Finding a positive definite symmetric bilinear function . [closed]

0

$begingroup$

Could anyone give me a hint about solving this problem please?

EDIT:
I am adding this for the comments of @David, I think he is speaking about this theorem:

linear-algebra functional-analysis operator-theory representation-theory

share|cite|improve this question

edited Feb 2 at 0:40

Idonotknow

asked Feb 1 at 16:58

IdonotknowIdonotknow

638

$endgroup$

closed as off-topic by David Hill, mrtaurho, José Carlos Santos, ancientmathematician, Adrian Keister Feb 6 at 14:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – David Hill, mrtaurho, José Carlos Santos, ancientmathematician, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The usual thing to do is average over the group: Take $(cdot,cdot)$ to be the dot product and define $$langle v, wrangle=frac{1}{3}((v,w)+(Tv,Tw)+(T^2v,T^2w))$$
  $endgroup$
  – David Hill
  Feb 1 at 18:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know that the elements of $Z_{3}$ are {1,2,3} ..... then how will I use the given matrix .... the generator of $Z$ is 1 ..... what does it mean for the generator to go into the linear operator ...... why you wrote the inner product of v and w like this ? ..... I will add the formula for the averaging that I know in the question above (which I actually do not understand how to use it)@DavidHill
  $endgroup$
  – Idonotknow
  Feb 1 at 21:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know that for the reals with orthonormal basis the positive definite symmetric bilinear function is just the dot product.... but how can I use this information ? @DavidHill
  $endgroup$
  – Idonotknow
  Feb 1 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you understand what a representation of $mathbb{Z}_3$ is? Do you understand what a $T$-invariant means?
  $endgroup$
  – David Hill
  Feb 1 at 21:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I took $T$ to mean $T(1)=begin{pmatrix}0&-1\1&-1end{pmatrix}$. You should interpret $T^2=T(2)=begin{pmatrix}0&-1\1&-1end{pmatrix}^2$. Take $v=begin{pmatrix}a\bend{pmatrix}$ and $w=begin{pmatrix}c\dend{pmatrix}$. Though, you should be able to give a proof that $langlecdot,cdotrangle$ is $T$-invariant without choosing coordinates. You will, of course, need to check that $I=T(0)=T(3)=begin{pmatrix}0&-1\1&-1end{pmatrix}^3$.
  $endgroup$
  – David Hill
  Feb 2 at 1:06
  

  
  
  
  

 | 
show 3 more comments

0

$begingroup$

Could anyone give me a hint about solving this problem please?

EDIT:
I am adding this for the comments of @David, I think he is speaking about this theorem:

linear-algebra functional-analysis operator-theory representation-theory

share|cite|improve this question

edited Feb 2 at 0:40

Idonotknow

asked Feb 1 at 16:58

IdonotknowIdonotknow

638

$endgroup$

closed as off-topic by David Hill, mrtaurho, José Carlos Santos, ancientmathematician, Adrian Keister Feb 6 at 14:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – David Hill, mrtaurho, José Carlos Santos, ancientmathematician, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The usual thing to do is average over the group: Take $(cdot,cdot)$ to be the dot product and define $$langle v, wrangle=frac{1}{3}((v,w)+(Tv,Tw)+(T^2v,T^2w))$$
  $endgroup$
  – David Hill
  Feb 1 at 18:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know that the elements of $Z_{3}$ are {1,2,3} ..... then how will I use the given matrix .... the generator of $Z$ is 1 ..... what does it mean for the generator to go into the linear operator ...... why you wrote the inner product of v and w like this ? ..... I will add the formula for the averaging that I know in the question above (which I actually do not understand how to use it)@DavidHill
  $endgroup$
  – Idonotknow
  Feb 1 at 21:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know that for the reals with orthonormal basis the positive definite symmetric bilinear function is just the dot product.... but how can I use this information ? @DavidHill
  $endgroup$
  – Idonotknow
  Feb 1 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you understand what a representation of $mathbb{Z}_3$ is? Do you understand what a $T$-invariant means?
  $endgroup$
  – David Hill
  Feb 1 at 21:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I took $T$ to mean $T(1)=begin{pmatrix}0&-1\1&-1end{pmatrix}$. You should interpret $T^2=T(2)=begin{pmatrix}0&-1\1&-1end{pmatrix}^2$. Take $v=begin{pmatrix}a\bend{pmatrix}$ and $w=begin{pmatrix}c\dend{pmatrix}$. Though, you should be able to give a proof that $langlecdot,cdotrangle$ is $T$-invariant without choosing coordinates. You will, of course, need to check that $I=T(0)=T(3)=begin{pmatrix}0&-1\1&-1end{pmatrix}^3$.
  $endgroup$
  – David Hill
  Feb 2 at 1:06
  

  
  
  
  

 | 
show 3 more comments

0

0

0

$begingroup$

Could anyone give me a hint about solving this problem please?

