Show that $ lim_{jtoinfty}prod_{n=j}^{mj}frac{pi}{2tan^{-1}(kn)}=m^{frac{2}{kpi}}$
$begingroup$
MathWorld states that
$displaystyle lim_{jtoinfty}prod_{n=j}^{2j}frac{pi}{2tan^{-1}n}=4^{frac{1}{pi}}$
(equation 130)
With a calculator, I find a general formula:
$displaystyle lim_{jtoinfty}prod_{n=j}^{mj}frac{pi}{2tan^{-1}(kn)}=m^{frac{2}{kpi}}$
But I have not found a proof for this. Any proof would be appreciated.
sequences-and-series
asked Aug 26 '18 at 23:15
LarryLarry
2,55031131
$endgroup$
add a comment |
$begingroup$
MathWorld states that
$displaystyle lim_{jtoinfty}prod_{n=j}^{2j}frac{pi}{2tan^{-1}n}=4^{frac{1}{pi}}$
(equation 130)
With a calculator, I find a general formula:
$displaystyle lim_{jtoinfty}prod_{n=j}^{mj}frac{pi}{2tan^{-1}(kn)}=m^{frac{2}{kpi}}$
But I have not found a proof for this. Any proof would be appreciated.
sequences-and-series
asked Aug 26 '18 at 23:15
LarryLarry
2,55031131
$endgroup$
add a comment |
$begingroup$
MathWorld states that
$displaystyle lim_{jtoinfty}prod_{n=j}^{2j}frac{pi}{2tan^{-1}n}=4^{frac{1}{pi}}$
(equation 130)
With a calculator, I find a general formula:
$displaystyle lim_{jtoinfty}prod_{n=j}^{mj}frac{pi}{2tan^{-1}(kn)}=m^{frac{2}{kpi}}$
But I have not found a proof for this. Any proof would be appreciated.
sequences-and-series
asked Aug 26 '18 at 23:15
LarryLarry
2,55031131
$endgroup$
MathWorld states that
$displaystyle lim_{jtoinfty}prod_{n=j}^{2j}frac{pi}{2tan^{-1}n}=4^{frac{1}{pi}}$
(equation 130)
With a calculator, I find a general formula:
$displaystyle lim_{jtoinfty}prod_{n=j}^{mj}frac{pi}{2tan^{-1}(kn)}=m^{frac{2}{kpi}}$
But I have not found a proof for this. Any proof would be appreciated.
sequences-and-series
sequences-and-series
asked Aug 26 '18 at 23:15
LarryLarry
2,55031131
asked Aug 26 '18 at 23:15
LarryLarry
2,55031131
edited Dec 21 '18 at 22:54
Larry
asked Aug 26 '18 at 23:15
LarryLarry
2,55031131
asked Aug 26 '18 at 23:15
LarryLarry
2,55031131
asked Aug 26 '18 at 23:15
LarryLarry
2,55031131
2,55031131
add a comment |
add a comment |
1 Answer
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$begingroup$
It is easier to analyze the behavior of the limit if you take logarithm. Indeed, write
$$ P_j = P_j(m,k) = prod_{n=j}^{mj} frac{pi}{2arctan(kn)} $$
for the expression inside the limit. Then using the relation $arctan(x) = frac{pi}{2} - arctan(frac{1}{x})$ which holds for $x > 0$, we find that
$$ logleft(frac{pi}{2arctan(x)}right)
= -logleft(1 - tfrac{2}{pi}arctanleft(tfrac{1}{x}right) right)
= tfrac{2}{pi x} + mathcal{O}left(x^{-2}right)
quad text{as} quad x to infty.$$
So it follows that
$$
log P_j
= sum_{n=j}^{mj} left( logfrac{pi}{2} - logarctan(kn)right)
= sum_{n=j}^{mj} left( frac{2}{pi kn} + mathcal{O}(n^{-2}) right).
$$
By noting that this is a Riemann sum with a vanishing error term, we realize that $log P_n$ converges to $int_{1}^{m} frac{2}{pi kx} , dx
= frac{2}{pi k}log m$ as $ntoinfty$. Exponentiation then yields $ P_n to m^{frac{2}{pi k}} $ as expected.
answered Aug 26 '18 at 23:37
Sangchul LeeSangchul Lee
96.6k12173283
$endgroup$
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
$begingroup$
It is easier to analyze the behavior of the limit if you take logarithm. Indeed, write
$$ P_j = P_j(m,k) = prod_{n=j}^{mj} frac{pi}{2arctan(kn)} $$
for the expression inside the limit. Then using the relation $arctan(x) = frac{pi}{2} - arctan(frac{1}{x})$ which holds for $x > 0$, we find that
$$ logleft(frac{pi}{2arctan(x)}right)
= -logleft(1 - tfrac{2}{pi}arctanleft(tfrac{1}{x}right) right)
= tfrac{2}{pi x} + mathcal{O}left(x^{-2}right)
quad text{as} quad x to infty.$$
So it follows that
$$
log P_j
= sum_{n=j}^{mj} left( logfrac{pi}{2} - logarctan(kn)right)
= sum_{n=j}^{mj} left( frac{2}{pi kn} + mathcal{O}(n^{-2}) right).
