$f(t)=(textdet(A+tB))^frac1d$ is concave if $f(t)leq f(0)+tf'(0)$?

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Let $A$, $B$ be $d$ by $d$ matrices where $A$, $A+B$ are symmetric positive semi-definite matrices and $textdetB=0$. $f(t)=(textdet(A+tB))^frac1d$
is my function where $tin[0,1]$.




If $f(t)leq f(0)+tf'(0)$, then $f$ is concave over $[0,1]$.




I just asked To be concave on $[0,1]$, $f(t)leq f(0)+tf'(0)$ is enough?
for general function, but turns out there is a counterexample.



$f$ is special here, which is a composition of exponent function
and determinent, it could be true. Thanks in advance.



My trial : $f(t)=(textdet(A+tB))^frac1d$
looks like a exponent $frac1d$ of a polynomial function of order
$d$. then $f(t)=(a_1t^d+cdots)^frac1d$.




Second trial : Let $f(t)=(a_dt^d+cdots+a_0)^1/d$.



$tf(1/t)=(det(tA+B))^1/d$, LHS$=(a_d+cdots+a_0t^d)^1/d$
take $t=0$, then $a_d=textdetB=0$.



So $f(t)=(a_d-1t^d-1+cdots+a_0)^1/d$



We can see $f$ is smooth. Let's claim $f$ doesn't have zero on $(0,1)$.



Suppose not, i.e. $f(z)=0$ for some $zin(0,1)$. We know $f(1)=left(textdet(A+B)right)^1/dgeq0$.



Then, we can change $A$ into $A+zB$ then the condition $tildef(t)leqtildef(0)+ttildef'(0)$
doesn't make sence. So $f$
doesn't have any zero on $(0,1)$.



Here, I can't go farther.










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  • you need special matrices $A,B$ if you want the determinant to be non-negative or positive
    – Calvin Khor
    Sep 6 at 8:45










  • @CalvinKhor Thanks for comment, I edited my question.
    – kayak
    Sep 6 at 8:48










  • Can you provide more context, and the textbook? also either the claim that $f(0) in mathbb R^+$ is false, or in fact A is more than pos. semi-def(or I don't like your notation $mathbb R^+$).
    – Calvin Khor
    Sep 7 at 18:29











  • What do you mean by changing $A$ into $A+zB$, and how did you use the inequality? also, can you tell us what textbook you are using?
    – Calvin Khor
    Sep 9 at 13:31






  • 1




    found it on the arxiv arxiv.org/abs/1705.00331
    – Calvin Khor
    Sep 9 at 13:40














up vote
1
down vote

favorite












Let $A$, $B$ be $d$ by $d$ matrices where $A$, $A+B$ are symmetric positive semi-definite matrices and $textdetB=0$. $f(t)=(textdet(A+tB))^frac1d$
is my function where $tin[0,1]$.




If $f(t)leq f(0)+tf'(0)$, then $f$ is concave over $[0,1]$.




I just asked To be concave on $[0,1]$, $f(t)leq f(0)+tf'(0)$ is enough?
for general function, but turns out there is a counterexample.



$f$ is special here, which is a composition of exponent function
and determinent, it could be true. Thanks in advance.



My trial : $f(t)=(textdet(A+tB))^frac1d$
looks like a exponent $frac1d$ of a polynomial function of order
$d$. then $f(t)=(a_1t^d+cdots)^frac1d$.




Second trial : Let $f(t)=(a_dt^d+cdots+a_0)^1/d$.



$tf(1/t)=(det(tA+B))^1/d$, LHS$=(a_d+cdots+a_0t^d)^1/d$
take $t=0$, then $a_d=textdetB=0$.



So $f(t)=(a_d-1t^d-1+cdots+a_0)^1/d$



We can see $f$ is smooth. Let's claim $f$ doesn't have zero on $(0,1)$.



Suppose not, i.e. $f(z)=0$ for some $zin(0,1)$. We know $f(1)=left(textdet(A+B)right)^1/dgeq0$.



Then, we can change $A$ into $A+zB$ then the condition $tildef(t)leqtildef(0)+ttildef'(0)$
doesn't make sence. So $f$
doesn't have any zero on $(0,1)$.



