Thông tin
Quiz

Bài tập ôn tập Toán Kinh Tế Phần 2 - Toán Kinh Tế | Đại học Tôn Đức Thắng

Forward: f’(0) h =0.51f’(x) = f(x0 +h) -f(x0) /hf’(0) = f(0+0.51)-f(0)/0.51 = (186 – 154)/0.51 = 62.745Backward: f’(3.82) h=0.58f’(x) = f(x0) – f(x0 – h)/h f’(3.82) = f(3.82) – f(3.82 – 0.58)/0.58 = 274 – 272 /0.58 = 3.4482. Tài liệu được sưu tầm và soạn thảo dưới dạng file PDF để gửi tới các bạn sinh viên cùng tham khảo, ôn tập đầy đủ kiến thức, chuẩn bị cho các buổi học thật tốt. Mời bạn đọc đón xem!

Môn:

Toán Kinh Tế (C01120) 73 tài liệu

Trường:

Đại học Tôn Đức Thắng 3.5 K tài liệu

Thông tin:

7 trang 4 tháng trước

Tác giả:

Mai Nguyệt

docx.com.vn

Bình luận

Vui lòng đăng nhập hoặc đăng ký để gửi bình luận.

Bài tập ôn tập Toán Kinh Tế Phần 2 - Toán Kinh Tế | Đại học Tôn Đức Thắng

41 21 lượt tải Tải xuống

t (s) 0 0.51 1.03 1.74 2.36 3.24 3.82

x (m) 154 186 209 250 262 272 274

Tính vận tốc tính gia tốc

Vận tốc f’()

Gia tốc f”()

h 0.51 0.52 0.71 0.62 0.88 0.58

Forward: f’(0) h =0.51

f’(x) = f(x0 +h) -f(x0) /h

 f’(0) = f(0+0.51)-f(0)/0.51 = (186 – 154)/0.51 = 62.745

Backward: f’(3.82) h=0.58

f’(x) = f(x0) – f(x0 – h)/h

 f’(3.82) = f(3.82) – f(3.82 – 0.58)/0.58 = 274 – 272 /0.58 = 3.4482

Central: h1 h2

f’’(x) = (h2*f(x0-h1) + h1*f(x0 +h2) – (h1+h2)*f(x0) )/ (h1^2*h2/2 + h2^2 * h1 /2)

. At t = 0.51 => h2 = 0.52, h1 = 0.51

f’’(0.51) = (0.52*154 + 0.51*209 –(0.51+0.52)*186)/(0.51^2*0.52/2 + 0.52^2*0.51/2)

= -35.9501

.At t = 1.03 => h2 = 0.71, h1 = 0.52

x 0 2 4 6 8 10

y 0 14 20 49 16 23

h = 2

f’(2) hay f”(2)

Central:

f’(x) = f(x0+h) – f(x0-h) /2h

f’’(x) = f(x0+h) -2f(x0) + f(x0-h)/h^2

f’(x) hoặc f”(x) từ những giá trị x, x+h, x-h, x+2h,….. or f(x), f(x+h), f(x-h),….

f’(x) = Af(x) + Bf(x-h) + Cf(x+h)

f’’(x) = Af(x) + Bf(x-h) + Cf(x+h)

Taylor’s theorem gives:

f(x+h) = f(x) + hf’(x) + h^2/2!f”(x) + h^3/3! f”’(xi) + ….

f(x-h) = f(x) - hf’(x) + h^2/2!f”(x) - h^3/3! f”’(xi) + ….

A+B+C = 0

Truncation error: o(h^n)

Exercise:

f”(x) = Af(x) + Bf(x-h) + Cf(x-2h) (1)

Taylor’s theorem gives:

f(x-h) = f(x) -hf’(x) + h^2/2!f”(x) - h^3/3!f”’(C1) (2)

f(x-2h) = f(x) -2hf’(x) + 4h^2/2! f”(x) - 8h^3/3!f”’(C2) (3)

f”(x) = (A+B+C)f(x) + (-Bh -2Ch)f’(x) +1/2*(Bh^2 + 4Ch^2)f”(x) + h^3/3!(-B f”’(C1) – 8C f”’(C2))

A+B+C =0

-Bh – 2Ch = 0

½*Bh^2 + ½*4Ch^2 = 1

 A = 1/h^2

 B =-2/h^2

 C = 1/h^2

f”(x) = 1/h^2f(x) -2/h^2f(x-h) + 1/h^2f(x-2h) + h/3(f”’(C1) – 4f”’(C2))

