Mixing
Have these “calculus-based” consecutive summations been discovered yet?
$begingroup$
Consider the sum-of-powers series
$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$
For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdotint_0^{n}y_2(x),dx cdot frac{y_2^prime(n)+1}{y_2^prime(n)} tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}\[4pt]
end{align}$$
The only thing is that I'm not quite sure why these relations can include derivatives and integrals and somehow even work at all - they were simply a result of an extremely tiresome trial and error process. And can this "template" be extended to any $S_i(n)$ series or even more unorthodox variants?
Credits for the condensation and formatting of my answer goes to Blue (in the comments)
calculus sequences-and-series
asked Jan 4 at 11:41
Jainam ShahJainam Shah
213
$endgroup$
add a comment |
$begingroup$
Consider the sum-of-powers series
$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$
For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdotint_0^{n}y_2(x),dx cdot frac{y_2^prime(n)+1}{y_2^prime(n)} tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}\[4pt]
end{align}$$
The only thing is that I'm not quite sure why these relations can include derivatives and integrals and somehow even work at all - they were simply a result of an extremely tiresome trial and error process. And can this "template" be extended to any $S_i(n)$ series or even more unorthodox variants?
Credits for the condensation and formatting of my answer goes to Blue (in the comments)
calculus sequences-and-series
asked Jan 4 at 11:41
Jainam ShahJainam Shah
213
$endgroup$
2
$begingroup$
The Euler–Maclaurin formula might be of interest in this context, it describes the connection between the sum $sum_{i=0}^n f(i)$ and the integral $int_0^n f(x) dx$
$endgroup$
– Martin R
Jan 4 at 12:24
$begingroup$
@MartinR I just had a quick look and I think I might need to spend more time decrypting the logic, since the terminology is much more complex than I can possibly understand. But thanks for pointing me in the right direction!
$endgroup$
– Jainam Shah
Jan 4 at 12:37
$begingroup$
By the way, the name for the general formula for these sums is called Faulhaber's formula. en.wikipedia.org/wiki/Faulhaber%27s_formula
$endgroup$
– Deepak
Jan 5 at 10:05
add a comment |
$begingroup$
Consider the sum-of-powers series
$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$
For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdotint_0^{n}y_2(x),dx cdot frac{y_2^prime(n)+1}{y_2^prime(n)} tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}\[4pt]
end{align}$$
The only thing is that I'm not quite sure why these relations can include derivatives and integrals and somehow even work at all - they were simply a result of an extremely tiresome trial and error process. And can this "template" be extended to any $S_i(n)$ series or even more unorthodox variants?
Credits for the condensation and formatting of my answer goes to Blue (in the comments)
calculus sequences-and-series
asked Jan 4 at 11:41
Jainam ShahJainam Shah
213
$endgroup$
Consider the sum-of-powers series
$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$
For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdotint_0^{n}y_2(x),dx cdot frac{y_2^prime(n)+1}{y_2^prime(n)} tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}\[4pt]
end{align}$$
The only thing is that I'm not quite sure why these relations can include derivatives and integrals and somehow even work at all - they were simply a result of an extremely tiresome trial and error process. And can this "template" be extended to any $S_i(n)$ series or even more unorthodox variants?
Credits for the condensation and formatting of my answer goes to Blue (in the comments)
calculus sequences-and-series
calculus sequences-and-series
asked Jan 4 at 11:41
Jainam ShahJainam Shah
213
asked Jan 4 at 11:41
Jainam ShahJainam Shah
213
edited Jan 5 at 10:49
Jainam Shah
asked Jan 4 at 11:41
Jainam ShahJainam Shah
213
asked Jan 4 at 11:41
Jainam ShahJainam Shah
213
asked Jan 4 at 11:41
Jainam ShahJainam Shah
213
213
2
$begingroup$
The Euler–Maclaurin formula might be of interest in this context, it describes the connection between the sum $sum_{i=0}^n f(i)$ and the integral $int_0^n f(x) dx$
$endgroup$
– Martin R
Jan 4 at 12:24
$begingroup$
@MartinR I just had a quick look and I think I might need to spend more time decrypting the logic, since the terminology is much more complex than I can possibly understand. But thanks for pointing me in the right direction!
$endgroup$
– Jainam Shah
Jan 4 at 12:37
$begingroup$
By the way, the name for the general formula for these sums is called Faulhaber's formula. en.wikipedia.org/wiki/Faulhaber%27s_formula
$endgroup$
– Deepak
Jan 5 at 10:05
add a comment |
2
$begingroup$
The Euler–Maclaurin formula might be of interest in this context, it describes the connection between the sum $sum_{i=0}^n f(i)$ and the integral $int_0^n f(x) dx$
$endgroup$
– Martin R
Jan 4 at 12:24
$begingroup$
@MartinR I just had a quick look and I think I might need to spend more time decrypting the logic, since the terminology is much more complex than I can possibly understand. But thanks for pointing me in the right direction!