Here, I can't go farther.










share|cite|improve this question























  • you need special matrices $A,B$ if you want the determinant to be non-negative or positive
    – Calvin Khor
    Sep 6 at 8:45










  • @CalvinKhor Thanks for comment, I edited my question.
    – kayak
    Sep 6 at 8:48










  • Can you provide more context, and the textbook? also either the claim that $f(0) in mathbb R^+$ is false, or in fact A is more than pos. semi-def(or I don't like your notation $mathbb R^+$).
    – Calvin Khor
    Sep 7 at 18:29











  • What do you mean by changing $A$ into $A+zB$, and how did you use the inequality? also, can you tell us what textbook you are using?
    – Calvin Khor
    Sep 9 at 13:31






  • 1




    found it on the arxiv arxiv.org/abs/1705.00331
    – Calvin Khor
    Sep 9 at 13:40












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $A$, $B$ be $d$ by $d$ matrices where $A$, $A+B$ are symmetric positive semi-definite matrices and $textdetB=0$. $f(t)=(textdet(A+tB))^frac1d$
is my function where $tin[0,1]$.




If $f(t)leq f(0)+tf'(0)$, then $f$ is concave over $[0,1]$.




I just asked To be concave on $[0,1]$, $f(t)leq f(0)+tf'(0)$ is enough?
for general function, but turns out there is a counterexample.



$f$ is special here, which is a composition of exponent function
and determinent, it could be true. Thanks in advance.



My trial : $f(t)=(textdet(A+tB))^frac1d$
looks like a exponent $frac1d$ of a polynomial function of order
$d$. then $f(t)=(a_1t^d+cdots)^frac1d$.




Second trial : Let $f(t)=(a_dt^d+cdots+a_0)^1/d$.



$tf(1/t)=(det(tA+B))^1/d$, LHS$=(a_d+cdots+a_0t^d)^1/d$
take $t=0$, then $a_d=textdetB=0$.



So $f(t)=(a_d-1t^d-1+cdots+a_0)^1/d$



We can see $f$ is smooth. Let's claim $f$ doesn't have zero on $(0,1)$.



Suppose not, i.e. $f(z)=0$ for some $zin(0,1)$. We know $f(1)=left(textdet(A+B)right)^1/dgeq0$.



Then, we can change $A$ into $A+zB$ then the condition $tildef(t)leqtildef(0)+ttildef'(0)$
doesn't make sence. So $f$
doesn't have any zero on $(0,1)$.



Here, I can't go farther.










share|cite|improve this question















Let $A$, $B$ be $d$ by $d$ matrices where $A$, $A+B$ are symmetric positive semi-definite matrices and $textdetB=0$. $f(t)=(textdet(A+tB))^frac1d$
is my function where $tin[0,1]$.




If $f(t)leq f(0)+tf'(0)$, then $f$ is concave over $[0,1]$.




I just asked To be concave on $[0,1]$, $f(t)leq f(0)+tf'(0)$ is enough?
for general function, but turns out there is a counterexample.



$f$ is special here, which is a composition of exponent function
and determinent, it could be true. Thanks in advance.



My trial : $f(t)=(textdet(A+tB))^frac1d$
looks like a exponent $frac1d$ of a polynomial function of order
$d$. then $f(t)=(a_1t^d+cdots)^frac1d$.




Second trial : Let $f(t)=(a_dt^d+cdots+a_0)^1/d$.



$tf(1/t)=(det(tA+B))^1/d$, LHS$=(a_d+cdots+a_0t^d)^1/d$
take $t=0$, then $a_d=textdetB=0$.



So $f(t)=(a_d-1t^d-1+cdots+a_0)^1/d$



We can see $f$ is smooth. Let's claim $f$ doesn't have zero on $(0,1)$.



Suppose not, i.e. $f(z)=0$ for some $zin(0,1)$. We know $f(1)=left(textdet(A+B)right)^1/dgeq0$.



Then, we can change $A$ into $A+zB$ then the condition $tildef(t)leqtildef(0)+ttildef'(0)$
doesn't make sence. So $f$
doesn't have any zero on $(0,1)$.