Truncation: h/3(f”’(C1) – 4f”’(C2))

Order: O(h)

Middle-point:

Im = h*sigma(i:0->n) f(xi + xi+1/2)

Em = |I(f) -Im| = M/24 (b-a) *h^2

Order: O(h^2)

i 0 1 2 3

xi 2 4 8 10

Xi + xi+1/2 3 6 9

Xi 3 6 9

f(xi)

Exercise: f(x) = tp(0->1) e^-x^2 dx

h = 0.25 => n = 4

f(x) = e^-x^2

i 0 1 2 3 4

Xi 0 0.25 0.5 0.75 1

Xi + xi+1/2 0.125 0.375 0.625 0.875

f 0.9844 0.8688 0.6766 0.465

Im = h*sigma(i:0->n) f(xi + xi+1/2)

= 0.25*(0.9844 + 0.8688 + 0.6766 + 0.465)

= 0.7487

b. e = 10^-3

Em = M/24 (b-a) *h^2 <= 10^-3

f(x) = e^-x^2

 f’(x) = -2x. e^-x^2

 f’’(x) =

M = max(0,1) |f”(x)| = 2

 2/24 *1*(1/n)^2 <= 10^-3

 n >= 9.1287

We choose n at least 10

Trapezoidal:

It = h/2 *(f(a) + f(b) + 2*sigma(i:1->n-1)f(xi))

Et = |I(f) – It| = M/12 (b-a) *h^2

Order: O(h^2)

I 0 1 2 3

xi 0 0.25 0.5 0.75

It = h/2 *(f(0) + f(0.75) + 2*(f(x1) + f(x2)))

Exercise: f(x) = tp(0->1) e^-x^2 dx

h = 0.25 => n = 4

I 0 1 2 3 4

xi 0 0.25 0.5 0.75 1

f 1 0.9394 0.7788 0.5697 0.3678

It = h/2 *(f(0) + f(0.75) + 2*(f(x1) + f(x2)+f(x3)))

= 0.25/2 * (1 + 0.3678 + 2*(0.9394 + 0.7788 + 0.5697 ))

= 0.74295

e = 10^-3

Et = M/12 (b-a) *h^2 <= 10^-3

f(x) = e^-x^2

 f’(x) = -2x. e^-x^2

 f’’(x) =

M = max(0,1) |f”(x)| = 2

2/12*(1/n)^2 <= 10^-3

 n >= 12.9099

We chose n at least 13

Simpson’s 1/3:

Is1/3 = h/3*(f(a) + f(b) + 2sigma chẵn + 4 sigma lẻ)

Es1/3 = |I(f) – Is1/3| = M/180(b-a)*h^4

M = max|f””(x)|

Exercise: f(x) = tp(0->1) e^-x^2 dx

h = 0.25 => n = 4

I 0 1 2 3 4

xi 0 0.25 0.5 0.75 1

f 1 0.9394 0.7788 0.5697 0.3678

It = h/3 *(f(0) + f(0.75) + 2*f(x2) + 4*(f(x1) + f(x3)))

= 0.25/3 * (1 + 0.3678 + 2*(0.7788) + 4*(0.9394+0.5697))

= 0.7468

e = 10^-3

Et = M/180 (b-a) *h^4 <= 10^-3

f(x) = e^-x^2

 f’(x) = -2x. e^-x^2

 f’’(x) = -2.e^-x^2 + 4x^2.e^-x^2

 f”’(x) = -8x^3*e^-x^2 + 12xe^-x^2

M = max(0,1) |f””(x)| = 12

12/180*(1/n)^4 <= 10^-3

 n >= 2.8574

We chose n at least 4

Euler:

IVP

Dy/dt = f(x,y(t)), t in [a,b]

Y() = a

Let w = (w0, w1,…,wn) be a approx solution:

W0 = a

Wi+1 = wi +hf(ti, wi)

Error = sqrt( sigma Yexact – wi)

Stepsize small h:

M.h/2L * (e^(L(ti-a))-1)

M = max|f”(y(t))|

Y(t)

1. Exact

2. D^2y/dt^2

L = max|df/dy|

F=dy/dt

Exercise: (1)

Y’ + 4y/t = t^4 , 1<= t <=2

Y(1) = 1

F = t^4 – 4y/t

h=0.25 => n = 4

i 0 1 2 3 4

ti 1 1.25 1.5 1.75 2

Let w = (w0, w1,…,w4) be approx solution of (1)

w0 = 1

w1 = w0 +hf(t0,w0) = 0.25

w2 = 0.6603

w3 = 1.4857

w4 = 2.9814

Solution w = (0.25, 0.6603, 1.4857, 2.9814)

