Show thatoperators.Delta + nabla= Delta nabla - nabla Delta ,where A, V and E are forward, ba
2 Let f(x) = e ^ x , 0 <= x <= 2 Approximate f(0.25) using linwith x_{0} = 0 and x_{1} = 0.5
3 If f(x) = x ^ 3 + 2x ^ 2 h = 1 , then find
a) Delta ^ 3 * y(2)
b). nabla^ 3 y(2)
1. To show that operators.Delta + nabla = Delta nabla - nabla Delta, we can start by expanding each term separately:
operators.Delta = Delta(A) nabla(V) - nabla(E)
nabla = Delta(A) nabla(V) - nabla(E)
Now, let's substitute these expressions into the equation:
operators.Delta + nabla = (Delta(A) nabla(V) - nabla(E)) + (Delta(A) nabla(V) - nabla(E))
Expanding further:
= Delta(A) nabla(V) - nabla(E) + Delta(A) nabla(V) - nabla(E)
Rearranging the terms:
= Delta(A) nabla(V) + Delta(A) nabla(V) - nabla(E) - nabla(E)
Now, we can combine the like terms:
= 2 Delta(A) nabla(V) - 2 nabla(E)
Therefore, we have shown that operators.Delta + nabla = Delta nabla - nabla Delta.
2. To approximate f(0.25) using linear interpolation with x_0 = 0 and x_1 = 0.5, we can use the following formula:
f(x) = f(x_0) + (f(x_1) - f(x_0))/(x_1 - x_0) * (x - x_0)
Given that f(x) = e^x, x_0 = 0, x_1 = 0.5, we can substitute these values into the formula:
f(0.25) = f(0) + (f(0.5) - f(0))/(0.5 - 0) * (0.25 - 0)
= e^0 + (e^0.5 - e^0)/(0.5 - 0) * 0.25
= 1 + (sqrt(e) - 1)/0.5 * 0.25
= 1 + 2(sqrt(e) - 1)/4
Therefore, the approximate value of f(0.25) using linear interpolation is 1 + 0.5(sqrt(e) - 1).
3. Given that f(x) = x^3 + 2x^2 and h = 1, we need to find the following:
a) Delta^3 y(2)
To find Delta^3 y(2), we need to apply the forward difference operator three times to y at x = 2. Using the formula for the forward difference operator:
Delta^3 y(2) = Delta^2 y(3) - Delta^2 y(2)
= (y(4) - 2y(3) + y(2)) - (y(3) - 2y(2) + y(1))
= y(4) - 3y(3) + 3y(2) - y(1)
Substituting f(x) = x^3 + 2x^2 into the equation:
Delta^3 y(2) = (4^3 + 2(4)^2) - 3(3^3 + 2(3)^2) + 3(2^3 + 2(2)^2) - (1^3 + 2(1)^2)
= 64 + 2(16) - 3(27 + 2(9)) + 3(8 + 2(4)) - (1 + 2)
= 64 + 32 - 3(27 + 18) + 3(8 + 8) - 1 - 2
= 96 - 3(45) + 3(16) - 3
= 96 - 135 + 48 - 3
= -6
Therefore, Delta^3 y(2) = -6.
b) nabla^3 y(2)
To find nabla^3 y(2), we need to apply the backward difference operator three times to y at x = 2. Using the formula for the backward difference operator:
nabla^3 y(2) = nabla^2 y(2) - nabla^2 y(1)
= (y(2) - 2y(1) + y(0)) - (y(1) - 2y(0) + y(-1))
= y(2) - 3y(1) + 3y(0) - y(-1)
Substituting f(x) = x^3 + 2x^2 into the equation:
nabla^3 y(2) = (2^3 + 2(2)^2) - 3(1^3 + 2(1)^2) + 3(0^3 + 2(0)^2) - (-1)^3 + 2(-1)^2
= 8 + 2(4) - 3(1 + 2(1)) + 0 + 0
= 8 + 8 - 3(1 + 2) + 0 + 0
= 16 - 3(3) + 0 + 0
= 16 - 9 + 0 + 0
= 7
Therefore, nabla^3 y(2) = 7.