Here, I can't go farther.







convex-analysis






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edited Sep 9 at 10:28

























asked Sep 6 at 8:42









kayak

578318




578318











  • you need special matrices $A,B$ if you want the determinant to be non-negative or positive
    – Calvin Khor
    Sep 6 at 8:45










  • @CalvinKhor Thanks for comment, I edited my question.
    – kayak
    Sep 6 at 8:48










  • Can you provide more context, and the textbook? also either the claim that $f(0) in mathbb R^+$ is false, or in fact A is more than pos. semi-def(or I don't like your notation $mathbb R^+$).
    – Calvin Khor
    Sep 7 at 18:29











  • What do you mean by changing $A$ into $A+zB$, and how did you use the inequality? also, can you tell us what textbook you are using?
    – Calvin Khor
    Sep 9 at 13:31






  • 1




    found it on the arxiv arxiv.org/abs/1705.00331
    – Calvin Khor
    Sep 9 at 13:40
















  • you need special matrices $A,B$ if you want the determinant to be non-negative or positive
    – Calvin Khor
    Sep 6 at 8:45










  • @CalvinKhor Thanks for comment, I edited my question.
    – kayak
    Sep 6 at 8:48










  • Can you provide more context, and the textbook? also either the claim that $f(0) in mathbb R^+$ is false, or in fact A is more than pos. semi-def(or I don't like your notation $mathbb R^+$).
    – Calvin Khor
    Sep 7 at 18:29











  • What do you mean by changing $A$ into $A+zB$, and how did you use the inequality? also, can you tell us what textbook you are using?
    – Calvin Khor
    Sep 9 at 13:31






  • 1




    found it on the arxiv arxiv.org/abs/1705.00331
    – Calvin Khor
    Sep 9 at 13:40















you need special matrices $A,B$ if you want the determinant to be non-negative or positive
– Calvin Khor
Sep 6 at 8:45




you need special matrices $A,B$ if you want the determinant to be non-negative or positive
– Calvin Khor
Sep 6 at 8:45












@CalvinKhor Thanks for comment, I edited my question.
– kayak
Sep 6 at 8:48




@CalvinKhor Thanks for comment, I edited my question.
– kayak
Sep 6 at 8:48












Can you provide more context, and the textbook? also either the claim that $f(0) in mathbb R^+$ is false, or in fact A is more than pos. semi-def(or I don't like your notation $mathbb R^+$).
– Calvin Khor
Sep 7 at 18:29





Can you provide more context, and the textbook? also either the claim that $f(0) in mathbb R^+$ is false, or in fact A is more than pos. semi-def(or I don't like your notation $mathbb R^+$).
– Calvin Khor
Sep 7 at 18:29













What do you mean by changing $A$ into $A+zB$, and how did you use the inequality? also, can you tell us what textbook you are using?
– Calvin Khor
Sep 9 at 13:31




What do you mean by changing $A$ into $A+zB$, and how did you use the inequality? also, can you tell us what textbook you are using?
– Calvin Khor
Sep 9 at 13:31




1




1




found it on the arxiv arxiv.org/abs/1705.00331
– Calvin Khor
Sep 9 at 13:40




found it on the arxiv arxiv.org/abs/1705.00331
– Calvin Khor
Sep 9 at 13:40










1 Answer
1






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oldest

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up vote
1
down vote













Some headway... Let $C(t) = A+tB$ for $tin[0,1]$. Note that $C(t)$ is a convex combination of positive semi-definite matrices $A$ and $A+B$, so it is positive semi-definite. Thus, $det C(t) ge 0$ and thus $f(t)$ is well defined for every $tin[0,1]$.



