Sigma notation for sum of $ln(x)^2$ from $2$ to $20$ with steps of $0.5$
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Is it possible to use sigma notation for non-integer steps, for example I want to sum $ln(x)^2$ from $2$ to $20$ with steps of $0.5$, is there a way I could write this in sigma notation or some other form of notation.
summation notation
edited Feb 11 at 15:37
user21820
39.3k543154
asked Feb 10 at 19:26
H.LinkhornH.Linkhorn
473113
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add a comment |
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Is it possible to use sigma notation for non-integer steps, for example I want to sum $ln(x)^2$ from $2$ to $20$ with steps of $0.5$, is there a way I could write this in sigma notation or some other form of notation.
summation notation
edited Feb 11 at 15:37
user21820
39.3k543154
asked Feb 10 at 19:26
H.LinkhornH.Linkhorn
473113
$endgroup$
add a comment |
$begingroup$
Is it possible to use sigma notation for non-integer steps, for example I want to sum $ln(x)^2$ from $2$ to $20$ with steps of $0.5$, is there a way I could write this in sigma notation or some other form of notation.
summation notation
edited Feb 11 at 15:37
user21820
39.3k543154
asked Feb 10 at 19:26
H.LinkhornH.Linkhorn
473113
$endgroup$
Is it possible to use sigma notation for non-integer steps, for example I want to sum $ln(x)^2$ from $2$ to $20$ with steps of $0.5$, is there a way I could write this in sigma notation or some other form of notation.
summation notation
summation notation
edited Feb 11 at 15:37
user21820
39.3k543154
asked Feb 10 at 19:26
H.LinkhornH.Linkhorn
473113
edited Feb 11 at 15:37
user21820
39.3k543154
asked Feb 10 at 19:26
H.LinkhornH.Linkhorn
473113
edited Feb 11 at 15:37
user21820
39.3k543154
edited Feb 11 at 15:37
user21820
39.3k543154
edited Feb 11 at 15:37
user21820
39.3k543154
39.3k543154
asked Feb 10 at 19:26
H.LinkhornH.Linkhorn
473113
asked Feb 10 at 19:26
H.LinkhornH.Linkhorn
473113
asked Feb 10 at 19:26
H.LinkhornH.Linkhorn
473113
473113
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
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In this particular case that you have the constant difference, I would go with the other answers; it's the simplest and most non-confusing manner to write what you want to convey. However, if you have any arbitrary set $S$, and wanted to sum based on its elements, you can write something like
$$sum_sin Sf(s).$$
In particular you might have $S=2,2.5,3,dots,20$.
answered Feb 10 at 22:38
YiFanYiFan
4,4091627
$endgroup$
$begingroup$
Arbitrary finite set (or else invoke topology and convergence conditions).
$endgroup$
– Marc van Leeuwen
Feb 11 at 10:17
1
$begingroup$
You could also index over the set $S$, to get $sum_i=0^nf(s_i)$, where $S=s_0,s_1,ldots,s_n$.
$endgroup$
– Jam
Feb 11 at 10:20
add a comment |
$begingroup$
Does $$sum_n = 0^36lnleft(2+dfrac12nright)^2$$ satisfy your requirements?
answered Feb 10 at 19:29
JimmyK4542JimmyK4542
41.2k245107
$endgroup$
add a comment |
$begingroup$
You could change $ln(x)^2$ into $lnleft(fracx2right)^2$ to achieve the steps of $0.5$ in this case. You want $fracx2$ to go from $2$ to $20$ as well (with steps of $0.5$), so $x$ must go from $4$ to $40$. Therefore, the sum becomes$sum_limitsx = 4^40lnleft(fracx2right)^2$.
answered Feb 10 at 19:32
KM101KM101
6,0701525
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add a comment |
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Just use $fracx2$ instead of $x$ in this example to get integers. This trick can always be used when we have to sum up finite many rationals or rationals with a limited denominator.
answered Feb 10 at 19:30
PeterPeter
48.3k1139133
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In this particular case that you have the constant difference, I would go with the other answers; it's the simplest and most non-confusing manner to write what you want to convey. However, if you have any arbitrary set $S$, and wanted to sum based on its elements, you can write something like
$$sum_sin Sf(s).$$
In particular you might have $S=2,2.5,3,dots,20$.
answered Feb 10 at 22:38
YiFanYiFan
4,4091627
$endgroup$
$begingroup$
Arbitrary finite set (or else invoke topology and convergence conditions).
