MA1132 2015-2016 Tutorial Sheet 11 PDF

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Module MA1132 (Frolov), Advanced Calculus Tutorial Sheet 11 To be solved during the tutorial session Thursday/Friday, 7/8 April 2016 Name: 1. Compute the integral Z In (A, B) =

∞

−∞

···

Z

∞

−∞

exp −

n X

xi Aij xj + 2

i,j=1

n X

Bi xi

i=1

!

dx1 · · · dxn ,

where A = (Aij ) is a positive-definite symmetric n × n matrix.

Solution: Let us introduce the column x with entries xi and the column B with entries Bi . Then we can write in matrix form n n X X Bi xi = −xt · A · x + 2B t · x , x A x + 2 − (1) i ij j i=1

i,j=1

where · denotes the matrix multiplication, and t denotes matrix transposition. Then using that At = A we can write (complete the square) −xt · A · x + 2B t · x = −(x − A−1 · B)t · A · (x − A−1 · B) + B t · A−1 · B = −y t · A · y + B t · A−1 · B ,

where y = x − A−1 · B. Thus, the integral In (A, B) is equal to Z ∞ Z ∞ t t −1 e−y ·A·y+B ·A ·B dy1 · · · dyn In (A, B) = ··· −∞ −∞ Z Z ∞ ∞ t t −1 e−y ·A·y dy1 · · · dyn . = eB ·A ·B ···

(2)

(3)

−∞

−∞

Since A is symmetric it can be diagonalised by an orthogonal matrix A = O · Λ · Ot ,

Λ = diag(λ1 , . . . , λn ) ,

Now, we make the following change of variables y = O·z

⇔

yi =

λi > 0 ∀i , n X

O : O · Ot = I .

Oij zj .

(4)

(5)

j=1

The Jacobian matrix is (∂yi /∂zj ) = Oij . Since O is orthogonal one gets J = det(∂yi /∂zj ) = ±1 . Thus the integral takes the form Z ∞ Z ∞ t t B t ·A−1 ·B e−y ·O··ΛO ·y dy1 · · · dyn In (A, B) = e ··· Z−∞ Z −∞ ∞ ∞ t t −1 e−z ··Λz dz1 · · · dzn = eB ·A ·B ··· Z−∞ Z −∞ ∞ ∞ Pn t −1 2 = eB ·A ·B ··· e− i=1 λiz i dz1 · · · dzn −∞ −∞ n t −1 B t ·A−1 ·B Z ∞ eB ·A ·B n/2 e −z 2 e dz π . =√ = √ λ1 · · · λn λ1 · · · λn −∞ 1

(6)

Since λ1 · · · λn = det A, we finally get In (A, B) = eB

t ·A−1 ·B

π n/2 . det A

(7)

2. Find the volume Vn,h of an n-simplex ∆ n ∆n :

x1 ≥ 0 , . . . , xn ≥ 0 ,

x1 + x2 + · · · + xn ≤ h .

(8)

Solution: From the definition of ∆n we get 0 ≤ xn ≤ h − x1 − x2 − · · · − xn−1 .

(9)

Thus, the projection of ∆n onto the hyperplane xn = 0 is the n − 1-simplex ∆n−1 , and Z h−x1 −···−xn−1  Z Z Z Z Vn,h = · · · dx1 · · · dxn = · · · dxn dx1 · · · dxn−1 . (10) ∆n

0

∆n−1

Repeating the procedure we get Z Z h Z h−x1 Vn,h = ··· 0

h−x1 −···−xn−1



(11)

dxn · · · dx2 dx1 .

0

0



Making the change x1 = hξ1 , x2 = hξ2 , . . . , xn = hξn ,

(12)

one gets n

Vn,h = h αn ,

αn ≡ Vn,1 =

Z 1 Z 0

1−ξ 1

···

0

Z

1−ξ 1 −···−ξ n−1 0





dξn · · · dξ2 dξ1 .

On the other hand comparing αn with (11), one sees that αn is equal to Z 1−ξ1 −···−ξn−1   Z 1 Z 1−ξ1 Z 1 ··· (Vn,1−ξ1 ) dξ1 αn = dξn · · · dξ2 dξ1 = 0 0 0 0 Z 1 αn−1 n−1 = αn−1 . (1 − ξ1 ) dξ1 = n 0

(13)

(14)

Taking into account that α1 = 1, one finds αn =

1 , n!

(15)

Vn,h =

hn . n!

(16)

and finally

2...