Yexact(t) = 1/9 t^5 + 8/9t^-4

The error doesn’t exceed 10^-3

Df/dy = 4/t

L = max|df/dy| = 4

Y(t) = 1/9 t^5 + 8/9t^-4

Y’(t) = 5/9 t^4 -32/9*t^-5

M = max|y”(t)| = 20

h.M/2L * (e^(L(ti-a))-1) <= 10^-3

N >= 133996,1677

We chose n at least 133997

Bấm Tải xuống để xem toàn bộ.

| 1/7

| | Tải xuống

Preview text:

t (s) 0 0.51 1.03 1.74 2.36 3.24 3.82 x (m) 154 186 209 250 262 272 274
Tính vận tốc tính gia tốc Vận tốc f’() Gia tốc f”() h 0.51 0.52 0.71 0.62 0.88 0.58 Forward: f’(0) h =0.51 f’(x) = f(x0 +h) -f(x0) /h
 f’(0) = f(0+0.51)-f(0)/0.51 = (186 – 154)/0.51 = 62.745 Backward: f’(3.82) h=0.58
f’(x) = f(x0) – f(x0 – h)/h
 f’(3.82) = f(3.82) – f(3.82 – 0.58)/0.58 = 274 – 272 /0.58 = 3.4482 Central: h1 h2
f’’(x) = (h2*f(x0-h1) + h1*f(x0 +h2) – (h1+h2)*f(x0) )/ (h1^2*h2/2 + h2^2 * h1 /2)
. At t = 0.51 => h2 = 0.52, h1 = 0.51
f’’(0.51) = (0.52*154 + 0.51*209 –(0.51+0.52)*186)/(0.51^2*0.52/2 + 0.52^2*0.51/2) = -35.9501
.At t = 1.03 => h2 = 0.71, h1 = 0.52 x 0 2 4 6 8 10 y 0 14 20 49 16 23 h = 2 f’(2) hay f”(2) Central:
f’(x) = f(x0+h) – f(x0-h) /2h
f’’(x) = f(x0+h) -2f(x0) + f(x0-h)/h^2
f’(x) hoặc f”(x) từ những giá trị x, x+h, x-h, x+2h,….. or f(x), f(x+h), f(x-h),….
f’(x) = Af(x) + Bf(x-h) + Cf(x+h)
f’’(x) = Af(x) + Bf(x-h) + Cf(x+h) Taylor’s theorem gives:
f(x+h) = f(x) + hf’(x) + h^2/2!f”(x) + h^3/3! f”’(xi) + ….
f(x-h) = f(x) - hf’(x) + h^2/2!f”(x) - h^3/3! f”’(xi) + …. A+B+C = 0 Truncation error: o(h^n) Exercise:
f”(x) = Af(x) + Bf(x-h) + Cf(x-2h) (1) Taylor’s theorem gives:
f(x-h) = f(x) -hf’(x) + h^2/2!f”(x) - h^3/3!f”’(C1) (2)
f(x-2h) = f(x) -2hf’(x) + 4h^2/2! f”(x) - 8h^3/3!f”’(C2) (3)
f”(x) = (A+B+C)f(x) + (-Bh -2Ch)f’(x) +1/2*(Bh^2 + 4Ch^2)f”(x) + h^3/3!(-B f”’(C1) – 8C f”’(C2)) A+B+C =0 -Bh – 2Ch = 0 ½*Bh^2 + ½*4Ch^2 = 1  A = 1/h^2  B =-2/h^2  C = 1/h^2
f”(x) = 1/h^2f(x) -2/h^2f(x-h) + 1/h^2f(x-2h) + h/3(f”’(C1) – 4f”’(C2))
Truncation: h/3(f”’(C1) – 4f”’(C2)) Order: O(h) Middle-point:
Im = h*sigma(i:0->n) f(xi + xi+1/2)
Em = |I(f) -Im| = M/24 (b-a) *h^2 Order: O(h^2) i 0 1 2 3 xi 2 4 8 10 Xi + xi+1/2 3 6 9 f Xi 3 6 9 f(xi)
Exercise: f(x) = tp(0->1) e^-x^2 dx h = 0.25 => n = 4 f(x) = e^-x^2 i 0 1 2 3 4 Xi 0 0.25 0.5 0.75 1 Xi + xi+1/2 0.125 0.375 0.625 0.875 f 0.9844 0.8688 0.6766 0.465