$$
By noting that this is a Riemann sum with a vanishing error term, we realize that $log P_n$ converges to $int_{1}^{m} frac{2}{pi kx} , dx
= frac{2}{pi k}log m$ as $ntoinfty$. Exponentiation then yields $ P_n to m^{frac{2}{pi k}} $ as expected.
answered Aug 26 '18 at 23:37
Sangchul LeeSangchul Lee
96.6k12173283
$endgroup$
add a comment |
$begingroup$
It is easier to analyze the behavior of the limit if you take logarithm. Indeed, write
$$ P_j = P_j(m,k) = prod_{n=j}^{mj} frac{pi}{2arctan(kn)} $$
for the expression inside the limit. Then using the relation $arctan(x) = frac{pi}{2} - arctan(frac{1}{x})$ which holds for $x > 0$, we find that
$$ logleft(frac{pi}{2arctan(x)}right)
= -logleft(1 - tfrac{2}{pi}arctanleft(tfrac{1}{x}right) right)
= tfrac{2}{pi x} + mathcal{O}left(x^{-2}right)
quad text{as} quad x to infty.$$
So it follows that
$$
log P_j
= sum_{n=j}^{mj} left( logfrac{pi}{2} - logarctan(kn)right)
= sum_{n=j}^{mj} left( frac{2}{pi kn} + mathcal{O}(n^{-2}) right).
$$
By noting that this is a Riemann sum with a vanishing error term, we realize that $log P_n$ converges to $int_{1}^{m} frac{2}{pi kx} , dx
= frac{2}{pi k}log m$ as $ntoinfty$. Exponentiation then yields $ P_n to m^{frac{2}{pi k}} $ as expected.
answered Aug 26 '18 at 23:37
Sangchul LeeSangchul Lee
96.6k12173283
$endgroup$
add a comment |
$begingroup$
It is easier to analyze the behavior of the limit if you take logarithm. Indeed, write
$$ P_j = P_j(m,k) = prod_{n=j}^{mj} frac{pi}{2arctan(kn)} $$
for the expression inside the limit. Then using the relation $arctan(x) = frac{pi}{2} - arctan(frac{1}{x})$ which holds for $x > 0$, we find that
$$ logleft(frac{pi}{2arctan(x)}right)
= -logleft(1 - tfrac{2}{pi}arctanleft(tfrac{1}{x}right) right)
= tfrac{2}{pi x} + mathcal{O}left(x^{-2}right)
quad text{as} quad x to infty.$$
So it follows that
$$
log P_j
= sum_{n=j}^{mj} left( logfrac{pi}{2} - logarctan(kn)right)
= sum_{n=j}^{mj} left( frac{2}{pi kn} + mathcal{O}(n^{-2}) right).
$$
By noting that this is a Riemann sum with a vanishing error term, we realize that $log P_n$ converges to $int_{1}^{m} frac{2}{pi kx} , dx
= frac{2}{pi k}log m$ as $ntoinfty$. Exponentiation then yields $ P_n to m^{frac{2}{pi k}} $ as expected.
answered Aug 26 '18 at 23:37
Sangchul LeeSangchul Lee
96.6k12173283
$endgroup$
It is easier to analyze the behavior of the limit if you take logarithm. Indeed, write
$$ P_j = P_j(m,k) = prod_{n=j}^{mj} frac{pi}{2arctan(kn)} $$
for the expression inside the limit. Then using the relation $arctan(x) = frac{pi}{2} - arctan(frac{1}{x})$ which holds for $x > 0$, we find that
$$ logleft(frac{pi}{2arctan(x)}right)
= -logleft(1 - tfrac{2}{pi}arctanleft(tfrac{1}{x}right) right)
= tfrac{2}{pi x} + mathcal{O}left(x^{-2}right)
quad text{as} quad x to infty.$$
So it follows that
$$
log P_j
= sum_{n=j}^{mj} left( logfrac{pi}{2} - logarctan(kn)right)
= sum_{n=j}^{mj} left( frac{2}{pi kn} + mathcal{O}(n^{-2}) right).
$$
By noting that this is a Riemann sum with a vanishing error term, we realize that $log P_n$ converges to $int_{1}^{m} frac{2}{pi kx} , dx
= frac{2}{pi k}log m$ as $ntoinfty$. Exponentiation then yields $ P_n to m^{frac{2}{pi k}} $ as expected.
answered Aug 26 '18 at 23:37
Sangchul LeeSangchul Lee
96.6k12173283
answered Aug 26 '18 at 23:37
Sangchul LeeSangchul Lee
96.6k12173283
answered Aug 26 '18 at 23:37
Sangchul LeeSangchul Lee
96.6k12173283
answered Aug 26 '18 at 23:37
Sangchul LeeSangchul Lee
96.6k12173283
96.6k12173283
add a comment |
add a comment |
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