $x > 1$. Then $|f_n(x)| le f(x)$ for each $n ge 2$ and $x ge 0$. Since $f in L^1$, we can apply DCT.
answered Jan 6 at 11:26
mathworker21mathworker21
9,4721929
answered Jan 6 at 11:26
mathworker21mathworker21
9,4721929
answered Jan 6 at 11:26
mathworker21mathworker21
9,4721929
answered Jan 6 at 11:26
mathworker21mathworker21
9,4721929
9,4721929
$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
add a comment |
$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
$begingroup$
Thank you for the answer, i think i see this now!
$endgroup$
– Montyro
Jan 6 at 11:43
add a comment |
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