Im = h*sigma(i:0->n) f(xi + xi+1/2)
= 0.25*(0.9844 + 0.8688 + 0.6766 + 0.465) = 0.7487 b. e = 10^-3
Em = M/24 (b-a) *h^2 <= 10^-3 f(x) = e^-x^2  f’(x) = -2x. e^-x^2  f’’(x) = M = max(0,1) |f”(x)| = 2
 2/24 *1*(1/n)^2 <= 10^-3  n >= 9.1287 We choose n at least 10 Trapezoidal:
It = h/2 *(f(a) + f(b) + 2*sigma(i:1->n-1)f(xi))
Et = |I(f) – It| = M/12 (b-a) *h^2 Order: O(h^2) I 0 1 2 3 xi 0 0.25 0.5 0.75
It = h/2 *(f(0) + f(0.75) + 2*(f(x1) + f(x2)))
Exercise: f(x) = tp(0->1) e^-x^2 dx h = 0.25 => n = 4 I 0 1 2 3 4 xi 0 0.25 0.5 0.75 1 f 1 0.9394 0.7788 0.5697 0.3678
It = h/2 *(f(0) + f(0.75) + 2*(f(x1) + f(x2)+f(x3)))
= 0.25/2 * (1 + 0.3678 + 2*(0.9394 + 0.7788 + 0.5697 )) = 0.74295 e = 10^-3
Et = M/12 (b-a) *h^2 <= 10^-3 f(x) = e^-x^2  f’(x) = -2x. e^-x^2  f’’(x) = M = max(0,1) |f”(x)| = 2 2/12*(1/n)^2 <= 10^-3  n >= 12.9099 We chose n at least 13 Simpson’s 1/3:
Is1/3 = h/3*(f(a) + f(b) + 2sigma chẵn + 4 sigma lẻ)
Es1/3 = |I(f) – Is1/3| = M/180(b-a)*h^4 M = max|f””(x)|
Exercise: f(x) = tp(0->1) e^-x^2 dx h = 0.25 => n = 4 I 0 1 2 3 4 xi 0 0.25 0.5 0.75 1 f 1 0.9394 0.7788 0.5697 0.3678
It = h/3 *(f(0) + f(0.75) + 2*f(x2) + 4*(f(x1) + f(x3)))
= 0.25/3 * (1 + 0.3678 + 2*(0.7788) + 4*(0.9394+0.5697)) = 0.7468 e = 10^-3
Et = M/180 (b-a) *h^4 <= 10^-3 f(x) = e^-x^2  f’(x) = -2x. e^-x^2
 f’’(x) = -2.e^-x^2 + 4x^2.e^-x^2
 f”’(x) = -8x^3*e^-x^2 + 12xe^-x^2 M = max(0,1) |f””(x)| = 12 12/180*(1/n)^4 <= 10^-3  n >= 2.8574 We chose n at least 4 Euler: IVP Dy/dt = f(x,y(t)), t in [a,b] Y() = a
Let w = (w0, w1,…,wn) be a approx solution: W0 = a Wi+1 = wi +hf(ti, wi)
Error = sqrt( sigma Yexact – wi) Stepsize small h: M.h/2L * (e^(L(ti-a))-1) M = max|f”(y(t))| Y(t) 1. Exact 2. D^2y/dt^2 L = max|df/dy| F=dy/dt Exercise: (1)
Y’ + 4y/t = t^4 , 1<= t <=2 Y(1) = 1 F = t^4 – 4y/t h=0.25 => n = 4 i 0 1 2 3 4 ti 1 1.25 1.5 1.75 2
Let w = (w0, w1,…,w4) be approx solution of (1) w0 = 1 w1 = w0 +hf(t0,w0) = 0.25 w2 = 0.6603 w3 = 1.4857 w4 = 2.9814
Solution w = (0.25, 0.6603, 1.4857, 2.9814) Yexact(t) = 1/9 t^5 + 8/9t^-4
The error doesn’t exceed 10^-3 Df/dy = 4/t L = max|df/dy| = 4 Y(t) = 1/9 t^5 + 8/9t^-4 Y’(t) = 5/9 t^4 -32/9*t^-5 M = max|y”(t)| = 20
h.M/2L * (e^(L(ti-a))-1) <= 10^-3 N >= 133996,1677 We chose n at least 133997