$endgroup$
– Marc van Leeuwen
Feb 11 at 10:17
1
$begingroup$
You could also index over the set $S$, to get $sum_i=0^nf(s_i)$, where $S=s_0,s_1,ldots,s_n$.
$endgroup$
– Jam
Feb 11 at 10:20
add a comment |
$begingroup$
In this particular case that you have the constant difference, I would go with the other answers; it's the simplest and most non-confusing manner to write what you want to convey. However, if you have any arbitrary set $S$, and wanted to sum based on its elements, you can write something like
$$sum_sin Sf(s).$$
In particular you might have $S=2,2.5,3,dots,20$.
answered Feb 10 at 22:38
YiFanYiFan
4,4091627
$endgroup$
$begingroup$
Arbitrary finite set (or else invoke topology and convergence conditions).
$endgroup$
– Marc van Leeuwen
Feb 11 at 10:17
1
$begingroup$
You could also index over the set $S$, to get $sum_i=0^nf(s_i)$, where $S=s_0,s_1,ldots,s_n$.
$endgroup$
– Jam
Feb 11 at 10:20
add a comment |
$begingroup$
In this particular case that you have the constant difference, I would go with the other answers; it's the simplest and most non-confusing manner to write what you want to convey. However, if you have any arbitrary set $S$, and wanted to sum based on its elements, you can write something like
$$sum_sin Sf(s).$$
In particular you might have $S=2,2.5,3,dots,20$.
answered Feb 10 at 22:38
YiFanYiFan
4,4091627
$endgroup$
In this particular case that you have the constant difference, I would go with the other answers; it's the simplest and most non-confusing manner to write what you want to convey. However, if you have any arbitrary set $S$, and wanted to sum based on its elements, you can write something like
$$sum_sin Sf(s).$$
In particular you might have $S=2,2.5,3,dots,20$.
answered Feb 10 at 22:38
YiFanYiFan
4,4091627
answered Feb 10 at 22:38
YiFanYiFan
4,4091627
answered Feb 10 at 22:38
YiFanYiFan
4,4091627
answered Feb 10 at 22:38
YiFanYiFan
4,4091627
4,4091627
$begingroup$
Arbitrary finite set (or else invoke topology and convergence conditions).
$endgroup$
– Marc van Leeuwen
Feb 11 at 10:17
1
$begingroup$
You could also index over the set $S$, to get $sum_i=0^nf(s_i)$, where $S=s_0,s_1,ldots,s_n$.
$endgroup$
– Jam
Feb 11 at 10:20
add a comment |
$begingroup$
Arbitrary finite set (or else invoke topology and convergence conditions).
$endgroup$
– Marc van Leeuwen
Feb 11 at 10:17
1
$begingroup$
You could also index over the set $S$, to get $sum_i=0^nf(s_i)$, where $S=s_0,s_1,ldots,s_n$.
$endgroup$
– Jam
Feb 11 at 10:20
$begingroup$
Arbitrary finite set (or else invoke topology and convergence conditions).
$endgroup$
– Marc van Leeuwen
Feb 11 at 10:17
$begingroup$
Arbitrary finite set (or else invoke topology and convergence conditions).
$endgroup$
– Marc van Leeuwen
Feb 11 at 10:17
1
1
$begingroup$
You could also index over the set $S$, to get $sum_i=0^nf(s_i)$, where $S=s_0,s_1,ldots,s_n$.
$endgroup$
– Jam
Feb 11 at 10:20
$begingroup$
You could also index over the set $S$, to get $sum_i=0^nf(s_i)$, where $S=s_0,s_1,ldots,s_n$.
