Mixing
Proof of AM GM theorem using Lagrangian
Given:
$prod_{i=1}^n x_i = 1$ leads to constraint function $G(x_1,x_2,...,x_n)=prod_{i=1}^n x_i-1$
($prod_{i=1}^n x_i =x_1 x_2...x_n$)
Task is to to find the minimum using conditional extrema of the following (the induction method that is most convinient is forbidden), if we proove this special case then the derivation can be generalised to prove the AMGM theorem:
- $F(x_1,x_2,...,x_n) = sum_{i=1}^n x_i$
($sum_{i=1}^n x_i =x_1+x_2+...+x_n$)
Idea is that it should be $sum_{i=1}^n x_i geqslant n$ afaik
And finally Using the derivation above prove the AM-GM theorem:
$frac{sum_{i=1}^n x_i}{n} geqslant sqrt[n]{prod_{i=1}^n x_i - 1}$
Solution so far:
what I come up with is writing down the Lagrangian:
$L = F(x_1,x_2,...,x_n) - lambda G(x_1,x_2,...,x_n) implies L = sum_{i=1}^n x_i - lambda (prod_{i=1}^n x_i-1)$
taking the partial derivatives
$frac{dL}{dx_1} = 1 - lambda frac{prod_{i=1}^n x_i}{x_1}=0 implies lambda frac{prod_{i=1}^n x_i}{x_1}=1 implies lambda frac{1}{x_1}=1 implies lambda = x_1$
$frac{dL}{dx_2} = 1 - lambda frac{prod_{i=1}^n x_i}{x_2}=0 implies lambda frac{prod_{i=1}^n x_i}{x_2}=1 implies lambda frac{1}{x_2}=1implies lambda = x_2$
...
$frac{dL}{dx_n} = 1 - lambda frac{prod_{i=1}^n x_i}{x_n}=0 implies lambda frac{prod_{i=1}^n x_i}{x_n}=1 implies lambda frac{1}{x_n}=1 implies lambda = x_n$
$frac{dL}{dlambda} = - prod_{i=1}^n x_i + 1= 0 implies$ $prod_{i=1}^n x_i = 1$
$ lambda = x_1 = x_2 = ...=x_n = 1$ is our critical point.
Taking the differential of our constraint
$dG(x_1,x_2,...,x_n) = 0$
$frac{partial G}{partial x_1}Delta x_1+frac{partial G}{partial x_2}Delta x_2+...+frac{partial G}{partial x_n}Delta x_n=0$
$frac{prod_{i=1}^n x_i}{x_1}Delta x_1 + frac{prod_{i=1}^n x_i}{x_2}Delta x_2+...+frac{prod_{i=1}^n x_i}{x_n}Delta x_n=0$
substituting the roots of critical point $x_1=x_2=...=x_n=1$ and $prod_{i=1}^n x_i=1$ leads to
$Delta x_1 + Delta x_2 +...+ Delta x_n = 0$
First question.
update, answered by Andreas below
The second order differential of Lagrangian has to be positive, but I’m getting negative sign
$d^2L = sum_{j=1}^nsum_{i=1}^n L_{x_j x_i} Delta x_j Delta x_i=-lambda (prod x_i )(frac{Delta x_i Delta x_j}{x_i x_j})<0$
Here I took the second order partial derivatives of L
$frac{d^2 L} {dx_1 dx_1} = -lambda frac{prod x_i}{x_1 x_1}$
$frac{d^2 L} {dx_1 dx_2} = -lambda frac{prod x_i}{x_1 x_2}$
...
$frac{d^2 L} {dx_n dx_n} = -lambda frac{prod x_i}{x_n x_n}$
Second question.
Are these reasonings correct?
It is needed to justify why local extrema is global as well.
If the second differential will become positive and therefore at the point (1,1,...,1) is local minima(since the function might be just like a cubic polynomial with no global minima), then we can check the function limit along axis direction (by sending all but one variable to infinity and getting the that one variable limit to zero) and thus we find out that the function F is located in the positive n-dimensional quadrant and therefore the extreme point has to be global minima.
I don't know it's just a feeling, was thinking of rotating the function 45 degrees towards the vertical axis and then stating that function goes to infinity in each direction.
Update. Probs solved the global issue part, gonna consilt with Professor and update the solution if my assumptions are right.
