The Chemistry Maths Book, Second Edition

20.4 Interpolation 567

Linear interpolation

The straight liney 1 = 1 p

1

(x) 1 = 1 a

0

1 + 1 a

1

xthrough the two

points (x

0

, y

0

) and (x

1

, y

1

), as in Figure 20.4, satisfies

the pair of linear equations

y

0

1 = 1 a

0

1 + 1 a

1

x

0

, y

1

1 = 1 a

0

1 + 1 a

1

x

1

with solution

Then

and this can be rearranged as

(20.17)

The linear interpolation formula is used to find an approximate value of a function at

a point xin the intervalx

0

tox

1

. The method is illustrated in the following example

for the (known) functione

x

.

EXAMPLE 20.7Linear interpolation

Two points on the graph ofy 1 = 1 e

x

are (to 4 decimal places in y)(x

0

, y

0

) 1 = 1 (0.80, 2.2255)

and(x

1

, y

1

) 1 = 1 (0.84, 2.3164). The linear interpolation formula (20.17) is then

Then, atx 1 = 1 0.832,

e

0.832

1 ≈ 1 2.2255 1 + 1 0.0727 1 = 1 2.2982

The true value is 2.2979 (to 4 decimal places).

0 Exercise 16

When linear interpolation is used between each pair of neighbouring points we have

piecewise linear interpolation, as illustrated in Figure 20.5 for the points listed in

Table 20.4 (compare with Figures 20.6 to 20.8).

ey x

x

≈=. + −.

.−.

.−.



2 2255 0 80

2 3164 2 2255

084 080

()











yy xx

yy

xx

=+−

−

−

















00

10

10

()

y

yx yx

xx

yy

xx

=

−

−

















• 
  

−

−

















01 10

10

10

10

xx

a

yx yx

xx

a

yy

xx

0

01 10

10

1

10

10

=

−

−

,=

−

−

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1

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Figure 20.4