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Homework set #8 solutions, Math 128A
J. Xia
Sec 4.7: 1a, 3a, 7
1a. 3a. (^) ∫
- 5 1
x^2 ln xdx =
− 1
t + 5 4
ln
t + 5 4
dt.
Let f (t) =
( (^) t+ 4
ln
( (^) t+ 4
, then use Gaussian quadrature formula together with the roots of the Legendre polynomials and the coefficients for Gaussian quadrature in Table 4.11 of the textbook. For n = 2, ∫ (^1). 5
1
x^2 ln xdx ≈
[f (0.5773502692) + f (− 0 .5773502692)]
= 0. 192268706.
For n = 4, ∫ (^1). 5
1
x^2 ln xdx ≈
[0. 3478548451 · f (0.8611363116) + 0. 6521451548 · f (0.3399810436)
- 6521451548 · f (− 0 .3399810436) + 0. 3478548451 · f (− 0 .8611363116)] = 0. 192259358.
The exact value is ∫ (^1). 5
1
x^2 ln xdx =
x^3 ln x −
x^3 9 |^11. 5 (integration by parts)
= 0. 1922593577.
We can see for n = 2, the |error| = 9. 3483 × 10 −^6. But for n = 4, the |error| = 3 × 10 −^10 and we get higher accuracy.
- Assume otherwise the formula
Q(P ) =
∑^ n
i=
ciP (xi) (1)
has degree of precision greater than 2n − 1. Then it must also be exact for any polynomial with degree 2n. Now we can easily find a counterexample. Let
P (x) = (x − x 1 )^2 · · · (x − xn)^2 ,
where xi, i = 1, · · · , n are the same as those in (1), i.e. the roots of the degree-n Legendre polynomial. Then the left hand side of (1)
Q(P ) =
− 1
P (x)dx =
− 1
(x − x 1 )^2 · · · (x − xn)^2 dx > 0.
(Here Q(P ) > 0 because of the following. All xi, i = 1, · · · , n are distinct points in (− 1 , 1), so there exists a point ¯x ∈ some (xi, xi+1) ⊂ (− 1 , 1) s.t. P (¯x) 6 = 0, and thus P (¯x) > 0. By the continuity of P (x) in (− 1 , 1), there exists a interval [¯x − ≤, ¯x + ≤] ⊂ (xi, xi+1) s.t. P (x) > 0 for all x ∈ [¯x − ≤, x¯ + ≤]. Again by continuity P (x) reaches its minimum in [¯x − ≤, x¯ + ≤] so P (x) > k > 0 for x ∈ [¯x − ≤, x¯ + ≤]. Thus
− 1 P^ (x)dx^ ≥^
∫ (^) x¯+≤ x ¯−≤ kdx^ = 2≤k >^ 0.) But the right hand side of (1)
∑^ n
i=
ciP (xi) =
∑^ n
i=
ci · 0 (because P (xi) = 0)
≡ 0. regardless of ci
Thus (1) cannot be true. Contradiction.
Sec 4.8: 1a, 5a, 7a(iv), 15a
1a. Algorithm 4.4(Simpson’s double integral) function J = dblsimp(a,b,c,d,f,m,n) J1 = 0; J2 = 0; J3 = 0; for i = 0:n x = a+ih; dx = d; cx = c; % Here c, d are constants. % When they are functions as in problem 3 % use dx=feval(d,x); cx=feval(c,x); HX = (dx-cx)/m; K1 = feval(f,x,cx)+feval(f,x,dx); K2 = 0; K3 = 0; for j = 1:m- y = cx+jHX; Q = feval(f,x,y); if mod(j,2) == 0 K2 = K2+Q; else K3 = K3+Q; end end L = (K1+2K2+4K3)HX/3; if i == 0 | i == n J1 = J1+L; elseif mod(i,2) == 0 J2 = J2+L; else J3 = J3+L; end end J = h(J1+2J2+4J3)/3; And define the function f. Run in matlab dblsimp(2.1,2.5,1.2,1.4,’f’,4,4)
We have
∫ (^1)
0
P 4 (x) x^1 /^4
dx =
0
x − x 3 6 x^1 /^4
dx =
0
(x^3 /^4 −
x^11 /^4 6 )dx =
Then (^) ∫ (^1)
0
x−^1 /^4 sin xdx =
0
P 4 (x) x^1 /^4
dx +
0
G(x)dx,
where
G(x) =
sin x−P 4 (x) x^1 /^4 ,^ when^0 < x <^1 0 , when x = 0.
Applying the Composite Simpson’s rule to integrate G(x) by hand or by the following code. function xi = cmpsimp(a,b,n,f) h = (b-a)/n; xi0 = feval(’f’,a)+feval(’f’,b); xi1 = 0; xi2 = 0; for i = 1:n- x = a+ih; if mod(i,2) == 0 xi2 = xi2+feval(’f’,x); else xi1 = xi1+feval(’f’,x); end end xi = h(xi0+2xi2+4xi1)/3; Function definition: function y = f(x) if x>0 & x <= 1 y = (sin(x).-x.+x.^3/6)./sqrt(sqrt(x)); elseif x == 0 y = 0; end We have (^) ∫ (^1)
0
P 4 (x) x^1 /^4
dx +
0
G(x)dx ≈ 0. 528416326.
3a. Let t = 1/x, then
∫ (^) ∞
1
x^2 + 9
dx =
1
1 /t^2 + 9
t^2
)dt =
0
1 + 9t^2
dt ≈ 0. 41126487.
The evaluation in the last step can be done with the previous code, Composite Simpson’s rule, in problem 1a.