Update. Just posted Proof to Wiki: MyProof
Idea was to simply use the Weierstrass theorem and apply it to any closed domain interval of function.
calculus multivariable-calculus lagrange-multiplier extreme-value-theorem extreme-value-analysis
asked Nov 21 '18 at 10:32
MotherLand
7218
add a comment |
Given:
$prod_{i=1}^n x_i = 1$ leads to constraint function $G(x_1,x_2,...,x_n)=prod_{i=1}^n x_i-1$
($prod_{i=1}^n x_i =x_1 x_2...x_n$)
Task is to to find the minimum using conditional extrema of the following (the induction method that is most convinient is forbidden), if we proove this special case then the derivation can be generalised to prove the AMGM theorem:
- $F(x_1,x_2,...,x_n) = sum_{i=1}^n x_i$
($sum_{i=1}^n x_i =x_1+x_2+...+x_n$)
Idea is that it should be $sum_{i=1}^n x_i geqslant n$ afaik
And finally Using the derivation above prove the AM-GM theorem:
$frac{sum_{i=1}^n x_i}{n} geqslant sqrt[n]{prod_{i=1}^n x_i - 1}$
Solution so far:
what I come up with is writing down the Lagrangian:
$L = F(x_1,x_2,...,x_n) - lambda G(x_1,x_2,...,x_n) implies L = sum_{i=1}^n x_i - lambda (prod_{i=1}^n x_i-1)$
taking the partial derivatives
$frac{dL}{dx_1} = 1 - lambda frac{prod_{i=1}^n x_i}{x_1}=0 implies lambda frac{prod_{i=1}^n x_i}{x_1}=1 implies lambda frac{1}{x_1}=1 implies lambda = x_1$
$frac{dL}{dx_2} = 1 - lambda frac{prod_{i=1}^n x_i}{x_2}=0 implies lambda frac{prod_{i=1}^n x_i}{x_2}=1 implies lambda frac{1}{x_2}=1implies lambda = x_2$
...
$frac{dL}{dx_n} = 1 - lambda frac{prod_{i=1}^n x_i}{x_n}=0 implies lambda frac{prod_{i=1}^n x_i}{x_n}=1 implies lambda frac{1}{x_n}=1 implies lambda = x_n$
$frac{dL}{dlambda} = - prod_{i=1}^n x_i + 1= 0 implies$ $prod_{i=1}^n x_i = 1$
$ lambda = x_1 = x_2 = ...=x_n = 1$ is our critical point.
Taking the differential of our constraint
$dG(x_1,x_2,...,x_n) = 0$
$frac{partial G}{partial x_1}Delta x_1+frac{partial G}{partial x_2}Delta x_2+...+frac{partial G}{partial x_n}Delta x_n=0$
$frac{prod_{i=1}^n x_i}{x_1}Delta x_1 + frac{prod_{i=1}^n x_i}{x_2}Delta x_2+...+frac{prod_{i=1}^n x_i}{x_n}Delta x_n=0$
substituting the roots of critical point $x_1=x_2=...=x_n=1$ and $prod_{i=1}^n x_i=1$ leads to
$Delta x_1 + Delta x_2 +...+ Delta x_n = 0$
First question.
update, answered by Andreas below
The second order differential of Lagrangian has to be positive, but I’m getting negative sign
$d^2L = sum_{j=1}^nsum_{i=1}^n L_{x_j x_i} Delta x_j Delta x_i=-lambda (prod x_i )(frac{Delta x_i Delta x_j}{x_i x_j})<0$
Here I took the second order partial derivatives of L
$frac{d^2 L} {dx_1 dx_1} = -lambda frac{prod x_i}{x_1 x_1}$
$frac{d^2 L} {dx_1 dx_2} = -lambda frac{prod x_i}{x_1 x_2}$
...
$frac{d^2 L} {dx_n dx_n} = -lambda frac{prod x_i}{x_n x_n}$
Second question.
Are these reasonings correct?
It is needed to justify why local extrema is global as well.
If the second differential will become positive and therefore at the point (1,1,...,1) is local minima(since the function might be just like a cubic polynomial with no global minima), then we can check the function limit along axis direction (by sending all but one variable to infinity and getting the that one variable limit to zero) and thus we find out that the function F is located in the positive n-dimensional quadrant and therefore the extreme point has to be global minima.
I don't know it's just a feeling, was thinking of rotating the function 45 degrees towards the vertical axis and then stating that function goes to infinity in each direction.
Update. Probs solved the global issue part, gonna consilt with Professor and update the solution if my assumptions are right.
