Show thatoperators.Delta + nabla= Delta nabla - nabla Delta ,where A, V and E are forward, ba
Let f(x) = e ^ x , 0 <= x <= 2 Approximate f(0.25) using linwith x_{0} = 0 and x_{1} = 0.5
If f(x) = x ^ 3 + 2x ^ 2 h = 1 , then find
a) Delta ^ 3 * y(2)
b). nabla^ 3 y(2)
To show that operators.Delta + nabla = Delta nabla - nabla Delta, we can start by expanding the left-hand side:
operators.Delta + nabla = (E - A) + (V - A)
= (E + V) - 2A
Now let's expand the right-hand side:
Delta nabla - nabla Delta = (E - A)(V - A) - (V - A)(E - A)
= EV - AE - AV + A^2 - VE + AE + VA - A^2
= EV - VE + VA - AV
= V(E - A) - (E - A)V
= V - A - E + A
= V - E
Therefore, we have:
operators.Delta + nabla = Delta nabla - nabla Delta
(E + V) - 2A = V - E
E + V - 2A = V - E
2E = 2A
E = A
So, we have shown that operators.Delta + nabla = Delta nabla - nabla Delta if and only if E = A.
Now let's approximate f(0.25) using linear interpolation with x0 = 0 and x1 = 0.5:
f(x) = e^x
Using linear interpolation formula:
f(x) = f(x0) + (x - x0) * (f(x1) - f(x0)) / (x1 - x0)
= f(0) + (0.25 - 0) * (f(0.5) - f(0)) / (0.5 - 0)
= e^0 + 0.25 * (e^0.5 - e^0) / 0.5
= 1 + 0.25 * (1.64872127 - 1) / 0.5
= 1 + 0.25 * 0.64872127 / 0.5
= 1 + 0.1621803175
= 1.1621803175
Therefore, f(0.25) ≈ 1.1621803175.
For part a) of the second question, we are given that f(x) = x^3 + 2x^2 and h = 1. To find Delta^3 * y(2), we can use the forward difference formula:
Delta^3 * y(2) = Delta(Delta^2 * y(2))
= Delta(y(2 + h) - y(2)) - Delta(y(2 + 2h) - y(2 + h))
= (y(2 + 2h) - 2y(2 + h) + y(2)) - ((y(2 + 3h) - 2y(2 + 2h) + y(2 + h)) - (y(2 + 2h) - 2y(2 + h) + y(2)))
= y(2 + 2h) - 4y(2 + h) + 6y(2) - 4y(2 - h) + y(2 - 2h)
Substituting the given values:
h = 1, x = 2
Delta^3 * y(2) = y(2 + 2) - 4y(2 + 1) + 6y(2) - 4y(2 - 1) + y(2 - 2)
= y(4) - 4y(3) + 6y(2) - 4y(1) + y(0)
For part b) we are asked to find nabla^3 * y(2), which is the backward difference operator. Using the backward difference formula:
nabla^3 * y(2) = nabla(nabla^2 * y(2))
= nabla(y(2) - y(2 - h)) - nabla(y(2 + h) - y(2))
= (y(2) - y(2 - h)) - ((y(2 + h) - y(2)) - (y(2) - y(2 - h)))
= y(2) - 2y(2 - h) + y(2 - 2h) - y(2 + h) + 2y(2) - y(2 + 2h)
= -y(2 + 2h) + 3y(2 + h) - 3y(2) + y(2 - h)
Substituting the given values:
h = 1, x = 2
nabla^3 * y(2) = -y(2 + 2) + 3y(2 + 1) - 3y(2) + y(2 - 1)
= -y(4) + 3y(3) - 3y(2) + y(1)