$endgroup$
– Jam
Feb 11 at 10:20
add a comment |
$begingroup$
Does $$sum_n = 0^36lnleft(2+dfrac12nright)^2$$ satisfy your requirements?
answered Feb 10 at 19:29
JimmyK4542JimmyK4542
41.2k245107
$endgroup$
add a comment |
$begingroup$
Does $$sum_n = 0^36lnleft(2+dfrac12nright)^2$$ satisfy your requirements?
answered Feb 10 at 19:29
JimmyK4542JimmyK4542
41.2k245107
$endgroup$
add a comment |
$begingroup$
Does $$sum_n = 0^36lnleft(2+dfrac12nright)^2$$ satisfy your requirements?
answered Feb 10 at 19:29
JimmyK4542JimmyK4542
41.2k245107
$endgroup$
Does $$sum_n = 0^36lnleft(2+dfrac12nright)^2$$ satisfy your requirements?
answered Feb 10 at 19:29
JimmyK4542JimmyK4542
41.2k245107
answered Feb 10 at 19:29
JimmyK4542JimmyK4542
41.2k245107
answered Feb 10 at 19:29
JimmyK4542JimmyK4542
41.2k245107
answered Feb 10 at 19:29
JimmyK4542JimmyK4542
41.2k245107
41.2k245107
add a comment |
add a comment |
$begingroup$
You could change $ln(x)^2$ into $lnleft(fracx2right)^2$ to achieve the steps of $0.5$ in this case. You want $fracx2$ to go from $2$ to $20$ as well (with steps of $0.5$), so $x$ must go from $4$ to $40$. Therefore, the sum becomes$sum_limitsx = 4^40lnleft(fracx2right)^2$.
answered Feb 10 at 19:32
KM101KM101
6,0701525
$endgroup$
add a comment |
$begingroup$
You could change $ln(x)^2$ into $lnleft(fracx2right)^2$ to achieve the steps of $0.5$ in this case. You want $fracx2$ to go from $2$ to $20$ as well (with steps of $0.5$), so $x$ must go from $4$ to $40$. Therefore, the sum becomes$sum_limitsx = 4^40lnleft(fracx2right)^2$.
answered Feb 10 at 19:32
KM101KM101
6,0701525
$endgroup$
add a comment |
$begingroup$
You could change $ln(x)^2$ into $lnleft(fracx2right)^2$ to achieve the steps of $0.5$ in this case. You want $fracx2$ to go from $2$ to $20$ as well (with steps of $0.5$), so $x$ must go from $4$ to $40$. Therefore, the sum becomes$sum_limitsx = 4^40lnleft(fracx2right)^2$.
answered Feb 10 at 19:32
KM101KM101
6,0701525
$endgroup$
You could change $ln(x)^2$ into $lnleft(fracx2right)^2$ to achieve the steps of $0.5$ in this case. You want $fracx2$ to go from $2$ to $20$ as well (with steps of $0.5$), so $x$ must go from $4$ to $40$. Therefore, the sum becomes$sum_limitsx = 4^40lnleft(fracx2right)^2$.
answered Feb 10 at 19:32
KM101KM101
6,0701525
edited Feb 10 at 19:37
answered Feb 10 at 19:32
KM101KM101
6,0701525
answered Feb 10 at 19:32
KM101KM101
6,0701525
answered Feb 10 at 19:32
KM101KM101
6,0701525
6,0701525
add a comment |
add a comment |
$begingroup$
Just use $fracx2$ instead of $x$ in this example to get integers. This trick can always be used when we have to sum up finite many rationals or rationals with a limited denominator.
answered Feb 10 at 19:30
PeterPeter
48.3k1139133
$endgroup$
add a comment |
$begingroup$
Just use $fracx2$ instead of $x$ in this example to get integers. This trick can always be used when we have to sum up finite many rationals or rationals with a limited denominator.
answered Feb 10 at 19:30
PeterPeter
48.3k1139133
$endgroup$
add a comment |
$begingroup$
Just use $fracx2$ instead of $x$ in this example to get integers. This trick can always be used when we have to sum up finite many rationals or rationals with a limited denominator.
answered Feb 10 at 19:30
PeterPeter
48.3k1139133
$endgroup$
Just use $fracx2$ instead of $x$ in this example to get integers. This trick can always be used when we have to sum up finite many rationals or rationals with a limited denominator.
answered Feb 10 at 19:30
PeterPeter
48.3k1139133
answered Feb 10 at 19:30
PeterPeter
48.3k1139133
answered Feb 10 at 19:30
PeterPeter
48.3k1139133
answered Feb 10 at 19:30
PeterPeter
48.3k1139133
48.3k1139133
add a comment |
add a comment |
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