Update. Just posted Proof to Wiki: MyProof
Idea was to simply use the Weierstrass theorem and apply it to any closed domain interval of function.
calculus multivariable-calculus lagrange-multiplier extreme-value-theorem extreme-value-analysis
asked Nov 21 '18 at 10:32
MotherLand
7218
add a comment |
Given:
$prod_{i=1}^n x_i = 1$ leads to constraint function $G(x_1,x_2,...,x_n)=prod_{i=1}^n x_i-1$
($prod_{i=1}^n x_i =x_1 x_2...x_n$)
Task is to to find the minimum using conditional extrema of the following (the induction method that is most convinient is forbidden), if we proove this special case then the derivation can be generalised to prove the AMGM theorem:
- $F(x_1,x_2,...,x_n) = sum_{i=1}^n x_i$
($sum_{i=1}^n x_i =x_1+x_2+...+x_n$)
Idea is that it should be $sum_{i=1}^n x_i geqslant n$ afaik
And finally Using the derivation above prove the AM-GM theorem:
$frac{sum_{i=1}^n x_i}{n} geqslant sqrt[n]{prod_{i=1}^n x_i - 1}$
Solution so far:
what I come up with is writing down the Lagrangian:
$L = F(x_1,x_2,...,x_n) - lambda G(x_1,x_2,...,x_n) implies L = sum_{i=1}^n x_i - lambda (prod_{i=1}^n x_i-1)$
taking the partial derivatives
$frac{dL}{dx_1} = 1 - lambda frac{prod_{i=1}^n x_i}{x_1}=0 implies lambda frac{prod_{i=1}^n x_i}{x_1}=1 implies lambda frac{1}{x_1}=1 implies lambda = x_1$
$frac{dL}{dx_2} = 1 - lambda frac{prod_{i=1}^n x_i}{x_2}=0 implies lambda frac{prod_{i=1}^n x_i}{x_2}=1 implies lambda frac{1}{x_2}=1implies lambda = x_2$
...
$frac{dL}{dx_n} = 1 - lambda frac{prod_{i=1}^n x_i}{x_n}=0 implies lambda frac{prod_{i=1}^n x_i}{x_n}=1 implies lambda frac{1}{x_n}=1 implies lambda = x_n$
$frac{dL}{dlambda} = - prod_{i=1}^n x_i + 1= 0 implies$ $prod_{i=1}^n x_i = 1$
$ lambda = x_1 = x_2 = ...=x_n = 1$ is our critical point.
Taking the differential of our constraint
$dG(x_1,x_2,...,x_n) = 0$
$frac{partial G}{partial x_1}Delta x_1+frac{partial G}{partial x_2}Delta x_2+...+frac{partial G}{partial x_n}Delta x_n=0$
$frac{prod_{i=1}^n x_i}{x_1}Delta x_1 + frac{prod_{i=1}^n x_i}{x_2}Delta x_2+...+frac{prod_{i=1}^n x_i}{x_n}Delta x_n=0$
substituting the roots of critical point $x_1=x_2=...=x_n=1$ and $prod_{i=1}^n x_i=1$ leads to
$Delta x_1 + Delta x_2 +...+ Delta x_n = 0$
First question.
update, answered by Andreas below
The second order differential of Lagrangian has to be positive, but I’m getting negative sign
$d^2L = sum_{j=1}^nsum_{i=1}^n L_{x_j x_i} Delta x_j Delta x_i=-lambda (prod x_i )(frac{Delta x_i Delta x_j}{x_i x_j})<0$
Here I took the second order partial derivatives of L
$frac{d^2 L} {dx_1 dx_1} = -lambda frac{prod x_i}{x_1 x_1}$
$frac{d^2 L} {dx_1 dx_2} = -lambda frac{prod x_i}{x_1 x_2}$
...
$frac{d^2 L} {dx_n dx_n} = -lambda frac{prod x_i}{x_n x_n}$
Second question.
Are these reasonings correct?
It is needed to justify why local extrema is global as well.
If the second differential will become positive and therefore at the point (1,1,...,1) is local minima(since the function might be just like a cubic polynomial with no global minima), then we can check the function limit along axis direction (by sending all but one variable to infinity and getting the that one variable limit to zero) and thus we find out that the function F is located in the positive n-dimensional quadrant and therefore the extreme point has to be global minima.
I don't know it's just a feeling, was thinking of rotating the function 45 degrees towards the vertical axis and then stating that function goes to infinity in each direction.
