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

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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!

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Danh sách Quiz
/7
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

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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

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