EDIT:
I am adding this for the comments of @David, I think he is speaking about this theorem:

linear-algebra functional-analysis operator-theory representation-theory

share|cite|improve this question

edited Feb 2 at 0:40

Idonotknow

asked Feb 1 at 16:58

IdonotknowIdonotknow

638

$endgroup$

Could anyone give me a hint about solving this problem please?

EDIT:
I am adding this for the comments of @David, I think he is speaking about this theorem:

linear-algebra functional-analysis operator-theory representation-theory

linear-algebra functional-analysis operator-theory representation-theory

share|cite|improve this question

edited Feb 2 at 0:40

Idonotknow

asked Feb 1 at 16:58

IdonotknowIdonotknow

638

share|cite|improve this question

edited Feb 2 at 0:40

Idonotknow

asked Feb 1 at 16:58

IdonotknowIdonotknow

638

share|cite|improve this question

share|cite|improve this question

edited Feb 2 at 0:40

Idonotknow

edited Feb 2 at 0:40

Idonotknow

edited Feb 2 at 0:40

Idonotknow

asked Feb 1 at 16:58

IdonotknowIdonotknow

638

asked Feb 1 at 16:58

IdonotknowIdonotknow

638

asked Feb 1 at 16:58

IdonotknowIdonotknow

638

638

closed as off-topic by David Hill, mrtaurho, José Carlos Santos, ancientmathematician, Adrian Keister Feb 6 at 14:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – David Hill, mrtaurho, José Carlos Santos, ancientmathematician, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by David Hill, mrtaurho, José Carlos Santos, ancientmathematician, Adrian Keister Feb 6 at 14:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – David Hill, mrtaurho, José Carlos Santos, ancientmathematician, Adrian Keister

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The usual thing to do is average over the group: Take $(cdot,cdot)$ to be the dot product and define $$langle v, wrangle=frac{1}{3}((v,w)+(Tv,Tw)+(T^2v,T^2w))$$
  $endgroup$
  – David Hill
  Feb 1 at 18:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know that the elements of $Z_{3}$ are {1,2,3} ..... then how will I use the given matrix .... the generator of $Z$ is 1 ..... what does it mean for the generator to go into the linear operator ...... why you wrote the inner product of v and w like this ? ..... I will add the formula for the averaging that I know in the question above (which I actually do not understand how to use it)@DavidHill
  $endgroup$
  – Idonotknow
  Feb 1 at 21:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know that for the reals with orthonormal basis the positive definite symmetric bilinear function is just the dot product.... but how can I use this information ? @DavidHill
  $endgroup$
  – Idonotknow
  Feb 1 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you understand what a representation of $mathbb{Z}_3$ is? Do you understand what a $T$-invariant means?
  $endgroup$
  – David Hill
  Feb 1 at 21:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I took $T$ to mean $T(1)=begin{pmatrix}0&-1\1&-1end{pmatrix}$. You should interpret $T^2=T(2)=begin{pmatrix}0&-1\1&-1end{pmatrix}^2$. Take $v=begin{pmatrix}a\bend{pmatrix}$ and $w=begin{pmatrix}c\dend{pmatrix}$. Though, you should be able to give a proof that $langlecdot,cdotrangle$ is $T$-invariant without choosing coordinates. You will, of course, need to check that $I=T(0)=T(3)=begin{pmatrix}0&-1\1&-1end{pmatrix}^3$.
  $endgroup$
  – David Hill
  Feb 2 at 1:06
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The usual thing to do is average over the group: Take $(cdot,cdot)$ to be the dot product and define $$langle v, wrangle=frac{1}{3}((v,w)+(Tv,Tw)+(T^2v,T^2w))$$
  $endgroup$
  – David Hill
  Feb 1 at 18:41
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know that the elements of $Z_{3}$ are {1,2,3} ..... then how will I use the given matrix .... the generator of $Z$ is 1 ..... what does it mean for the generator to go into the linear operator ...... why you wrote the inner product of v and w like this ? ..... I will add the formula for the averaging that I know in the question above (which I actually do not understand how to use it)@DavidHill
  $endgroup$
  – Idonotknow
  Feb 1 at 21:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I know that for the reals with orthonormal basis the positive definite symmetric bilinear function is just the dot product.... but how can I use this information ? @DavidHill
  $endgroup$
  – Idonotknow
  Feb 1 at 21:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you understand what a representation of $mathbb{Z}_3$ is? Do you understand what a $T$-invariant means?
  $endgroup$
  – David Hill
  Feb 1 at 21:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I took $T$ to mean $T(1)=begin{pmatrix}0&-1\1&-1end{pmatrix}$. You should interpret $T^2=T(2)=begin{pmatrix}0&-1\1&-1end{pmatrix}^2$. Take $v=begin{pmatrix}a\bend{pmatrix}$ and $w=begin{pmatrix}c\dend{pmatrix}$. Though, you should be able to give a proof that $langlecdot,cdotrangle$ is $T$-invariant without choosing coordinates. You will, of course, need to check that $I=T(0)=T(3)=begin{pmatrix}0&-1\1&-1end{pmatrix}^3$.
  $endgroup$
  – David Hill
  Feb 2 at 1:06
  