Update. Probs solved the global issue part, gonna consilt with Professor and update the solution if my assumptions are right.
Update. Just posted Proof to Wiki: MyProof
Idea was to simply use the Weierstrass theorem and apply it to any closed domain interval of function.
calculus multivariable-calculus lagrange-multiplier extreme-value-theorem extreme-value-analysis
asked Nov 21 '18 at 10:32
MotherLand
7218
Given:
$prod_{i=1}^n x_i = 1$ leads to constraint function $G(x_1,x_2,...,x_n)=prod_{i=1}^n x_i-1$
($prod_{i=1}^n x_i =x_1 x_2...x_n$)
Task is to to find the minimum using conditional extrema of the following (the induction method that is most convinient is forbidden), if we proove this special case then the derivation can be generalised to prove the AMGM theorem:
- $F(x_1,x_2,...,x_n) = sum_{i=1}^n x_i$
($sum_{i=1}^n x_i =x_1+x_2+...+x_n$)
Idea is that it should be $sum_{i=1}^n x_i geqslant n$ afaik
And finally Using the derivation above prove the AM-GM theorem:
$frac{sum_{i=1}^n x_i}{n} geqslant sqrt[n]{prod_{i=1}^n x_i - 1}$
Solution so far:
what I come up with is writing down the Lagrangian:
$L = F(x_1,x_2,...,x_n) - lambda G(x_1,x_2,...,x_n) implies L = sum_{i=1}^n x_i - lambda (prod_{i=1}^n x_i-1)$
taking the partial derivatives
$frac{dL}{dx_1} = 1 - lambda frac{prod_{i=1}^n x_i}{x_1}=0 implies lambda frac{prod_{i=1}^n x_i}{x_1}=1 implies lambda frac{1}{x_1}=1 implies lambda = x_1$
$frac{dL}{dx_2} = 1 - lambda frac{prod_{i=1}^n x_i}{x_2}=0 implies lambda frac{prod_{i=1}^n x_i}{x_2}=1 implies lambda frac{1}{x_2}=1implies lambda = x_2$
...
$frac{dL}{dx_n} = 1 - lambda frac{prod_{i=1}^n x_i}{x_n}=0 implies lambda frac{prod_{i=1}^n x_i}{x_n}=1 implies lambda frac{1}{x_n}=1 implies lambda = x_n$
$frac{dL}{dlambda} = - prod_{i=1}^n x_i + 1= 0 implies$ $prod_{i=1}^n x_i = 1$
$ lambda = x_1 = x_2 = ...=x_n = 1$ is our critical point.
Taking the differential of our constraint
$dG(x_1,x_2,...,x_n) = 0$
$frac{partial G}{partial x_1}Delta x_1+frac{partial G}{partial x_2}Delta x_2+...+frac{partial G}{partial x_n}Delta x_n=0$
$frac{prod_{i=1}^n x_i}{x_1}Delta x_1 + frac{prod_{i=1}^n x_i}{x_2}Delta x_2+...+frac{prod_{i=1}^n x_i}{x_n}Delta x_n=0$
substituting the roots of critical point $x_1=x_2=...=x_n=1$ and $prod_{i=1}^n x_i=1$ leads to
$Delta x_1 + Delta x_2 +...+ Delta x_n = 0$
First question.
update, answered by Andreas below
The second order differential of Lagrangian has to be positive, but I’m getting negative sign
$d^2L = sum_{j=1}^nsum_{i=1}^n L_{x_j x_i} Delta x_j Delta x_i=-lambda (prod x_i )(frac{Delta x_i Delta x_j}{x_i x_j})<0$
Here I took the second order partial derivatives of L
$frac{d^2 L} {dx_1 dx_1} = -lambda frac{prod x_i}{x_1 x_1}$
$frac{d^2 L} {dx_1 dx_2} = -lambda frac{prod x_i}{x_1 x_2}$
...
$frac{d^2 L} {dx_n dx_n} = -lambda frac{prod x_i}{x_n x_n}$
Second question.
Are these reasonings correct?
It is needed to justify why local extrema is global as well.
If the second differential will become positive and therefore at the point (1,1,...,1) is local minima(since the function might be just like a cubic polynomial with no global minima), then we can check the function limit along axis direction (by sending all but one variable to infinity and getting the that one variable limit to zero) and thus we find out that the function F is located in the positive n-dimensional quadrant and therefore the extreme point has to be global minima.