$endgroup$
– Jainam Shah
Jan 4 at 12:37
$begingroup$
By the way, the name for the general formula for these sums is called Faulhaber's formula. en.wikipedia.org/wiki/Faulhaber%27s_formula
$endgroup$
– Deepak
Jan 5 at 10:05
2
2
$begingroup$
The Euler–Maclaurin formula might be of interest in this context, it describes the connection between the sum $sum_{i=0}^n f(i)$ and the integral $int_0^n f(x) dx$
$endgroup$
– Martin R
Jan 4 at 12:24
$begingroup$
The Euler–Maclaurin formula might be of interest in this context, it describes the connection between the sum $sum_{i=0}^n f(i)$ and the integral $int_0^n f(x) dx$
$endgroup$
– Martin R
Jan 4 at 12:24
$begingroup$
@MartinR I just had a quick look and I think I might need to spend more time decrypting the logic, since the terminology is much more complex than I can possibly understand. But thanks for pointing me in the right direction!
$endgroup$
– Jainam Shah
Jan 4 at 12:37
$begingroup$
@MartinR I just had a quick look and I think I might need to spend more time decrypting the logic, since the terminology is much more complex than I can possibly understand. But thanks for pointing me in the right direction!
$endgroup$
– Jainam Shah
Jan 4 at 12:37
$begingroup$
By the way, the name for the general formula for these sums is called Faulhaber's formula. en.wikipedia.org/wiki/Faulhaber%27s_formula
$endgroup$
– Deepak
Jan 5 at 10:05
$begingroup$
By the way, the name for the general formula for these sums is called Faulhaber's formula. en.wikipedia.org/wiki/Faulhaber%27s_formula
$endgroup$
– Deepak
Jan 5 at 10:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
(Too long for a comment.) The reuse and ambiguity of symbols is a little confusing. I'd like to try to restate the problem.
Consider the sum-of-powers series
$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$
For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdot y_1^prime(n)cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdot frac{y_2^prime(n)+1}{y_2^prime(n)}cdotint_0^{n}y_2(x),dx tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$
[Can this technique be extended? Is it new? etc, etc, etc]
Does this capture the intent of the question?
(BTW: I think you need to be a bit more explicit about why the relations in $(2)$ hold.)
answered Jan 4 at 12:12
BlueBlue
47.8k870152
$endgroup$
$begingroup$
Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
$endgroup$
– Jainam Shah
Jan 4 at 12:16
$begingroup$
Sure. I just noticed you added the case for cubes. Give me a minute.
$endgroup$
– Blue
Jan 4 at 12:17
$begingroup$
You are amazing - thanks a lot!
$endgroup$
– Jainam Shah
Jan 4 at 12:17
4
$begingroup$
@JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
$endgroup$
– Blue
Jan 4 at 12:40
2
$begingroup$
Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
$endgroup$
– Jainam Shah
Jan 4 at 12:43
|
show 2 more comments
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$begingroup$
(Too long for a comment.) The reuse and ambiguity of symbols is a little confusing. I'd like to try to restate the problem.
Consider the sum-of-powers series
$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$
For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdot y_1^prime(n)cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdot frac{y_2^prime(n)+1}{y_2^prime(n)}cdotint_0^{n}y_2(x),dx tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$
[Can this technique be extended? Is it new? etc, etc, etc]
Does this capture the intent of the question?
(BTW: I think you need to be a bit more explicit about why the relations in $(2)$ hold.)
answered Jan 4 at 12:12
BlueBlue
47.8k870152
$endgroup$
$begingroup$
Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
$endgroup$
– Jainam Shah
Jan 4 at 12:16
$begingroup$
Sure. I just noticed you added the case for cubes. Give me a minute.
$endgroup$
– Blue
Jan 4 at 12:17
$begingroup$
You are amazing - thanks a lot!
$endgroup$
– Jainam Shah
Jan 4 at 12:17
4
$begingroup$
@JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
$endgroup$
– Blue
Jan 4 at 12:40
2
$begingroup$
Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
$endgroup$
– Jainam Shah
Jan 4 at 12:43
|
show 2 more comments
$begingroup$
(Too long for a comment.) The reuse and ambiguity of symbols is a little confusing. I'd like to try to restate the problem.
Consider the sum-of-powers series
$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$
For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdot y_1^prime(n)cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdot frac{y_2^prime(n)+1}{y_2^prime(n)}cdotint_0^{n}y_2(x),dx tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$
[Can this technique be extended? Is it new? etc, etc, etc]
Does this capture the intent of the question?