  
  
  
  

2

2

$begingroup$
The usual thing to do is average over the group: Take $(cdot,cdot)$ to be the dot product and define $$langle v, wrangle=frac{1}{3}((v,w)+(Tv,Tw)+(T^2v,T^2w))$$
$endgroup$
– David Hill
Feb 1 at 18:41

$begingroup$
The usual thing to do is average over the group: Take $(cdot,cdot)$ to be the dot product and define $$langle v, wrangle=frac{1}{3}((v,w)+(Tv,Tw)+(T^2v,T^2w))$$
$endgroup$
– David Hill
Feb 1 at 18:41

$begingroup$
I know that the elements of $Z_{3}$ are {1,2,3} ..... then how will I use the given matrix .... the generator of $Z$ is 1 ..... what does it mean for the generator to go into the linear operator ...... why you wrote the inner product of v and w like this ? ..... I will add the formula for the averaging that I know in the question above (which I actually do not understand how to use it)@DavidHill
$endgroup$
– Idonotknow
Feb 1 at 21:08

$begingroup$
I know that the elements of $Z_{3}$ are {1,2,3} ..... then how will I use the given matrix .... the generator of $Z$ is 1 ..... what does it mean for the generator to go into the linear operator ...... why you wrote the inner product of v and w like this ? ..... I will add the formula for the averaging that I know in the question above (which I actually do not understand how to use it)@DavidHill
$endgroup$
– Idonotknow
Feb 1 at 21:08

$begingroup$
I know that for the reals with orthonormal basis the positive definite symmetric bilinear function is just the dot product.... but how can I use this information ? @DavidHill
$endgroup$
– Idonotknow
Feb 1 at 21:25

$begingroup$
I know that for the reals with orthonormal basis the positive definite symmetric bilinear function is just the dot product.... but how can I use this information ? @DavidHill
$endgroup$
– Idonotknow
Feb 1 at 21:25

$begingroup$
Do you understand what a representation of $mathbb{Z}_3$ is? Do you understand what a $T$-invariant means?
$endgroup$
– David Hill
Feb 1 at 21:27

$begingroup$
Do you understand what a representation of $mathbb{Z}_3$ is? Do you understand what a $T$-invariant means?
$endgroup$
– David Hill
Feb 1 at 21:27

1

1

$begingroup$
I took $T$ to mean $T(1)=begin{pmatrix}0&-1\1&-1end{pmatrix}$. You should interpret $T^2=T(2)=begin{pmatrix}0&-1\1&-1end{pmatrix}^2$. Take $v=begin{pmatrix}a\bend{pmatrix}$ and $w=begin{pmatrix}c\dend{pmatrix}$. Though, you should be able to give a proof that $langlecdot,cdotrangle$ is $T$-invariant without choosing coordinates. You will, of course, need to check that $I=T(0)=T(3)=begin{pmatrix}0&-1\1&-1end{pmatrix}^3$.
$endgroup$
– David Hill
Feb 2 at 1:06

