﻿ 17074229 作业9 - 计算数学达人 - 专，学者，数值代数，微分方程数值解

### 17074229 作业9

5.1 Exercises

2. Use the three-point centered-difference formula to approximate f(0), where f(x)=ex, for (a) h=0.1 (b) h=0.01 (c) h=0.001.

(2)The three-point centered-difference formula evaluates to f(x)[ f(x + h) f(x h) ]/(2h) = (e^0.01-e^-0.01)/0.021.000016665.

(3)The three-point centered-difference formula evaluates to f(x)[ f(x + h) f(x h) ]/(2h) = (e^0.001-e^-0.001)/0.0020.100000015.

5.1 Computer Problems

2. Use the three-point centered-difference formula to approximate f(0), where f(x)=ex, for (a) h=0.1 (b) h=0.01 (c) h=0.001.

y=(f((x)+h)-f((x)-h))/2*h;

(a)>> y=tpcd(f,0,0.1)

y =

0.0100

(b)>> y=tpcd(f,0,0.01)

y =

1.0000e-04

(c)>> y=tpcd(f,0,0.001)

y =

1.0000e-06

5.2 Exercises

1. Apply the composite Trapezoid Rule with m=1,2, and 4 panels to approximate the integral. Compute the error by comparing with the exact value from calculus. (b)  ≈((π/2)/2)*[y0+y1]=π/4

The errors is -π^3*cos1/96.

For the composite Trapezoid Rule on[0,π/2], two panels means that h=π/4. The approximation is ≈((π/4)/2)*[y0+y2+2* ]=(π/8)*(1+2)

The errors is -π^3*cos1/384.

For the composite Trapezoid Rule on[0,π/2], four panels means that h=π/8. The approximation is ≈((π/4)/2)*[y0+y2+2* ]=(π/8)*(1+2+2*cos(π/8)+2*cos(3π/8))

The errors is -π^3*cos1/1536.

2. Apply the Composite Midpoint Rule with m=1,2, and 4 panels to approximate the integrals in Exercise 1, and report the errors.

f’’(x)=-cosx.

When m=1, the midpoints are π/4, the Composite Midpoint Rule delivers ≈(π/2)*f(π/4)=√2π/4.

h=π/2,

The errors is -π^3*cos1/192.

When m=2, the midpoints are π/8,3π/8, so the Composite Midpoint Rule delivers ≈(π/4)* =(π/4)*(cos(π/8)+cos(3π/8)).

h=π/4,

The errors is -π^3*cos1/768.

When m=4, the midpoints are π/16,3π/16,5π/16,7π/16, so the Composite Midpoint Rule delivers ≈(π/8)* =(π/8)(cos(π/16)+cos(3π/16)+cos(5π/16)+cos(7π/16)).

h=π/8,

The errors is -π^3*cos1/3072.

3. Apply the composite Simpsons Rule with m=1,2, and 4 panels to the integrals in Exercise 1, and report the errors.

(b) A one-panel Simpsons Rule sets h=π/2. The approximation is π/6*[y0+y2+4*y1]=π/6*[1+0+2*2^(1/2)]=π/6*[1+2*2^(1/2)]

The errors is π^5/5760.

A two-panel Simpsons Rule sets h=π/4. The approximation is π/12*[y0+y4+4*(y1+y3)+2*y2]=π/12*[1+0+4*(cos(π/8)+cos(3π/8))+2^(1/2)]=π/12*[1+4*(cos(π/8)+cos(3π/8))+2^(1/2)]

The errors is π^5/92160.

A four-panel Simpsons Rule sets h=π/8. The approximation is π/24*[y0+y8+4*(y1+y3+y5+y7)+2*(y2+y4+y6)]=π/24*[1+0+4*(cos(π/16)+cos(3π/16)+cos(7π/16))+2*(cos(π/8)+(2^(1/2))/2+cos(3π/8))]=π/24*[1+4*(cos(π/16)+cos(3π/16)+cos(7π/16))+2*(cos(π/8)+(2^(1/2))/2+cos(3π/8))]

The errors is π^5/1474560.

4. Apply the composite Simpsons Rule with m=1,2, and 4 panels to the integrals, and report the errors. (b)  ≈1/3*[y0+y2+4*y1]=1/3*[1+(1/2)+(16/5)]=47/30

The errors is 3/4.

A two-panel Simpsons Rule sets h=1/2. The approximation is ≈1/6*[y0+y4+4*(y1+y3)+2*y2]=1/6*[1+(1/2)+4*((16/17)+(16/25))+2^(4/5)]=195/34

The errors is 3/16.

