Converting an Unconventional Riemann Sum to an Indefinite Integral
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I am trying to convert the following Riemann Sum into an indefinite integral:
$$lim_{n to infty} Bigl(nx_{n+1}^m +sum_{k=1}^n -x_k^m Bigr)$$
where $m$ is an arbitrary constant. I am wondering if this sum could be converted to an integral in the first place, given that pesky $nx_{n+1}^m$ term before the summation.
I have tried to look at other Riemann Sum-Integral conversions - trapezoidal approximation, center of mass integrals - but none of them seem to help in converting this particular limit. If anyone could help me out here, it would be much appreciated. Thanks!
calculus limits
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add a comment |
$begingroup$
I am trying to convert the following Riemann Sum into an indefinite integral:
$$lim_{n to infty} Bigl(nx_{n+1}^m +sum_{k=1}^n -x_k^m Bigr)$$
where $m$ is an arbitrary constant. I am wondering if this sum could be converted to an integral in the first place, given that pesky $nx_{n+1}^m$ term before the summation.
I have tried to look at other Riemann Sum-Integral conversions - trapezoidal approximation, center of mass integrals - but none of them seem to help in converting this particular limit. If anyone could help me out here, it would be much appreciated. Thanks!
calculus limits
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This equality might be of help $nx_{n+1}^m-sum_{k=1}^nx_k^m=sum_{k=1}^n(n-k+1)(x_{k+1}^m-x_k^m)$.
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– John_Wick
Dec 20 '18 at 19:45
1
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@John_Wick: I get $sum_{k=1}^nk!left(x_{k+1}^m-x_k^mright)$
$endgroup$
– robjohn♦
Dec 20 '18 at 20:27
add a comment |
$begingroup$
I am trying to convert the following Riemann Sum into an indefinite integral:
$$lim_{n to infty} Bigl(nx_{n+1}^m +sum_{k=1}^n -x_k^m Bigr)$$
where $m$ is an arbitrary constant. I am wondering if this sum could be converted to an integral in the first place, given that pesky $nx_{n+1}^m$ term before the summation.
I have tried to look at other Riemann Sum-Integral conversions - trapezoidal approximation, center of mass integrals - but none of them seem to help in converting this particular limit. If anyone could help me out here, it would be much appreciated. Thanks!
calculus limits
$endgroup$
I am trying to convert the following Riemann Sum into an indefinite integral:
$$lim_{n to infty} Bigl(nx_{n+1}^m +sum_{k=1}^n -x_k^m Bigr)$$
where $m$ is an arbitrary constant. I am wondering if this sum could be converted to an integral in the first place, given that pesky $nx_{n+1}^m$ term before the summation.
I have tried to look at other Riemann Sum-Integral conversions - trapezoidal approximation, center of mass integrals - but none of them seem to help in converting this particular limit. If anyone could help me out here, it would be much appreciated. Thanks!
calculus limits
calculus limits
edited Dec 20 '18 at 19:45
chaad
asked Dec 20 '18 at 19:39
chaadchaad
11
11
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This equality might be of help $nx_{n+1}^m-sum_{k=1}^nx_k^m=sum_{k=1}^n(n-k+1)(x_{k+1}^m-x_k^m)$.
$endgroup$
– John_Wick
Dec 20 '18 at 19:45
1
$begingroup$
@John_Wick: I get $sum_{k=1}^nk!left(x_{k+1}^m-x_k^mright)$
$endgroup$
– robjohn♦
Dec 20 '18 at 20:27
add a comment |
$begingroup$
This equality might be of help $nx_{n+1}^m-sum_{k=1}^nx_k^m=sum_{k=1}^n(n-k+1)(x_{k+1}^m-x_k^m)$.
$endgroup$
– John_Wick
Dec 20 '18 at 19:45
1
$begingroup$
@John_Wick: I get $sum_{k=1}^nk!left(x_{k+1}^m-x_k^mright)$
$endgroup$
– robjohn♦
Dec 20 '18 at 20:27
$begingroup$
This equality might be of help $nx_{n+1}^m-sum_{k=1}^nx_k^m=sum_{k=1}^n(n-k+1)(x_{k+1}^m-x_k^m)$.
$endgroup$
– John_Wick
Dec 20 '18 at 19:45
$begingroup$
This equality might be of help $nx_{n+1}^m-sum_{k=1}^nx_k^m=sum_{k=1}^n(n-k+1)(x_{k+1}^m-x_k^m)$.
$endgroup$
– John_Wick
Dec 20 '18 at 19:45
1
1
$begingroup$
@John_Wick: I get $sum_{k=1}^nk!left(x_{k+1}^m-x_k^mright)$
$endgroup$
– robjohn♦
Dec 20 '18 at 20:27
$begingroup$
@John_Wick: I get $sum_{k=1}^nk!left(x_{k+1}^m-x_k^mright)$
$endgroup$
– robjohn♦
Dec 20 '18 at 20:27
add a comment |
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$begingroup$
This equality might be of help $nx_{n+1}^m-sum_{k=1}^nx_k^m=sum_{k=1}^n(n-k+1)(x_{k+1}^m-x_k^m)$.
$endgroup$
– John_Wick
Dec 20 '18 at 19:45
1
$begingroup$
@John_Wick: I get $sum_{k=1}^nk!left(x_{k+1}^m-x_k^mright)$
$endgroup$
– robjohn♦
Dec 20 '18 at 20:27