One can attempt to employ the highschool second derivative test. Define $s := 1/d$, and $g(t):= det C(t)$ for notational brevity. Then we have



$$ f'(t) = fracddt (g(t)^s ) =sg(t)^s-1 g'(t)$$
and
$$ f''(t) = s(s-1) g(t)^s-2 g'(t)^2 + sg(t)^s-1 g''(t)$$



Now observe that



  • $fracddt [detA(t)]=detA(t) operatornametr[A^-1(t), fracddt A(t)]$

  • $fracd^2dt^2 det(A(t)) = det left(Aright) left( left(operatornametrleft(A^-1 A_tright)right)^2 + operatornametrleft(A^-1 A_ttright)-operatornametrleft(A^-1A_t A^-1 A_tright)right)$

so
beginalignf''(t)
&= s(s-1)g(t)^s [operatornametr(C^-1 B)]^2 + s g(t)^sleft([operatornametr(C^-1 B)]^2
+0-operatornametr(C^-1BC^-1B) right) \
&= sg(t)^s left( s [operatornametr(C^-1 B)]^2
-operatornametr((C^-1B)^2) right) endalign



Similarly to the linked question, it seems that the given inequality $ f(t) le f(0) + tf'(0)$,
i.e.,
$$g(t)^s le g(0)^s + tsg(0)^soperatornametr(A^-1B)$$
is not helpful. It seems like we need to know an inequality between traces of $D:=C^-1B$ and $D^2$. As the trace is a sum of eigenvalues, it may be useful to note that

$$ s(operatornametr D)^2 - operatornametr(D^2) = s(sum_i lambda_i )^2 - (sum_i lambda_i^2) $$



If one assumes that $A,A+B$ are better behaved, namely positive definite, then $ f > 0$ is smooth on $(0,1)$, so the above has a chance of working. If in addition the dimension is 2, then there is the equality
$$ det M = frac12((operatornametrM)^2 - operatornametr(M^2) ) $$
so we can notice that
$$ f''(t) = frac12g(t)^1/2 left( -frac12operatornametr((C^-1B)^2)right) < 0$$
which proves the result true in this special case, without using the inequality $f(t) le f(0) + tf'(0)$. There are also similar inequalities in dimensions $>2$ but I don't know how to use them.






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    up vote
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    down vote













    Some headway... Let $C(t) = A+tB$ for $tin[0,1]$. Note that $C(t)$ is a convex combination of positive semi-definite matrices $A$ and $A+B$, so it is positive semi-definite. Thus, $det C(t) ge 0$ and thus $f(t)$ is well defined for every $tin[0,1]$.



    One can attempt to employ the highschool second derivative test. Define $s := 1/d$, and $g(t):= det C(t)$ for notational brevity. Then we have



    $$ f'(t) = fracddt (g(t)^s ) =sg(t)^s-1 g'(t)$$
    and
    $$ f''(t) = s(s-1) g(t)^s-2 g'(t)^2 + sg(t)^s-1 g''(t)$$



    Now observe that



    • $fracddt [detA(t)]=detA(t) operatornametr[A^-1(t), fracddt A(t)]$

    • $fracd^2dt^2 det(A(t)) = det left(Aright) left( left(operatornametrleft(A^-1 A_tright)right)^2 + operatornametrleft(A^-1 A_ttright)-operatornametrleft(A^-1A_t A^-1 A_tright)right)$

    so
    beginalignf''(t)
    &= s(s-1)g(t)^s [operatornametr(C^-1 B)]^2 + s g(t)^sleft([operatornametr(C^-1 B)]^2
    +0-operatornametr(C^-1BC^-1B) right) \
    &= sg(t)^s left( s [operatornametr(C^-1 B)]^2
    -operatornametr((C^-1B)^2) right) endalign



    Similarly to the linked question, it seems that the given inequality $ f(t) le f(0) + tf'(0)$,
    i.e.,
    $$g(t)^s le g(0)^s + tsg(0)^soperatornametr(A^-1B)$$
    is not helpful. It seems like we need to know an inequality between traces of $D:=C^-1B$ and $D^2$. As the trace is a sum of eigenvalues, it may be useful to note that

    $$ s(operatornametr D)^2 - operatornametr(D^2) = s(sum_i lambda_i )^2 - (sum_i lambda_i^2) $$



    If one assumes that $A,A+B$ are better behaved, namely positive definite, then $ f > 0$ is smooth on $(0,1)$, so the above has a chance of working. If in addition the dimension is 2, then there is the equality
    $$ det M = frac12((operatornametrM)^2 - operatornametr(M^2) ) $$
    so we can notice that
    $$ f''(t) = frac12g(t)^1/2 left( -frac12operatornametr((C^-1B)^2)right) < 0$$
    which proves the result true in this special case, without using the inequality $f(t) le f(0) + tf'(0)$. There are also similar inequalities in dimensions $>2$ but I don't know how to use them.






    share|cite|improve this answer


























      up vote
      1
      down vote













      Some headway... Let $C(t) = A+tB$ for $tin[0,1]$. Note that $C(t)$ is a convex combination of positive semi-definite matrices $A$ and $A+B$, so it is positive semi-definite. Thus, $det C(t) ge 0$ and thus $f(t)$ is well defined for every $tin[0,1]$.