A four-panel Simpsons Rule sets h=1/4. The approximation is ≈1/12*[y0+y8+4*(y1+y3+y5+y7)+2*(y2+y4+y6)]=1/12*[1+(1/2)+4*((64/65)+(64/73)+(1/64/89)+(64/113))+2*((16/17)+(4/5)+(16/25))]=22.688291286

The errors is 3/64.

5.2 Computer Problems

1. Use the composite Trapezoid Rule with m=16 and 32 panels to approximate the denite integral. Compare with the correct integral and report the two errors.

(d) (h) h=(b-a)/m;

for i=1:m+1

x(i)=a+(i-1)*h;

end

y=f(x(1))+f(x(m+1));

for i=2:m

y=y+2*f(x(i));

end

y=y*h/2;

>>  f=@(x) x^2*exp(x);

>> y=CTR(f,16,1,3)

y =

98.0909

>> y=CTR(f,32,1,3)

y =

97.8048

(h)

>> f=@(x) x/(x^4+1)^(1/2);

>> y=CTR(f,16,0,1)

y =

0.4404

>> y=CTR(f,32,0,1)

y =

0.4406

2. Apply the composite Simpsons Rule to the integrals in Computer Problem 1. Use m=16 and 32, and report errors.

h=(b-a)/(2*m);
for i=1:2*m+1
x(i)=a+(i-1)*h;
end
y=f(x(1))+f(x(2*m+1));
for i=2:2:2*m
y=y+4*f(x(i));
end
for i=3:2:2*m+1
y=y+2*f(x(i));
end
y=y*h/3;

>> f=@(x) x^2*exp(x);

>> y=CSR(f,1,3,16)

y =

105.2415

>> y=CSR(f,1,3,32)

y =

101.4754

h>> f=@(x) x/(x^4+1)^(1/2);

>> y=CSR(f,0,1,16)

y =

0.4554

>> y=CSR(f,0,1,32)

y =

0.4481

3. Use the composite Trapezoid Rule with m=16 and 32 panels to approximate the denite integral.

(a) (f) h=(b-a)/m;

for i=1:m+1

x(i)=a+(i-1)*h;

end

y=f(x(1))+f(x(m+1));

for i=2:m

y=y+2*f(x(i));

end

y=y*h/2;

(a)>> f=@(x) exp(x^2);

>> y=CTR(f,16,0,1)

y =

1.4644

>> y=CTR(f,32,0,1)

y =

1.4631

(f)>> f=@(x) cos(x)*exp(x);

>> y=CTR(f,16,0,pi)

y =

-12.1480

>> y=CTR(f,32,0,pi)

y =

-12.0897

4. Apply the composite Simpsons Rule to the integrals of Computer Problem 3, using m=16

h=(b-a)/(2*m);
for i=1:2*m+1
x(i)=a+(i-1)*h;
end
y=f(x(1))+f(x(2*m+1));
for i=2:2:2*m
y=y+4*f(x(i));
end
for i=3:2:2*m+1
y=y+2*f(x(i));
end
y=y*h/3;

(a)>> f=@(x) exp(x^2);

>> y=CSR(f,0,1,16)

y =

1.5193

>> y=CSR(f,0,1,32)

y =

1.4910

(f)>> f=@(x) cos(x)*exp(x);

>> y=CSR(f,0,pi,16)

y =

-13.5849

>> y=CSR(f,0,pi,32)

y =

-12.8276

5.3 Exercises

1. Apply Romberg Integration to nd R33 for the integrals.

(c) h1=b-a=1;

h2=(1/2)*(b-a)=1/2;

h3=(1/4)*(b-a)=1/4;

R11=h1/[f(a)+f(b)]=e+1;

R21=h2/[f(a)+f(b)+2*f((a+b)/2))]=(e+1)/2+e^(1/2);

R31=(1/2)*R21+h3* =(1/4)*(e+1+2*e^(1/2)+e^(1/4)+e^(3/4));

R22=[(2^2)R21-R11]/3=(1/3)*[e+1+4*e^(1/2)]

R32=[(2^2)R31-R21]/3=(1/2)*[e+1+2*e^(1/2)]+e^(1/4)+e^(3/4);

R33=[(4^2)*R32-R22]/[(4^2)-1]=[23*(e+1)+44*e^(1/2)]/45+[16*(e^(1/4)+e^(3/4)]/15;

6.1 Exercises

3. Use separation of variables to ﬁnd solutions of the IVP given by y(0)=1 and the following differential equations: (a) y′=t (b) y′=(t^2)*y (c) y′=2*(t + 1)*y (d) y′=5*(t^4)*y (e) y′=1/(y^2) (f) y′=t^3/y^2

（a） Separate variables, y′=dy/dt=t

dy=tdt

y=(1/2)*t^2+c, c is a constant.

