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Relation between roots and coefficients - manipulation of identities
Symmetric polynomials and the Newton identitiesRelation betwen coefficients and roots of a polynomialExpressing polynomial roots expression in terms of coefficientsPolynomials such that roots=coefficientsFind polynomial whose root is sum of roots of other polynomialsRelation between the roots and the coefficients of a polynomialFinding roots of cubic equationRoots of cubic equationWhat is the sum of all possible values of $gamma$How to prove relationship between coefficients and roots of a cubic (or) general polynomialRoots of polynomials and their formulae relating to coefficients
$begingroup$
The polynomial $x^3+3x^2-2x+1$ has roots $alpha, beta, gamma$ . Find $$alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)$$
I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.
polynomials
edited 2 hours ago
Eevee Trainer
6,41811237
asked 2 hours ago
MmloilerMmloiler
161
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Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
The polynomial $x^3+3x^2-2x+1$ has roots $alpha, beta, gamma$ . Find $$alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)$$
I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.
polynomials
edited 2 hours ago
Eevee Trainer
6,41811237
asked 2 hours ago
MmloilerMmloiler
161
New contributor
Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The polynomial $x^3+3x^2-2x+1$ has roots $alpha, beta, gamma$ . Find $$alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)$$
I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.
polynomials
edited 2 hours ago
Eevee Trainer
6,41811237
asked 2 hours ago
MmloilerMmloiler
161
New contributor
Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The polynomial $x^3+3x^2-2x+1$ has roots $alpha, beta, gamma$ . Find $$alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)$$
I tried finding the relation using $-b/a$, $c/a$ and $-d/a$. I couldn’t seem to find anything. I also tried solving for one root but it gave me back the polynomial but with the root as the variable. Also the polynomial can not be factorised.
polynomials
polynomials
edited 2 hours ago
Eevee Trainer
6,41811237
asked 2 hours ago
MmloilerMmloiler
161
New contributor
Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Eevee Trainer
6,41811237
asked 2 hours ago
MmloilerMmloiler
161
New contributor
Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
Eevee Trainer
6,41811237
edited 2 hours ago
Eevee Trainer
6,41811237
edited 2 hours ago
Eevee Trainer
6,41811237
6,41811237
asked 2 hours ago
MmloilerMmloiler
161
New contributor
Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
MmloilerMmloiler
161
asked 2 hours ago
MmloilerMmloiler
161
161
New contributor
Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mmloiler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since the polynomial has three roots and its highest degree is 3, we can write
$$
p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
$$
It then follows from
$$
x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
$$
that
$$
alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
alphabeta gamma = -1.
$$
Note that
$$
-2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
$$
Thus $alpha^2(beta + gamma)=1-2alpha$.
Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
Therefore,
begin{align}
alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
&= (1-2alpha) + (1-2beta) + (1-2gamma) \
&= 3 -2(alpha+beta+gamma) = 3 +6 =9.
end{align}
answered 1 hour ago
induction601induction601
1,226314
$endgroup$
add a comment |
$begingroup$
Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
$$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$
--
In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.
answered 1 hour ago
MacavityMacavity
35.5k52554
$endgroup$
$begingroup$
Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
$endgroup$
– Macavity
1 hour ago
add a comment |
$begingroup$
$a,b,c$ are the three roots.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$
edited 28 mins ago
Brahadeesh
6,46942363
answered 1 hour ago
Wentao WangWentao Wang
11
New contributor
Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
$endgroup$
– John Omielan
1 hour ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since the polynomial has three roots and its highest degree is 3, we can write
$$
p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
$$
It then follows from
$$
x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
$$
that
$$
alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
alphabeta gamma = -1.
$$
Note that
$$
-2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
$$
Thus $alpha^2(beta + gamma)=1-2alpha$.
Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
Therefore,
begin{align}
alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
&= (1-2alpha) + (1-2beta) + (1-2gamma) \
&= 3 -2(alpha+beta+gamma) = 3 +6 =9.
end{align}
answered 1 hour ago
induction601induction601
1,226314
$endgroup$
add a comment |
$begingroup$
Since the polynomial has three roots and its highest degree is 3, we can write
$$
p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
$$
It then follows from
$$
x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
$$
that
$$
alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
alphabeta gamma = -1.
$$
Note that
$$
-2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
$$
Thus $alpha^2(beta + gamma)=1-2alpha$.
Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
Therefore,
begin{align}
alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
&= (1-2alpha) + (1-2beta) + (1-2gamma) \
&= 3 -2(alpha+beta+gamma) = 3 +6 =9.
end{align}
answered 1 hour ago
induction601induction601
1,226314
$endgroup$
add a comment |
$begingroup$
Since the polynomial has three roots and its highest degree is 3, we can write
$$
p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
$$
It then follows from
$$
x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
$$
that
$$
alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
alphabeta gamma = -1.
$$
Note that
$$
-2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
$$
Thus $alpha^2(beta + gamma)=1-2alpha$.
Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
Therefore,
begin{align}
alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
&= (1-2alpha) + (1-2beta) + (1-2gamma) \
&= 3 -2(alpha+beta+gamma) = 3 +6 =9.
end{align}
answered 1 hour ago
induction601induction601
1,226314
$endgroup$
Since the polynomial has three roots and its highest degree is 3, we can write
$$
p(x) = (x-alpha)(x-beta)(x-gamma) = x^3 +3x^2 - 2x + 1.
$$
It then follows from
$$
x^3 - (alpha+beta+gamma)x^2 + (alphabeta + beta gamma + gamma alpha)x - alphabeta gamma = x^3+3x^2-2x + 1
$$
that
$$
alpha+beta+gamma = -3, quad alphabeta + beta gamma + gamma alpha = -2, quad
alphabeta gamma = -1.
$$
Note that
$$
-2alpha = alpha(alphabeta + beta gamma + gamma alpha) = alpha^2(beta+gamma) +alphabetagamma = alpha^2(beta+gamma) - 1.
$$
Thus $alpha^2(beta + gamma)=1-2alpha$.
Similarly, $beta^2(alpha + gamma) = 1-2beta$ and $gamma^2(alpha + beta) = 1-2gamma$.
Therefore,
begin{align}
alpha^2(beta + gamma) + beta^2(alpha + gamma) + gamma^2(alpha + beta)
&= (1-2alpha) + (1-2beta) + (1-2gamma) \
&= 3 -2(alpha+beta+gamma) = 3 +6 =9.
end{align}
answered 1 hour ago
induction601induction601
1,226314
answered 1 hour ago
induction601induction601
1,226314
answered 1 hour ago
induction601induction601
1,226314
answered 1 hour ago
induction601induction601
1,226314
1,226314
add a comment |
add a comment |
$begingroup$
Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
$$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$
--
In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.
answered 1 hour ago
MacavityMacavity
35.5k52554
$endgroup$
$begingroup$
Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
$endgroup$
– Macavity
1 hour ago
add a comment |
$begingroup$
Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
$$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$
--
In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.
answered 1 hour ago
MacavityMacavity
35.5k52554
$endgroup$
$begingroup$
Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
$endgroup$
– Macavity
1 hour ago
add a comment |
$begingroup$
Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
$$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$
--
In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.
answered 1 hour ago
MacavityMacavity
35.5k52554
$endgroup$
Any symmetric (polynomial) function of the roots can be expressed in terms of the Vieta coefficients. Here, check the hint:
$$sum alpha^2(beta+gamma) = (alpha+beta+gamma)(alphabeta+betagamma+gammaalpha)-3alphabetagamma$$
--
In case you want a systematic method to express in terms of elementary symmetric polynomials, check this answer for Gauss' algorithm.