      One can attempt to employ the highschool second derivative test. Define $s := 1/d$, and $g(t):= det C(t)$ for notational brevity. Then we have



      $$ f'(t) = fracddt (g(t)^s ) =sg(t)^s-1 g'(t)$$
      and
      $$ f''(t) = s(s-1) g(t)^s-2 g'(t)^2 + sg(t)^s-1 g''(t)$$



      Now observe that



      • $fracddt [detA(t)]=detA(t) operatornametr[A^-1(t), fracddt A(t)]$

      • $fracd^2dt^2 det(A(t)) = det left(Aright) left( left(operatornametrleft(A^-1 A_tright)right)^2 + operatornametrleft(A^-1 A_ttright)-operatornametrleft(A^-1A_t A^-1 A_tright)right)$

      so
      beginalignf''(t)
      &= s(s-1)g(t)^s [operatornametr(C^-1 B)]^2 + s g(t)^sleft([operatornametr(C^-1 B)]^2
      +0-operatornametr(C^-1BC^-1B) right) \
      &= sg(t)^s left( s [operatornametr(C^-1 B)]^2
      -operatornametr((C^-1B)^2) right) endalign



      Similarly to the linked question, it seems that the given inequality $ f(t) le f(0) + tf'(0)$,
      i.e.,
      $$g(t)^s le g(0)^s + tsg(0)^soperatornametr(A^-1B)$$
      is not helpful. It seems like we need to know an inequality between traces of $D:=C^-1B$ and $D^2$. As the trace is a sum of eigenvalues, it may be useful to note that

      $$ s(operatornametr D)^2 - operatornametr(D^2) = s(sum_i lambda_i )^2 - (sum_i lambda_i^2) $$



      If one assumes that $A,A+B$ are better behaved, namely positive definite, then $ f > 0$ is smooth on $(0,1)$, so the above has a chance of working. If in addition the dimension is 2, then there is the equality
      $$ det M = frac12((operatornametrM)^2 - operatornametr(M^2) ) $$
      so we can notice that
      $$ f''(t) = frac12g(t)^1/2 left( -frac12operatornametr((C^-1B)^2)right) < 0$$
      which proves the result true in this special case, without using the inequality $f(t) le f(0) + tf'(0)$. There are also similar inequalities in dimensions $>2$ but I don't know how to use them.






      share|cite|improve this answer
























        up vote
        1
        down vote










        up vote
        1
        down vote









        Some headway... Let $C(t) = A+tB$ for $tin[0,1]$. Note that $C(t)$ is a convex combination of positive semi-definite matrices $A$ and $A+B$, so it is positive semi-definite. Thus, $det C(t) ge 0$ and thus $f(t)$ is well defined for every $tin[0,1]$.



        One can attempt to employ the highschool second derivative test. Define $s := 1/d$, and $g(t):= det C(t)$ for notational brevity. Then we have



        $$ f'(t) = fracddt (g(t)^s ) =sg(t)^s-1 g'(t)$$
        and
        $$ f''(t) = s(s-1) g(t)^s-2 g'(t)^2 + sg(t)^s-1 g''(t)$$



        Now observe that



        • $fracddt [detA(t)]=detA(t) operatornametr[A^-1(t), fracddt A(t)]$

        • $fracd^2dt^2 det(A(t)) = det left(Aright) left( left(operatornametrleft(A^-1 A_tright)right)^2 + operatornametrleft(A^-1 A_ttright)-operatornametrleft(A^-1A_t A^-1 A_tright)right)$

        so
        beginalignf''(t)
        &= s(s-1)g(t)^s [operatornametr(C^-1 B)]^2 + s g(t)^sleft([operatornametr(C^-1 B)]^2
        +0-operatornametr(C^-1BC^-1B) right) \
        &= sg(t)^s left( s [operatornametr(C^-1 B)]^2
        -operatornametr((C^-1B)^2) right) endalign