(BTW: I think you need to be a bit more explicit about why the relations in $(2)$ hold.)
answered Jan 4 at 12:12
BlueBlue
47.8k870152
$endgroup$
$begingroup$
Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
$endgroup$
– Jainam Shah
Jan 4 at 12:16
$begingroup$
Sure. I just noticed you added the case for cubes. Give me a minute.
$endgroup$
– Blue
Jan 4 at 12:17
$begingroup$
You are amazing - thanks a lot!
$endgroup$
– Jainam Shah
Jan 4 at 12:17
4
$begingroup$
@JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
$endgroup$
– Blue
Jan 4 at 12:40
2
$begingroup$
Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
$endgroup$
– Jainam Shah
Jan 4 at 12:43
|
show 2 more comments
$begingroup$
(Too long for a comment.) The reuse and ambiguity of symbols is a little confusing. I'd like to try to restate the problem.
Consider the sum-of-powers series
$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$
For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdot y_1^prime(n)cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdot frac{y_2^prime(n)+1}{y_2^prime(n)}cdotint_0^{n}y_2(x),dx tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$
[Can this technique be extended? Is it new? etc, etc, etc]
Does this capture the intent of the question?
(BTW: I think you need to be a bit more explicit about why the relations in $(2)$ hold.)
answered Jan 4 at 12:12
BlueBlue
47.8k870152
$endgroup$
(Too long for a comment.) The reuse and ambiguity of symbols is a little confusing. I'd like to try to restate the problem.
Consider the sum-of-powers series
$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$
For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdot y_1^prime(n)cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdot frac{y_2^prime(n)+1}{y_2^prime(n)}cdotint_0^{n}y_2(x),dx tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$
[Can this technique be extended? Is it new? etc, etc, etc]
Does this capture the intent of the question?
(BTW: I think you need to be a bit more explicit about why the relations in $(2)$ hold.)
answered Jan 4 at 12:12
BlueBlue
47.8k870152
edited Jan 4 at 12:22
answered Jan 4 at 12:12
BlueBlue
47.8k870152
answered Jan 4 at 12:12
BlueBlue
47.8k870152
answered Jan 4 at 12:12
BlueBlue
47.8k870152
47.8k870152
$begingroup$
Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
$endgroup$
– Jainam Shah
Jan 4 at 12:16
$begingroup$
Sure. I just noticed you added the case for cubes. Give me a minute.
$endgroup$
– Blue
Jan 4 at 12:17
$begingroup$
You are amazing - thanks a lot!
$endgroup$
– Jainam Shah
Jan 4 at 12:17
4
$begingroup$
@JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
$endgroup$
– Blue
Jan 4 at 12:40
2
$begingroup$
Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
$endgroup$
– Jainam Shah
Jan 4 at 12:43
|
show 2 more comments
$begingroup$
Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
$endgroup$
– Jainam Shah
Jan 4 at 12:16
$begingroup$
Sure. I just noticed you added the case for cubes. Give me a minute.
$endgroup$
– Blue
Jan 4 at 12:17
$begingroup$
You are amazing - thanks a lot!
$endgroup$
– Jainam Shah
Jan 4 at 12:17
4
$begingroup$
@JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
$endgroup$
– Blue
Jan 4 at 12:40
2
$begingroup$
Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
$endgroup$
– Jainam Shah
Jan 4 at 12:43
$begingroup$
Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
$endgroup$
– Jainam Shah
Jan 4 at 12:16
$begingroup$
Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
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– Jainam Shah
Jan 4 at 12:16
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Sure. I just noticed you added the case for cubes. Give me a minute.
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– Blue
Jan 4 at 12:17
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Sure. I just noticed you added the case for cubes. Give me a minute.
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– Blue
Jan 4 at 12:17
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You are amazing - thanks a lot!
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– Jainam Shah
Jan 4 at 12:17
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You are amazing - thanks a lot!
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– Jainam Shah
Jan 4 at 12:17
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@JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
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– Blue
Jan 4 at 12:40
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@JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
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– Blue
Jan 4 at 12:40
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Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
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– Jainam Shah
Jan 4 at 12:43
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Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
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– Jainam Shah
Jan 4 at 12:43
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The Euler–Maclaurin formula might be of interest in this context, it describes the connection between the sum $sum_{i=0}^n f(i)$ and the integral $int_0^n f(x) dx$
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– Martin R
Jan 4 at 12:24
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@MartinR I just had a quick look and I think I might need to spend more time decrypting the logic, since the terminology is much more complex than I can possibly understand. But thanks for pointing me in the right direction!
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– Jainam Shah
Jan 4 at 12:37
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By the way, the name for the general formula for these sums is called Faulhaber's formula. en.wikipedia.org/wiki/Faulhaber%27s_formula
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– Deepak
Jan 5 at 10:05