First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 169
This formula gives an obvious way to estimate the derivative by
f ′ (x 0 ) ≈ f(x 0 + h) − f(x 0 )
h
for small h. What this formula lacks, however, is it does not give any information about the
error of the approximation.
We will try another approach. Similar to Newton-Cotes quadrature, we will construct
the interpolating polynomial for f, and then use the derivative of the polynomial as an
approximation for the derivative of f.
Let’s assume f ∈ C 2 (a, b),x 0 ∈ (a, b), and x 0 + h ∈ (a, b). Construct the linear Lagrange
interpolating polynomial p 1 (x) for the data (x 0 ,f(x 0 )), (x 1 ,f(x 1 )) = (x 0 + h, f(x 0 + h)).
From Theorem 51, wehave
f(x) = x − x 1
f(x 0 )+ x − x 0
f(x 1 ) + f ′′ (ξ(x))
(x − x 0 )(x − x 1 )
x 0 − x 1 x 1 − x
} {{ 0 2!
} } {{ }
p 1 (x)
interpolation error
= x − (x 0 + h)
x 0 − (x 0 + h) f(x 0)+ x − x 0
f(x 0 + h)+ f ′′ (ξ(x))
(x − x 0 )(x − x 0 − h)
x 0 + h − x 0 2!
= x − x 0 − h
f(x 0 )+ x − x 0
f(x 0 + h)+ f ′′ (ξ(x))
(x − x 0 )(x − x 0 − h).
−h
h
2!
Now let’s differentiate f(x) :
f ′ (x) =− f(x 0)
h
= f(x 0 + h) − f(x 0 )
h
+ f(x 0 + h)
h
+ f ′′ (ξ(x)) d
dx
+ (x − x 0)(x − x 0 − h)
2
+ 2x − 2x 0 − h
2
[ ]
(x − x0 )(x − x 0 − h)
2
d
dx [f ′′ (ξ(x))]
f ′′ (ξ(x)) + (x − x 0)(x − x 0 − h)
f ′′′ (ξ(x))ξ ′ (x).
2
In the above equation, we know ξ is between x 0 and x 0 + h, however, we have no knowledge
about ξ ′ (x), which appears in the last term. Fortunately, if we set x = x 0 , the term with
ξ ′ (x) vanishes and we get:
f ′ (x 0 )= f(x 0 + h) − f(x 0 )
h
− h 2 f ′′ (ξ(x 0 )).
This formula is called the forward-difference formula if h>0 and backward-difference