I don't know it's just a feeling, was thinking of rotating the function 45 degrees towards the vertical axis and then stating that function goes to infinity in each direction.
Update. Probs solved the global issue part, gonna consilt with Professor and update the solution if my assumptions are right.
Update. Just posted Proof to Wiki: MyProof
Idea was to simply use the Weierstrass theorem and apply it to any closed domain interval of function.
calculus multivariable-calculus lagrange-multiplier extreme-value-theorem extreme-value-analysis
calculus multivariable-calculus lagrange-multiplier extreme-value-theorem extreme-value-analysis
asked Nov 21 '18 at 10:32
MotherLand
7218
asked Nov 21 '18 at 10:32
MotherLand
7218
edited Nov 27 '18 at 8:47
asked Nov 21 '18 at 10:32
MotherLand
7218
asked Nov 21 '18 at 10:32
MotherLand
7218
asked Nov 21 '18 at 10:32
MotherLand
7218
7218
add a comment |
add a comment |
1 Answer
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Taking the second derivatives and evaluating at $lambda = x_1 = x_2 = ...=x_n = 1$ gives
$$ L = sum_{i=1}^n x_i - lambda (prod_{i=1}^n x_i-1)\
frac{dL}{dx_k} = 1 - lambda frac{prod_{i=1}^n x_i}{x_k}\
frac{d^2L}{dx_k^2} = 0\
frac{d^2L}{dx_k d x_m} = - lambda frac{prod_{i=1}^n x_i}{x_k x_m} = -1 (k ne m)
$$
Now you need that for any vector $Delta x$ with $Delta x_1 + Delta x_2 +...+ Delta x_n = 0$ holds: $$sum_{i=1}^nsum_{j=1}^n frac{d^2L}{dx_i d x_j} Delta x_i Delta x_j > 0$$. We have
$$sum_{i=1}^nsum_{j=1}^n frac{d^2L}{dx_i d x_j} Delta x_i Delta x_j = - sum_{i=1}^nsum_{j neq i} Delta x_i Delta x_j \
= - sum_{i=1}^n Delta x_i sum_{j neq i} Delta x_j \
= - sum_{i=1}^n Delta x_i (0 -Delta x_i ) = sum_{i=1}^n Delta x_i^2 > 0$$.
answered Nov 21 '18 at 13:42
Andreas
7,8281037
Thanks! Figured out as well. Now left with the local to global part to think of.
– MotherLand
Nov 21 '18 at 14:50
add a comment |
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Taking the second derivatives and evaluating at $lambda = x_1 = x_2 = ...=x_n = 1$ gives
$$ L = sum_{i=1}^n x_i - lambda (prod_{i=1}^n x_i-1)\
frac{dL}{dx_k} = 1 - lambda frac{prod_{i=1}^n x_i}{x_k}\
frac{d^2L}{dx_k^2} = 0\
frac{d^2L}{dx_k d x_m} = - lambda frac{prod_{i=1}^n x_i}{x_k x_m} = -1 (k ne m)
$$
Now you need that for any vector $Delta x$ with $Delta x_1 + Delta x_2 +...+ Delta x_n = 0$ holds: $$sum_{i=1}^nsum_{j=1}^n frac{d^2L}{dx_i d x_j} Delta x_i Delta x_j > 0$$. We have
$$sum_{i=1}^nsum_{j=1}^n frac{d^2L}{dx_i d x_j} Delta x_i Delta x_j = - sum_{i=1}^nsum_{j neq i} Delta x_i Delta x_j \
= - sum_{i=1}^n Delta x_i sum_{j neq i} Delta x_j \
= - sum_{i=1}^n Delta x_i (0 -Delta x_i ) = sum_{i=1}^n Delta x_i^2 > 0$$.
answered Nov 21 '18 at 13:42
Andreas
7,8281037
Thanks! Figured out as well. Now left with the local to global part to think of.
