Download MATH313 Preliminary Exam: Matrix Inversion, Gaussian Elimination, Quadrature and more Exams Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity!
MATH313 – Preliminary Examination.
August 20, 2007
Instructions: Answer two out of the four questions.You do not have to prove results which you rely upon, just state them clearly!
Q1) (a) Suppose that A(1)^ = A is an invertible n × n matrix and that the Gaussian elimination algorithm with partial pivoting applied to A(1)^ produces the upper triangular matrix A(n). As usual, let A(k)^ be the renamed A(k) following any necessary row intrechanges before the k–th major step of the elimination so that
a( i,jk+1) =
a( i,jk) , when i ≤ k, 1 ≤ j ≤ n, 0 when i ≥ k + 1, 1 ≤ j ≤ k, a( i,jk) − a( i,kk) a( k,jk) /a( k,kk), when i, j ≥ k + 1.
Show that the total number of multiplication and division operations needed to reduce A(1)^ to A(n)^ is (n^3 − n)/3. [Hint: Recall that
∑n i=1 i
(^2) = n(n +
1)(2n + 1)/6.]
b) Suppose that all the leading principal minors of A are positive. Show that A has an LU–factorization with unit diagonal entries in L and positive diagonal entries in U.
( c) Suppose now that no partial pivoting is necessary and that^ A(1)^ = a(1) i,j
) is tridiagonal, that is, a(1) i,j = 0 when |i − j| > 1 , 1 ≤ i, j ≤ n. Show
that each of A(1),... , A(n)^ is tridiagonal.
d) Suppose that A is an n × n invertible matrix which admits an LU– factorization without pivoting. Partition A into:
A =
A 1 , 1 A 1 , 2
A 2 , 1 A 2 , 2
,
with A 1 , 1 being a (k − 1) × (k − 1) matrix. Knowing that A 1 , 1 is invertible (why?), show that the current active array which is the (n−k+1)×(n−k−1) matrix Ak =
( a( i,jk)
) , i, j = k,... , n is given by:
Ak = A 2 , 2 − A 2 , 1 A− 2 ,^12 A 1 , 2.
Assume now that in addition to A being invertible, A is Hermitian. Use this formula to deduce that Ak is also Hermitian, k = 1,... , n.
Q3) (a) Prove: A qudrature formula In(f ) =
∑n k=0 αkf^ (xk) that uses the n+1 distinct nodes x 0 ,... , xn and is exact of order at least n is interpolatory, that is,
αk =
∫ (^) b
a
Lk(x)dx, k = 0,... , n,
where
Lk(x) =
∏n j= j 6 =k
(x − xj ) ∏n j= j 6 =k
(xk − xj )
, k = 0,... , n.
(b) The Legendre polynomial of degree n is defined by
Pn(x) =
2 nn!
dn dxn
( x^2 − 1
)n ,
with P 0 (x) ≡ 1. Calculate explicitly P 1 ,... , P 4. Prove (verify) that for k = 0, 1 ,... , n − 1, (^) ∫ 1 − 1
xkPn(x)dx = 0.
(c) Use part (b) to conclude that
∫ (^1) − 1 Pn(x)Pm(x)dx^ = 0, when^ m^6 =^ n, and that
∫ (^1) − 1 P^
2 n (x)dx^ = 2/(2n^ + 1).