Because of y(0)=1, c=1, to sum up y=(1/2)*t^2+1.

（b） Separate variables, y′=dy/dt= y′=(t^2)*y

dy/y=t^2dt

ln|y|=(1/3)*(t^3)+c, c is a constant.

|y|=e^[(1/3)*(t^3)+c]

Because of y(0)=1, c=0, to sum up y=e^[(1/3)*(t^3)].

（c）  Separate variables, y′=dy/dt= y′=2*(t + 1)*y

dy/y=2*(t+1)dt

ln|y|=t^2+2*t+c, c is a constant.

|y|=e^[t^2+2*t+c]

Because of y(0)=1, c=0, to sum up y=e^[t^2+2*t].

（d） Separate variables, y′=dy/dt= y′=5*(t^4)*y

dy/y=5*(t^4)dt

ln|y|=t^5+c, c is a constant.

|y|=e^[t^5+c]

Because of y(0)=1, c=0, to sum up y=e^[t^5].

e of y(0)=1, c=0, to sum up y=e^[t^2+2*t].

（e） Separate variables, y′=dy/dt= y′=1/(y^2)

y^2dy=dt

(1/3)y^3=t+c, c is a constant.

y=(3*(t+c))^(1/3)

Because of y(0)=1, c=1/3, to sum up y=(3*t+1)^(1/3).

（f） Separate variables, y′=dy/dt= y′=t^3/y^2

y^2dy=t^3dt

(1/3)y^3=(1/4)*(t^4)+c, c is a constant.

y=[(3/4)*(t^4)+(3*c)]^(1/3)

Because of y(0)=1, c=1/3, to sum up y=[(3/4)*(t^4)+1]^(1/3).

5. Apply Euler’s Method with step size h=1/4 to the IVPs in Exercise 3 on the interval [0,1]. List the wi,i =0,...,4, and ﬁnd the error at t =1 by comparing with the correct solution.

(a) The right-hand side of the differential equation is f(t,y)=t. Therefore, Eulers Method will be the iteration

w0=1, w(i+1)=wi + h*ti. Using the grid with step size h=1/4, we calculate the approximate solution iteratively . The values wi given by Eulers Method and plotted in Figure are compared with the true values yi in the following table:

Step     ti     wi        yi          ei

0        0     1         1          0

1       1/4    1+(1/16)   1+(1/32)   1/32

2       2/4    1+(3/16)   1+(1/8)    1/16

3       3/4    1+(6/16)   1+(9/32)   3/32

4        1     1+(10/16)  1+(1/2  1/8

When t=1, errors is 1/8.

(b) The right-hand side of the differential equation is f(t,y)=t. Therefore, Eulers Method will be the iteration

w0=1, w(i+1)=wi + h*ti. Using the grid with step size h=1/4, we calculate the approximate solution iteratively . The values wi given by Eulers Method and plotted in Figure are compared with the true values yi in the following table:

Step  ti   wi                         yi          ei

0     0   1                          1          0

1    1/4  1+(1/4)*e^(1/192)            e^(1/192)  1-(3/4)*e^(1/192)

2    2/4  1+(1/4)*[e^(1/24)+e^(1/192)]  e^(1/24)   1-(3/4)*e^(1/24)+(1/4)*e^(192)

3    3/4  1+(1/4)*[e^(1/24)+e^(1/192)+  e^(9/64)  1-(3/4)*e^(9/64)+(1/4)*[e^(1/24)+

e^(9/64)]                              e^(1/192)]

4     1   1+(1/4)*[e^(1/24)+e^(1/192)+  e^(1/3)   1-(3/4)*e^(1/3)+(1/4)*[e^(1/24)+

e^(9/64)+e^(1/3)]                       e^(9/64) +e^(1/192)]

When t=1, errors is 1-(3/4)*e^(1/3)+(1/4)*[e^(1/24)+ e^(9/64) +e^(1/192)].

(c) The right-hand side of the differential equation is f(t,y)=t. Therefore, Eulers Method will be the iteration

w0=1, w(i+1)=wi + h*ti. Using the grid with step size h=1/4, we calculate the approximate solution iteratively . The values wi given by Eulers Method and plotted in Figure are compared with the true values yi in the following table:

Step  ti   wi                         yi          ei

0     0   1                          1          0

1    1/4  1+(1/4)*e^(1/192)            e^(1/192)  1-(3/4)*e^(1/192)

2    2/4  1+(1/4)*[e^(1/24)+e^(1/192)]  e^(1/24)   1-(3/4)*e^(1/24)+(1/4)*e^(192)

3    3/4  1+(1/4)*[e^(1/24)+e^(1/192)+  e^(9/64)  1-(3/4)*e^(9/64)+(1/4)*[e^(1/24)+

e^(9/64)]                              e^(1/192)]

4     1   1+(1/4)*[e^(1/24)+e^(1/192)+  e^(1/3)   1-(3/4)*e^(1/3)+(1/4)*[e^(1/24)+

e^(9/64)+e^(1/3)]                       e^(9/64) +e^(1/192)]

When t=1, errors is 1-(3/4)*e^(1/3)+(1/4)*[e^(1/24)+ e^(9/64) +e^(1/192)].

