DIFFERENTIAL GEOMETRY OF CURVES 71

p(x) = a 0 + a 1 x+a 2 x^2 (3.9)

with coefficients a 0 ,a 1 and a 2 unknown. At x = xi, the ordinate value from Eq. (3.9) is p(xi) and

therefore the error in the ordinate values is δii = – (yaaxax 01 + i + (^2) i^2 ). Squaring and adding the
error values for all the data points, we get

Δ = ΣΣ = [ – ( + + )]

=0

–1

2

=0

–1

01 2

22

i

n

i i

n

δ yaaxaxiii (3.10)

The unknowns can be determined by minimizing Δ by differentiating Eq. (3.10) with respect to ar,
r = 0, 1 and 2, that is

∂

∂

Δ

a

xy a ax ax

r i

n

i

r ii

= –2=0 [ – ( + + i)] = 0

–1

Σ 01 22 (3.11)

which after slight rearrangement leads to a symmetric 3 × 3 system of linear equations.

aaxaxyi

n

i

n

i i

n

i i

n

(^0) =0 i
–1
(^1) =0
–1
(^2) =0
–1
2
=0
–1

ΣΣ ΣΣ 1 + + =

axaxax xyi

n

i i

n

i i

n

i i

n

(^0) =0 ii
–1
(^1) =0
–1
2
(^2) =0
–1
3
=0
–1

ΣΣ ΣΣ + + =

axaxax xy

i

n

i i

n

i i

n

i i

n

(^0) =0 i i
–1
2
(^1) =0
–1
3
(^2) =0
–1
4
=0
–1

ΣΣΣΣ + + =^2 (3.12)

Example 3.2.With the four data points in Example 3.1, i.e. (0, 0), (1, 2), (3, 2) and (6, −1), obtain
the best quadratic fit through them. Comment on the change in curve shape when point (3, 2) is
moved to (1.5, 4).
Using Eq. (3.12), we have

ΣΣ Σ Σ

iii i i i i

xx y

=0

3

=0

3

=0

(^32)
=0
3
1 = 4, = 10, = 46, = 3

i=0ΣΣΣ Σxxyxi i ii i i i xyi i

(^33)
=0
3
=0
(^34)
=0
(^32)
= 244, = 2, = 1378, = – 16
The 3 × 3 system becomes
410 46
10 46 244
46 244 1378

3
2
–16
0
1
2
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
a
a
a
or =
0.27
1.55

• 0.30

0

1

2

a

a

a

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

⎡

⎣

⎢

⎢

⎢

⎤

⎦

⎥

⎥

⎥

and the quadratic is y = 0.27 + 1.55x – 0.30x^2 , the plot of which is shown in Figure 3.3 along
with data points (solid lines). Although the curve does not pass through the data points, note its
proximity with the latter. For a new set of data points, namely, (0, 0), (1, 2), (1.5, 4) and (6, −1),
we have