$begingroup$
I took $T$ to mean $T(1)=begin{pmatrix}0&-1\1&-1end{pmatrix}$. You should interpret $T^2=T(2)=begin{pmatrix}0&-1\1&-1end{pmatrix}^2$. Take $v=begin{pmatrix}a\bend{pmatrix}$ and $w=begin{pmatrix}c\dend{pmatrix}$. Though, you should be able to give a proof that $langlecdot,cdotrangle$ is $T$-invariant without choosing coordinates. You will, of course, need to check that $I=T(0)=T(3)=begin{pmatrix}0&-1\1&-1end{pmatrix}^3$.
$endgroup$
– David Hill
Feb 2 at 1:06

 | 
show 3 more comments

1 Answer
1

active

oldest

votes

2

+50

$begingroup$

Let $(cdot,cdot):mathbb{R}^2tomathbb{R}$ be the usual dot product and define
$$
langlecdot,cdotrangle:mathbb{R}^2tomathbb{R}$$
by
$$langle v, wrangle=frac{1}{3}(langle T(0)v, T(0)wrangle+langle T(1)v, T(1)wrangle+langle T(2)v, T(2)wrangle).
$$
Note that
begin{align}
langle T(i)v, T(i)wrangle&=frac{1}{3}(langle T(0)T(i)v, T(0)T(i)wrangle+langle T(1)T(i)v, T(1)T(i)wrangle+langle T(2)T(i)v, T(2)T(i)wrangle)\
&=frac{1}{3}(langle T(0+i)v, T(0+i)wrangle+langle T(1+i)v, T(1+i)wrangle+langle T(2+i)v, T(2+i)wrangle)\
&=frac{1}{3}(langle T(0)v, T(0)wrangle+langle T(1)v, T(1)wrangle+langle T(2)v, T(2)wrangle)\
&=langle v,wrangle.
end{align}
It follows that $langlecdot,cdotrangle$ is $T$-invariant. To see that it is nondegenerate, let $displaystyle v=begin{pmatrix}a\bend{pmatrix}$ and compute directly that $langle v,vrangle=frac{2}{3}(a^2+b^2+(a-b)^2)$. Therefore, $langle v,vrangle >0$ unless $a=b=0$.

share|cite|improve this answer

answered Feb 4 at 18:53

David HillDavid Hill

9,5461619

$endgroup$

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

+50

$begingroup$

Let $(cdot,cdot):mathbb{R}^2tomathbb{R}$ be the usual dot product and define
$$
langlecdot,cdotrangle:mathbb{R}^2tomathbb{R}$$
by
$$langle v, wrangle=frac{1}{3}(langle T(0)v, T(0)wrangle+langle T(1)v, T(1)wrangle+langle T(2)v, T(2)wrangle).
$$
Note that
begin{align}
langle T(i)v, T(i)wrangle&=frac{1}{3}(langle T(0)T(i)v, T(0)T(i)wrangle+langle T(1)T(i)v, T(1)T(i)wrangle+langle T(2)T(i)v, T(2)T(i)wrangle)\
&=frac{1}{3}(langle T(0+i)v, T(0+i)wrangle+langle T(1+i)v, T(1+i)wrangle+langle T(2+i)v, T(2+i)wrangle)\
&=frac{1}{3}(langle T(0)v, T(0)wrangle+langle T(1)v, T(1)wrangle+langle T(2)v, T(2)wrangle)\
&=langle v,wrangle.
end{align}
It follows that $langlecdot,cdotrangle$ is $T$-invariant. To see that it is nondegenerate, let $displaystyle v=begin{pmatrix}a\bend{pmatrix}$ and compute directly that $langle v,vrangle=frac{2}{3}(a^2+b^2+(a-b)^2)$. Therefore, $langle v,vrangle >0$ unless $a=b=0$.