answered 1 hour ago
MacavityMacavity
35.5k52554
edited 1 hour ago
answered 1 hour ago
MacavityMacavity
35.5k52554
answered 1 hour ago
MacavityMacavity
35.5k52554
answered 1 hour ago
MacavityMacavity
35.5k52554
35.5k52554
$begingroup$
Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
$endgroup$
– Macavity
1 hour ago
add a comment |
$begingroup$
Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
$endgroup$
– Macavity
1 hour ago
$begingroup$
Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
$endgroup$
– Macavity
1 hour ago
$begingroup$
Another often useful approach is to transform the polynomial (and / or) the result using the given polynomial. For e.g. $sum alpha^2(beta+gamma) = sumalpha^2(-3-alpha) = sum (-alpha^3-3alpha^2) = sum (-2alpha+1)=-2cdot(-3)+3=9 $
$endgroup$
– Macavity
1 hour ago
add a comment |
$begingroup$
$a,b,c$ are the three roots.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$
edited 28 mins ago
Brahadeesh
6,46942363
answered 1 hour ago
Wentao WangWentao Wang
11
New contributor
Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
$endgroup$
– John Omielan
1 hour ago
add a comment |
$begingroup$
$a,b,c$ are the three roots.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$
edited 28 mins ago
Brahadeesh
6,46942363
answered 1 hour ago
Wentao WangWentao Wang
11
New contributor
Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
5
$begingroup$
Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
$endgroup$
– John Omielan
1 hour ago
add a comment |
$begingroup$
$a,b,c$ are the three roots.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$
edited 28 mins ago
Brahadeesh
6,46942363
answered 1 hour ago
Wentao WangWentao Wang
11
New contributor
Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$a,b,c$ are the three roots.
$$
begin{align}
&a^2*(b+c)+b^2*(a+c)+c^2*(a+b)\
={}&(a+b+c)*(a^2+b^2+c^2)-(a^3+b^3+c^3)\
={}&(-3)*(a^2+b^2+c^2)-(a+b+c)^3\
={}&(-3)*((a+b+c)^2-2ab-2ac-2bc)-(a+b+c) * (a^2+b^2+c^2) + ab(a+b) + ac(a+c) + bc(b+c)\
={}& (-3)*(9-2*(-2))-(-3)*(9-2*(-2)) + ab(a+b+c-c) + ac(a+b+c-b) + bc(a+b+c-a)\
={}&(a+b+c)(ab+ac+bc)-3abc\
={}&(-3)*(-2)-3*(-1)\
={}&6-(-3)\
={}&9
end{align}
$$
edited 28 mins ago
Brahadeesh
6,46942363
answered 1 hour ago
Wentao WangWentao Wang
11
New contributor
Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 28 mins ago
Brahadeesh
6,46942363
edited 28 mins ago
Brahadeesh
6,46942363
edited 28 mins ago
Brahadeesh
6,46942363
6,46942363
answered 1 hour ago
Wentao WangWentao Wang
11
New contributor
Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
Wentao WangWentao Wang
11
answered 1 hour ago
Wentao WangWentao Wang
11
11
New contributor
Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Wentao Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
5
$begingroup$
Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
$endgroup$
– John Omielan
1 hour ago
add a comment |
5
$begingroup$
Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
$endgroup$
– John Omielan
1 hour ago
5
5
$begingroup$
Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
$endgroup$
– John Omielan
1 hour ago
$begingroup$
Welcome to MSE. Note that I, at least, found what you wrote hard to read. To help make your future math formatting look better, I suggest you read & use what it says in MathJax basic tutorial and quick reference.
$endgroup$
– John Omielan
1 hour ago
add a comment |
Mmloiler is a new contributor. Be nice, and check out our Code of Conduct.
Mmloiler is a new contributor. Be nice, and check out our Code of Conduct.
Mmloiler is a new contributor. Be nice, and check out our Code of Conduct.
Mmloiler is a new contributor. Be nice, and check out our Code of Conduct.
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