Have these “calculus-based” consecutive summations been discovered yet?












4












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















4












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














4












4








4


2



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 5 at 10:49







Jainam Shah

















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














  • 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










1 Answer
1






active

oldest

votes


















3












$begingroup$

(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.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

(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.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















3












$begingroup$

(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.)






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














3












3








3





$begingroup$

(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.)






share|cite|improve this answer











$endgroup$



(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.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 4 at 12:22

























answered Jan 4 at 12:12









BlueBlue

48.1k870153




48.1k870153












  • $begingroup$
    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












  • $begingroup$
    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!
$endgroup$
– 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




$begingroup$
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




$begingroup$
You are amazing - thanks a lot!
$endgroup$
– Jainam Shah
Jan 4 at 12:17




4




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






$begingroup$
@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




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




$begingroup$
Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
$endgroup$
– Jainam Shah
Jan 4 at 12:43


















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