share|cite|improve this answer

answered Feb 4 at 18:53

David HillDavid Hill

9,5461619

$endgroup$

add a comment | 

2

+50

$begingroup$

Let $(cdot,cdot):mathbb{R}^2tomathbb{R}$ be the usual dot product and define
$$
langlecdot,cdotrangle:mathbb{R}^2tomathbb{R}$$
by
$$langle v, wrangle=frac{1}{3}(langle T(0)v, T(0)wrangle+langle T(1)v, T(1)wrangle+langle T(2)v, T(2)wrangle).
$$
Note that
begin{align}
langle T(i)v, T(i)wrangle&=frac{1}{3}(langle T(0)T(i)v, T(0)T(i)wrangle+langle T(1)T(i)v, T(1)T(i)wrangle+langle T(2)T(i)v, T(2)T(i)wrangle)\
&=frac{1}{3}(langle T(0+i)v, T(0+i)wrangle+langle T(1+i)v, T(1+i)wrangle+langle T(2+i)v, T(2+i)wrangle)\
&=frac{1}{3}(langle T(0)v, T(0)wrangle+langle T(1)v, T(1)wrangle+langle T(2)v, T(2)wrangle)\
&=langle v,wrangle.
end{align}
It follows that $langlecdot,cdotrangle$ is $T$-invariant. To see that it is nondegenerate, let $displaystyle v=begin{pmatrix}a\bend{pmatrix}$ and compute directly that $langle v,vrangle=frac{2}{3}(a^2+b^2+(a-b)^2)$. Therefore, $langle v,vrangle >0$ unless $a=b=0$.

share|cite|improve this answer

answered Feb 4 at 18:53

David HillDavid Hill

9,5461619

$endgroup$

add a comment | 

2

+50

2

+50

2

+50

$begingroup$

Let $(cdot,cdot):mathbb{R}^2tomathbb{R}$ be the usual dot product and define
$$
langlecdot,cdotrangle:mathbb{R}^2tomathbb{R}$$
by
$$langle v, wrangle=frac{1}{3}(langle T(0)v, T(0)wrangle+langle T(1)v, T(1)wrangle+langle T(2)v, T(2)wrangle).
$$
Note that
begin{align}
langle T(i)v, T(i)wrangle&=frac{1}{3}(langle T(0)T(i)v, T(0)T(i)wrangle+langle T(1)T(i)v, T(1)T(i)wrangle+langle T(2)T(i)v, T(2)T(i)wrangle)\
&=frac{1}{3}(langle T(0+i)v, T(0+i)wrangle+langle T(1+i)v, T(1+i)wrangle+langle T(2+i)v, T(2+i)wrangle)\
&=frac{1}{3}(langle T(0)v, T(0)wrangle+langle T(1)v, T(1)wrangle+langle T(2)v, T(2)wrangle)\
&=langle v,wrangle.
end{align}
It follows that $langlecdot,cdotrangle$ is $T$-invariant. To see that it is nondegenerate, let $displaystyle v=begin{pmatrix}a\bend{pmatrix}$ and compute directly that $langle v,vrangle=frac{2}{3}(a^2+b^2+(a-b)^2)$. Therefore, $langle v,vrangle >0$ unless $a=b=0$.

share|cite|improve this answer

answered Feb 4 at 18:53

David HillDavid Hill

9,5461619

$endgroup$

Let $(cdot,cdot):mathbb{R}^2tomathbb{R}$ be the usual dot product and define
$$
langlecdot,cdotrangle:mathbb{R}^2tomathbb{R}$$
by
$$langle v, wrangle=frac{1}{3}(langle T(0)v, T(0)wrangle+langle T(1)v, T(1)wrangle+langle T(2)v, T(2)wrangle).
$$
Note that
begin{align}
langle T(i)v, T(i)wrangle&=frac{1}{3}(langle T(0)T(i)v, T(0)T(i)wrangle+langle T(1)T(i)v, T(1)T(i)wrangle+langle T(2)T(i)v, T(2)T(i)wrangle)\
&=frac{1}{3}(langle T(0+i)v, T(0+i)wrangle+langle T(1+i)v, T(1+i)wrangle+langle T(2+i)v, T(2+i)wrangle)\
&=frac{1}{3}(langle T(0)v, T(0)wrangle+langle T(1)v, T(1)wrangle+langle T(2)v, T(2)wrangle)\
&=langle v,wrangle.
end{align}
It follows that $langlecdot,cdotrangle$ is $T$-invariant. To see that it is nondegenerate, let $displaystyle v=begin{pmatrix}a\bend{pmatrix}$ and compute directly that $langle v,vrangle=frac{2}{3}(a^2+b^2+(a-b)^2)$. Therefore, $langle v,vrangle >0$ unless $a=b=0$.

share|cite|improve this answer

answered Feb 4 at 18:53

David HillDavid Hill

9,5461619

share|cite|improve this answer

share|cite|improve this answer

answered Feb 4 at 18:53

David HillDavid Hill

9,5461619

answered Feb 4 at 18:53

David HillDavid Hill

9,5461619

answered Feb 4 at 18:53

David HillDavid Hill

9,5461619

9,5461619

add a comment | 

add a comment | 

This page is only for reference, If you need detailed information, please check here