First Semester in Numerical Analysis with Julia, 2020a

CHAPTER 4. NUMERICAL QUADRATURE AND DIFFERENTIATION 173
f(x 0 + h) =f(x 0 )+hf ′ (x 0 )+ h2
2 f ′′ (x 0 )+ h3
6 f (3) (x 0 )+ h4
24 f (4) (ξ + )
f(x 0 − h) =f(x 0 ) − hf ′ (x 0 )+ h2
2 f ′′ (x 0 ) − h3
6 f (3) (x 0 )+ h4
24 f (4) (ξ − )
where ξ + is between x 0 and x 0 + h, and ξ − is between x 0 and x 0 − h. Add the equations to
get
f(x 0 + h)+f(x 0 − h) =2f(x 0 )+h 2 f ′′ (x 0 )+ h4 [
f (4) (ξ + )+f (4) (ξ − ) ] .
24
Solving for f ′′ (x 0 ) gives
f ′′ (x 0 )= f(x 0 + h) − 2f(x 0 )+f(x 0 − h)
h 2
− h2 [
f (4) (ξ + )+f (4) (ξ − ) ] .
24
Note that f (4) (ξ + )+f (4) (ξ − )
2
is a number between f (4) (ξ + ) and f (4) (ξ − ), so from the Intermediate
Value Theorem 6, we can conclude there exists some ξ between ξ − and ξ + so that
Then the above formula simplifies as
f (4) (ξ) = f (4) (ξ + )+f (4) (ξ − )
.
2
f ′′ (x 0 )= f(x 0 + h) − 2f(x 0 )+f(x 0 − h)
h 2
for some ξ between x 0 − h and x 0 + h.
− h2
12 f (4) (ξ)