– MotherLand
Nov 21 '18 at 14:50
add a comment |
Taking the second derivatives and evaluating at $lambda = x_1 = x_2 = ...=x_n = 1$ gives
$$ L = sum_{i=1}^n x_i - lambda (prod_{i=1}^n x_i-1)\
frac{dL}{dx_k} = 1 - lambda frac{prod_{i=1}^n x_i}{x_k}\
frac{d^2L}{dx_k^2} = 0\
frac{d^2L}{dx_k d x_m} = - lambda frac{prod_{i=1}^n x_i}{x_k x_m} = -1 (k ne m)
$$
Now you need that for any vector $Delta x$ with $Delta x_1 + Delta x_2 +...+ Delta x_n = 0$ holds: $$sum_{i=1}^nsum_{j=1}^n frac{d^2L}{dx_i d x_j} Delta x_i Delta x_j > 0$$. We have
$$sum_{i=1}^nsum_{j=1}^n frac{d^2L}{dx_i d x_j} Delta x_i Delta x_j = - sum_{i=1}^nsum_{j neq i} Delta x_i Delta x_j \
= - sum_{i=1}^n Delta x_i sum_{j neq i} Delta x_j \
= - sum_{i=1}^n Delta x_i (0 -Delta x_i ) = sum_{i=1}^n Delta x_i^2 > 0$$.
answered Nov 21 '18 at 13:42
Andreas
7,8281037
Thanks! Figured out as well. Now left with the local to global part to think of.
– MotherLand
Nov 21 '18 at 14:50
add a comment |
Taking the second derivatives and evaluating at $lambda = x_1 = x_2 = ...=x_n = 1$ gives
$$ L = sum_{i=1}^n x_i - lambda (prod_{i=1}^n x_i-1)\
frac{dL}{dx_k} = 1 - lambda frac{prod_{i=1}^n x_i}{x_k}\
frac{d^2L}{dx_k^2} = 0\
frac{d^2L}{dx_k d x_m} = - lambda frac{prod_{i=1}^n x_i}{x_k x_m} = -1 (k ne m)
$$
Now you need that for any vector $Delta x$ with $Delta x_1 + Delta x_2 +...+ Delta x_n = 0$ holds: $$sum_{i=1}^nsum_{j=1}^n frac{d^2L}{dx_i d x_j} Delta x_i Delta x_j > 0$$. We have
$$sum_{i=1}^nsum_{j=1}^n frac{d^2L}{dx_i d x_j} Delta x_i Delta x_j = - sum_{i=1}^nsum_{j neq i} Delta x_i Delta x_j \
= - sum_{i=1}^n Delta x_i sum_{j neq i} Delta x_j \
= - sum_{i=1}^n Delta x_i (0 -Delta x_i ) = sum_{i=1}^n Delta x_i^2 > 0$$.
answered Nov 21 '18 at 13:42
Andreas
7,8281037
Taking the second derivatives and evaluating at $lambda = x_1 = x_2 = ...=x_n = 1$ gives
$$ L = sum_{i=1}^n x_i - lambda (prod_{i=1}^n x_i-1)\
frac{dL}{dx_k} = 1 - lambda frac{prod_{i=1}^n x_i}{x_k}\
frac{d^2L}{dx_k^2} = 0\
frac{d^2L}{dx_k d x_m} = - lambda frac{prod_{i=1}^n x_i}{x_k x_m} = -1 (k ne m)
$$
Now you need that for any vector $Delta x$ with $Delta x_1 + Delta x_2 +...+ Delta x_n = 0$ holds: $$sum_{i=1}^nsum_{j=1}^n frac{d^2L}{dx_i d x_j} Delta x_i Delta x_j > 0$$. We have
$$sum_{i=1}^nsum_{j=1}^n frac{d^2L}{dx_i d x_j} Delta x_i Delta x_j = - sum_{i=1}^nsum_{j neq i} Delta x_i Delta x_j \
= - sum_{i=1}^n Delta x_i sum_{j neq i} Delta x_j \
= - sum_{i=1}^n Delta x_i (0 -Delta x_i ) = sum_{i=1}^n Delta x_i^2 > 0$$.
answered Nov 21 '18 at 13:42
Andreas
7,8281037
answered Nov 21 '18 at 13:42
Andreas
7,8281037
answered Nov 21 '18 at 13:42
Andreas
7,8281037
answered Nov 21 '18 at 13:42
Andreas
7,8281037
7,8281037
Thanks! Figured out as well. Now left with the local to global part to think of.
– MotherLand
Nov 21 '18 at 14:50
add a comment |
Thanks! Figured out as well. Now left with the local to global part to think of.
– MotherLand
Nov 21 '18 at 14:50
Thanks! Figured out as well. Now left with the local to global part to think of.
– MotherLand
Nov 21 '18 at 14:50
Thanks! Figured out as well. Now left with the local to global part to think of.
– MotherLand
Nov 21 '18 at 14:50
add a comment |
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