Lab 01b Summary
M. M. Sussman
$Id: summary.txt,v 1.10 2016/12/22 21:03:18 mike Exp $
{@(#) Wed Dec 30 16:16:05 2015 }

EXERCISE 1
1. q_elt.m
2. following are zero or roundoff
   q_elt(@(x,y) 1,0,0,1)-1
   q_elt(@(x,y) 4*x*y,0,0,1)-1
   q_elt(@(x,y) 6*x^2*y,0,0,1)-1
   q_elt(@(x,y) 9*x^2*y^2,0,0,1)-1
   q_elt(@(x,y) 16*x^3*y^3,0,0,1)-1
3. q_elt(@(x,y) 25*x^4*y^4,0,0,1)-1=-0.054784 .NE. 0
   CHECK they have the correct integral!

EXERCISE 2
1. qerr_elt.m (using q_elt.m)
2. [q,errest]=qerr_elt(inline('16*x^3*y^3'),0,0,1);q-1,errest both zero
3. [q,errest]=qerr_elt(inline('25*x^4*y^4'),0,0,1);
      q-1=-0.0034692, errest=-0.0034210, rel diff=1.4%
      CHECK pct diff in errors

EXERCISE 3
1. q_total.m with:
   h=H/n;
   elt(elt_count).x=x+(k-1)*h;
   elt(elt_count).y=y+(j-1)*h;
   elt(elt_count).h=h;
2. area [0,1]X[0,1]=1 using 10 per side
3. area [-1,1]X[-1,1]=4 using 13 per side
4. add qerr_elt calls for q, errest
5. q=1, errest=roundoff for @(x,y)9*x^2*y^2 over [0,1]X[0,1] and n=1
6. q=4, errest=roundoff for @(x,y)9*x^2*y^2 over [-1,1]X[-1,1] and n=1
7. q=1, errest=roundoff for @(x,y)16*x^3*y^3 and n=2
8. n   integral    estimated error   true error
   2   0.99978       2.1681e-04      -2.1700e-04
   4   0.99999       1.3563e-05      -1.3563e-05 
   8   1.00000       8.4771e-07      -8.4771e-07
  16   1.00000       5.2982e-08      -5.2982e-08
9. 8.4771e-07/5.2982e-8=16: yes, O(h^4).

EXERCISE 4
1. copy three_peaks.m
2.   n   integral    estimated error   true error
    10   1.7553      1.4612e-04         5.4743e-05
    20   1.7552      4.6100e-06         7.2122e-08
    40   1.7552      4.2841e-07         1.3603e-09
    80   1.7552      2.6741e-08         8.5084e-11
   160   1.7552      1.6985e-09         5.3602e-12

3. 8.5084e-11/5.3602e-12=15.9, consistent with O(h^4)
4. q_total_noabs.m w/o abs: for n=80, esterr=8.5015e-11, almost same as true

EXERCISE 5
1. q_adaptive.m
2. 16*x^3*y^3 over [0,1]X[0,1] exactly 1
3. 9*x^2*y^2 over [-1,1]X[-1,1] exactly 4
4. 25*x^4*y^4 over [0,1]X[0,1]: q =  0.99978 errest =  2.1681e-04
   q-1 = -2.1700e-04, length(elt)=4
5. 25*x^4*y^4 over [0,1]X[0,1], tol=2.e-4: 
   q-1 =-1.184587e-04   errest = 1.18317e-04 length(elt)=7

EXERCISE 6
1. Download plotelt.m
2. 25*x^4*y^4, tolerance=1.e-6: q-1 = -9.5015e-07 q =  0.99999
   errest =  9.5001e-07 length(elt) =  55
3. plot: red at UR corner, some red along top and right
   two big reds near lower left corner
4. tol=9.e-7 two big reds are gone, refined once
5. tol=5.e-7, many smaller blocks along top and right are refined.

EXERCISE 7
Integrate three_peaks over [-1,1]X[-1,1], tolerance=1.e-5
q=1.7552, errest=9.7671e-06, q-1.75522375591726=-5.2826e-6, 145 elts, PLOT
  
EXERCISE 8
1. A*Ainv=identity
2. B*Binv, upper 5x5 is 5 reps of the line:
     4194304     4194304      131072           0           0

  A(1,:)*Ainv(:,1)=1
  B(1,:)*Binv(:,1)=4194304

 A(1,6)*Ainv(6,1)=-120
 A(1,5)*Ainv(5,1)= 240
 A(1,4)*Ainv(4,1)=-180
 A(1,3)*Ainv(3,1)=  80
 A(1,2)*Ainv(2,1)= -25
 A(1,1)*Ainv(1,1)=   6
           sum   =   1

 B(1,24)*Binv(24,1)=-2.5852e+22
 B(1,23)*Binv(23,1)= 5.1704e+22
 B(1,22)*Binv(22,1)=-3.8778e+22
 B(1,21)*Binv(21,1)= 1.7235e+22
 B(1,20)*Binv(20,1)=-5.3858e+21
 B(1,16)*Binv(16,1)=-5.7705e+18
 B(1,11)*Binv(11,1)= 5.8122e+13
 B(1,6) *Binv(6,1) =-76719720
 B(1,1) *Binv(1,1) = 24

EXERCISE 9
1. copy exer9.m: CHECK comments
2. exer9(160) error=0.055250 + plot
3. exer9(10)  error=84.772 + plot
4.
  nsteps      error       ratio
    10      84.772164   122.3186     
    20       0.693044     2.6850     
    40       0.258120     2.2281     
    80       0.115845     2.0967
   160       0.055250
first order: p=1

EXERCISE 10, extra credit
1. x=single(0.9999)
2. S=formula for sum= 9997.8877
3. a=9994.9707
4. b=9997.8994
5. a and S agree to 3 sig digits, b and S to 6 or 7
6. rel err a=2.9176e-04, rel err b=1.1721e-06
7. a1000=952.6, x^1001=.90
8. Why is b better?