6.2 Exercises

1. Using initial condition y(0)=1 and step size h=1/4, calculate the Trapezoid Method approximation w0,...,w4 on the interval[0,1]. Find the error at t =1 by comparing with the correct solution found in Exercise 6.1.3.  (a) y′=t (b) y′=(t^2)*y (c) y′=2*(t + 1)*y (d) y′=5*(t^4)*y (e) y′=1/(y^2) (f) y′=t^3/y^2

w0=y0;

w(i+1)=wi+h/2(f(ti,wi)+f(ti+h,wi+hf(ti,wi)))

(a)

w0=1

w1= w0+h/2(f(t0,w0)+f(t0+h,w0+hf(t0,w0)))=33/32

w2= w1+h/2(f(t1,w1)+f(t1+h,w1+hf(t1,w1)))=9/8

w3= w2+h/2(f(t2,w2)+f(t2+h,w2+hf(t2,w2)))=41/32

w4= w3+h/2(f(t3,w3)+f(t3+h,w3+hf(t3,w3)))=3/2

e=|y-w4|=0

(b)

w0=1

w1= w0+h/2(f(t0,w0)+f(t0+h,w0+hf(t0,w0)))=129/128

w2= w1+h/2(f(t1,w1)+f(t1+h,w1+hf(t1,w1)))=274641/262144

w3= w2+h/2(f(t2,w2)+f(t2+h,w2+hf(t2,w2)))=622061865/536870912

w4= w3+h/2(f(t3,w3)+f(t3+h,w3+hf(t3,w3)))=1.405352734454937

e=|y-w4|=0.009740309368848

(c)

w0=1

w1= w0+h/2(f(t0,w0)+f(t0+h,w0+hf(t0,w0)))=1.718750000000000

w2= w1+h/2(f(t1,w1)+f(t1+h,w1+hf(t1,w1)))=3.303222656250000

w3= w2+h/2(f(t2,w2)+f(t2+h,w2+hf(t2,w2)))=7.070960998535156

w4= w3+h/2(f(t3,w3)+f(t3+h,w3+hf(t3,w3)))=16.793532371520996

e=|y-w4|=3.292004551666672

(d)

w0=1

w1= w0+h/2(f(t0,w0)+f(t0+h,w0+hf(t0,w0)))=1.002441406250000

w2= w1+h/2(f(t1,w1)+f(t1+h,w1+hf(t1,w1)))=1.044237840920687

w3= w2+h/2(f(t2,w2)+f(t2+h,w2+hf(t2,w2)))=1.307663471185293

w4= w3+h/2(f(t3,w3)+f(t3+h,w3+hf(t3,w3)))=2.706793149522585

e= |y-w4|=0.011488678936460

(e)

w0=1

w1= w0+h/2(f(t0,w0)+f(t0+h,w0+hf(t0,w0)))=1.205000000000000

w2= w1+h/2(f(t1,w1)+f(t1+h,w1+hf(t1,w1)))=1.356993862239376

w3= w2+h/2(f(t2,w2)+f(t2+h,w2+hf(t2,w2)))=1.480971731811849

w4= w3+h/2(f(t3,w3)+f(t3+h,w3+hf(t3,w3)))=1.587101493145249

e=|y-w4|=2.995588229504076e-04

(f)

w0=1

w1= w0+h/2(f(t0,w0)+f(t0+h,w0+hf(t0,w0)))=1.001953125000000

w2= w1+h/2(f(t1,w1)+f(t1+h,w1+hf(t1,w1)))=1.019342601468375

w3= w2+h/2(f(t2,w2)+f(t2+h,w2+hf(t2,w2)))=1.082264953617726

w4= w3+h/2(f(t3,w3)+f(t3+h,w3+hf(t3,w3)))=1.218241931698513

e=|y-w4|=0.013170799610898 Copyright ©2019    计算数学达人 All Right Reserved.
 技术支持：自助建站 | 领地网站建设 |短信接口 |燕窝 版权所有 © 2005-2019 lingw.net.粤ICP备16125321号 -5