        Similarly to the linked question, it seems that the given inequality $ f(t) le f(0) + tf'(0)$,
        i.e.,
        $$g(t)^s le g(0)^s + tsg(0)^soperatornametr(A^-1B)$$
        is not helpful. It seems like we need to know an inequality between traces of $D:=C^-1B$ and $D^2$. As the trace is a sum of eigenvalues, it may be useful to note that

        $$ s(operatornametr D)^2 - operatornametr(D^2) = s(sum_i lambda_i )^2 - (sum_i lambda_i^2) $$



        If one assumes that $A,A+B$ are better behaved, namely positive definite, then $ f > 0$ is smooth on $(0,1)$, so the above has a chance of working. If in addition the dimension is 2, then there is the equality
        $$ det M = frac12((operatornametrM)^2 - operatornametr(M^2) ) $$
        so we can notice that
        $$ f''(t) = frac12g(t)^1/2 left( -frac12operatornametr((C^-1B)^2)right) < 0$$
        which proves the result true in this special case, without using the inequality $f(t) le f(0) + tf'(0)$. There are also similar inequalities in dimensions $>2$ but I don't know how to use them.






        share|cite|improve this answer














        Some headway... Let $C(t) = A+tB$ for $tin[0,1]$. Note that $C(t)$ is a convex combination of positive semi-definite matrices $A$ and $A+B$, so it is positive semi-definite. Thus, $det C(t) ge 0$ and thus $f(t)$ is well defined for every $tin[0,1]$.



        One can attempt to employ the highschool second derivative test. Define $s := 1/d$, and $g(t):= det C(t)$ for notational brevity. Then we have



        $$ f'(t) = fracddt (g(t)^s ) =sg(t)^s-1 g'(t)$$
        and
        $$ f''(t) = s(s-1) g(t)^s-2 g'(t)^2 + sg(t)^s-1 g''(t)$$



        Now observe that



        • $fracddt [detA(t)]=detA(t) operatornametr[A^-1(t), fracddt A(t)]$

        • $fracd^2dt^2 det(A(t)) = det left(Aright) left( left(operatornametrleft(A^-1 A_tright)right)^2 + operatornametrleft(A^-1 A_ttright)-operatornametrleft(A^-1A_t A^-1 A_tright)right)$

        so
        beginalignf''(t)
        &= s(s-1)g(t)^s [operatornametr(C^-1 B)]^2 + s g(t)^sleft([operatornametr(C^-1 B)]^2
        +0-operatornametr(C^-1BC^-1B) right) \
        &= sg(t)^s left( s [operatornametr(C^-1 B)]^2
        -operatornametr((C^-1B)^2) right) endalign



        Similarly to the linked question, it seems that the given inequality $ f(t) le f(0) + tf'(0)$,
        i.e.,
        $$g(t)^s le g(0)^s + tsg(0)^soperatornametr(A^-1B)$$
        is not helpful. It seems like we need to know an inequality between traces of $D:=C^-1B$ and $D^2$. As the trace is a sum of eigenvalues, it may be useful to note that

        $$ s(operatornametr D)^2 - operatornametr(D^2) = s(sum_i lambda_i )^2 - (sum_i lambda_i^2) $$



        If one assumes that $A,A+B$ are better behaved, namely positive definite, then $ f > 0$ is smooth on $(0,1)$, so the above has a chance of working. If in addition the dimension is 2, then there is the equality
        $$ det M = frac12((operatornametrM)^2 - operatornametr(M^2) ) $$
        so we can notice that
        $$ f''(t) = frac12g(t)^1/2 left( -frac12operatornametr((C^-1B)^2)right) < 0$$
        which proves the result true in this special case, without using the inequality $f(t) le f(0) + tf'(0)$. There are also similar inequalities in dimensions $>2$ but I don't know how to use them.







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        edited Sep 8 at 2:13

























        answered Sep 7 at 18:21









        Calvin Khor

        8,88121133




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