Question

Fundamental theorem for finite differences in numerical methods

Answer 1

Recall from The Antidifference Operator that F is said to be an antidifference of the real-valued function f is ΔF=f, or in using the antidifference operator Δ−1 we write F=Δ−1f.

We will now look at yet another theorem on antidifferences that is analogous to the Fundamental Theorem of Calculus.

Theorem 1 (The Fundamental Theorem of the Calculus of Finite Differences): Let f be a real-valued function and let a and b be integers such that a≤b. If F=Δ−1f then ∑bn=af(n)=F(b+1)−F(a).

Proof: Let f be a real-valued function and let a,b∈Z where a≤b.

Since F=Δ−1f we have that ΔF=f, i.e., ΔF(x)=F(x+1)−F(x)=f(x) and so:

(1)

∑n=abf(n)=∑n=ab(F(n+1)−F(n))∑n=abf(n)=(F(a+1)−F(a))+(F(a+2)−F(a+1))+...+(F(b)−F(b−1))+(F(b+1)−F(n))∑n=abf(n)=F(b+1)−F(a)■

Let's look at an example of applying the Fundamental Theorem of the Calculus of Finite Differences. Consider the function f(x)=2x+1. An antidifference of f is the function F(x)=x2. Suppose we want to evaluate the sum ∑6n=1f(n)=∑6n=1(2n+1). Then:

(2)

∑n=16(2n+1)=F(6+1)−F(1)=F(7)−F(1)=(7)2−(1)2=48

We will now look at a simply corollary to the Fundamental Theorem of the Calculus of Finite Differences

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Answer 2

Answer:

state and prove fundamental theorem of finite differences