科学计算作业 一

科学计算

陆一洲 5140309557

Page 45 - 1.7

double x=1;
for (int i=0; i<=16; i++){
    double  h=pow(10.0, -i),
            a=(tan(x+h)-tan(x))/h,
            b=1.0/sqr(cos(x));
    fprintf(FOUT, "%d\t%.25lf\n", i, fabs(a-b));
}
for (int i=0; i<=16; i++){
    double  h=pow(10.0, -i),
            a=(tan(x+h)-tan(x-h))/h/2,
            b=1.0/sqr(cos(x));
    fprintf(FOUT, "%d\t%.30lf\n", i, fabs(a-b));
}

(a)

当$n=8$时，误差取得最小值，高于标准误差$\sqrt{\epsilon _{mach}}$

(b)

当 $n=7$ 时，误差取得最小值，低于标准误差 $\sqrt{\epsilon _{mach}}$
由于，该倒数取的为区间的中点值，所以该倒数的斜率相对更具有代表性，所以精度得到提升。

Page46 - 1.9

(a)

double ans=0, x;
scanf("%lf", &x);
for (int i=0; ; i++){
    double inc=Power(x, i)/Factorial((double)i);
    if  (ans+inc==ans) break;
    ans+=inc;
}
printf("%.30lf\t%.30lf\t%.30lf\n", ans, exp(x), ans-exp(x));

(b)

进行不断累加，直到结果不再变动（由于精度问题不再变动）。

(c)

$x$	Tayler_exp($x$)	exp($x$)	绝对误差	相对误差
-20	8.359241e-009	2.061154e-009	6.298087e-009	7.534281e-001
-15	3.059036e-007	3.059023e-007	1.319961e-012	4.314957e-006
-10	4.539993e-005	4.539993e-005	1.241689e-014	2.735002e-010
-5	6.737947e-003	6.737947e-003	1.232607e-016	1.829350e-014
-1	3.678794e-001	3.678794e-001	6.795237e-017	1.847137e-016
1	2.718282e+000	2.718282e+000	2.994566e-016	1.101639e-016
5	1.484132e+002	1.484132e+002	3.191891e-014	2.150679e-016
10	2.202647e+004	2.202647e+004	5.902834e-012	2.679882e-016
15	3.269017e+006	3.269017e+006	3.092282e-011	9.459362e-018
20	4.851652e+008	4.851652e+008	1.199078e-007	2.471484e-016

在 $x>0$ 时，精度误差较小；

在 $x<0$ 时，精度误差随着 $x$ 的递减而增大。

(d)

$x$	Tayler_exp($x$)	exp($x$)	绝对误差	相对误差
-20	2.061154e-009	2.061154e-009	4.564034e-025	2.214310e-016
-15	3.059023e-007	3.059023e-007	2.481542e-024	8.112203e-018
-10	4.539993e-005	4.539993e-005	9.423242e-021	2.075607e-016
-5	6.737947e-003	6.737947e-003	1.639432e-018	2.433133e-016
-1	3.678794e-001	3.678794e-001	4.306993e-017	1.170762e-016

可以看出精度有了显著提升。

(e)

将所有计算出来的 $\frac{x^n}{n!}$ 都存入数组中，并将其按绝对值从小到大排序，再进行累加，以便得到更高的精度，以下为测试结果：

$x$	Tayler_exp($x$)	exp($x$)	绝对误差	相对误差
-20	2.061154e-009	2.061154e-009	4.281306e-026	2.077141e-017
-15	3.059023e-007	3.059023e-007	5.045802e-023	1.649481e-016
-10	4.539993e-005	4.539993e-005	1.090555e-020	2.402107e-016
-5	6.737947e-003	6.737947e-003	9.529121e-020	1.414247e-017
-1	3.678794e-001	3.678794e-001	1.244122e-017	3.381874e-017

发现精度并没有太大的变化，这主要原因为double本身精度的问题。

MATLEB TEST

1. (Matlab test) For n=500, 1000, 2000, 4000, 8000, generate an n*n random matrix, B, and an n*1 vector, b. Find the symmetric matrix A=B’*B.
   1) Using function “eig” to test the time cost for eigen decomposition;
   2) Using x=A\b and x=A^(-1)*b to test the time cost in finding the solution of Ax=b;
   3) Plot all the time costs as a function of n and the power law for the time costs.

(1)

$n$	$500$	$1000$	$2000$	$4000$	$8000$
用时($s$)	0.263475	1.279685	6.288209	38.239515	269.477866

(2)

$n$	$500$	$1000$	$2000$	$4000$	$8000$
$x=A \setminus b$	0.007135	0.037733	0.213802	1.326795	9.614620
$x=A^{-1}*b$	0.030764	0.134277	0.848402	5.948203	55.072248

(3)

General model Power1:
f(x) = a*x^b
Coefficients (with 95% confidence bounds):
a = 2.975e-09 (1.234e-09, 4.717e-09)
b = 2.807 (2.742, 2.872)
Goodness of fit:
SSE: 0.9548
R-square: 1
Adjusted R-square: 1
RMSE: 0.5641

General model Power1:
f(x) = a*x^b
Coefficients (with 95% confidence bounds):
a = 7.369e-11 (3.274e-11, 1.146e-10)
b = 2.848 (2.786, 2.91)
Goodness of fit:
SSE: 0.001032
R-square: 1
Adjusted R-square: 1
RMSE: 0.01855

General model Power1:
f(x) = a*x^b
Coefficients (with 95% confidence bounds):
a = 1.785e-11 (3.07e-12, 3.263e-11)
b = 3.2 (3.108, 3.292)
Goodness of fit:
SSE: 0.04505
R-square: 1
Adjusted R-square: 1
RMSE: 0.1225