NEWTON–GREGORY INTERPOLATING POLYNOMIALS
Difference Table
h = 0.3
i
xi
fi
∆fi
∆2 fi
∆3 fi
∆4 fi
0
1.0
+2.70
−0.04
−0.06
+0.02
+0.01
1
1.3
+2.66
−0.10
−0.04
+0.03
2
1.6
+2.56
−0.14
−0.01
3
1.9
+2.42
−0.15
4
2.2
+2.27
1. Let’s interpolate f (1.5) using polynomials of different degrees, each beginning with abscissa x0 = 1.0. So
1.5−1.0
0
s = x−x
= 5/3. Determine the binomial coefficients via its telescopic property:
h =
0.3
s
1
s
2
s
3
s
4
5
,
3
= s =
=
=
=
s s − 1
1
·
=
2
s s − 2
2
·
3
5
3
= 0.5 ·
3
s s − 3
·
5
·
3
−1
= 0.5 ,
2
5
3
−2
= −0.0617 283 950 ,
3
= −0.0617 283 950 ·
4
5
3
−3
= +0.0205 761 317 .
4
(a) Let’s use the 2nd degree interpolating polynomial containing points [0,1,2]:
s
s
∆2 f0 ,
1.0 ≤ x ≤ 1.6 ,
2
1
f (1.5) ≈ P[0−2] (s = 5/3)
5
= 2.70 + · (−0.04) + (0.5)(−0.06) = 2.600 000 000 .
3
P[0−2] (x) = f0 +
∆f0 +
(1)
(2)
(b) Let’s use the 3rd degree interpolating polynomial containing points [0,1,2,3]:
P[0−3] (x) = f0 +
|
s
1
∆f0 +
s
2
∆2 f0
{z
+
}
P[0−2] (x)
s
3
|
∆3 f0 ,
{z
1.0 ≤ x ≤ 1.9 .
(3)
}
“next term”
So
f (1.5) ≈ P[0−3] (s = 5/3)
= P[0−2] (s = 5/3) +
s
3
∆3 f0
= 2.6 + (−0.0617 283 950)(+0.02) ,
by Result (2)
= 2.6 − 0.00 123 456 790
(4)
2.5 987 654 320 .
(5)
=
c 1998–2017 Kevin G. TeBeest, Kettering University
Copyright 1
file diff.tex
04/20/2017
(c) Similarly, if we use the 4th degree interpolating polynomial containing points [0,1,2,3,4]:
f (1.5) ≈ P[0−4] (s = 5/3) ,
1.0 ≤ x ≤ 2.2
= P[0−3] (s = 5/3) +
s
∆4 f0
4
= 2.5 987 654 320 + (0.0205 761 317)(+0.01)
by Result (5)
= 2.5 987 654 320 + 0.0002 0576 1317
=
2.5989 7119 3416 .
(6)
2. Let’s interpolate f (1.5) using polynomials of different degrees, each beginning with abscissa x1 = 1.3. So
1.5−1.3
1
s = x−x
= 2/3. Let’s determine the binomial coefficients:
h =
0.3
s
1
s
2
s
3
2
,
3
= s =
=
=
s s − 1
1
·
2
s s − 2
2
·
2
3
2
·
3
=
−1
= −0.1 ,
2
2
3
= −0.1 ·
3
−2
= 0.049 382 716 .
3
(a) Let’s use the 2nd degree interpolating polynomial containing points [1,2,3]:
s
s
∆2 f1 ,
1.3 ≤ x ≤ 1.9 ,
1
2
f (1.5) ≈ P[1−3] (s = 5/3)
2
= 2.66 + · (−0.10) + (−0.1)(−0.04)
3
P[1−3] (x) = f1 +
=
∆f1 +
2.59 7 .
(7)
(8)
(b) Let’s use the 3rd degree interpolating polynomial containing points [1,2,3,4]:
P[1−4] (x) = P[1−3] (x) +
s
3
∆3 f1 .
f (1.5) ≈ P[1−3] (s = 2/3) +
s
3
1.3 ≤ x ≤ 2.2 ,
(9)
∆3 f1
= 2.59 7 + (0.049 382 716)(+0.03)
by Result (8)
= 2.59 7 + 0.00 148
=
2.59 925 .
c 1998–2017 Kevin G. TeBeest, Kettering University
Copyright (10)
file diff.tex
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NEXT TERM RULE: The next term can be used to estimate the interpolation error.
1. To estimate the error in using P[0−2] (x) to interpolate at x = 1.5, s =
Error ≈ “next term” =
So using points [0, 1, 2],
s
3
1.5 − 1.0
= 5/3:
0.3
∆3 f0 = (−0.0617 283 950)(+0.02) = −0.00 123 456 790 .
f (1.5) ≈ 2.6 with an error of about
Error ≈ −0.00 123 456 790.
2. To estimate the error in using P[0−3] (x) to interpolate at x = 1.5 (s = 5/3) :
Error ≈ “next term” =
So using points [0, 1, 2, 3],
s
4
∆4 f0 = (0.0205 761 317)(+0.01) = +0.0002 0576 1317 .
f (1.5) ≈ 2.5 987 654 320
with an error of about
3. To estimate the error in using P[1−3] (x) to interpolate at x = 1.5, so s =
Error ≈ “next term” =
So using points [1, 2, 3],
s
3
Error ≈ +0.0002 0576 1317.
1.5 − 1.3
= 2/3:
0.3
∆3 f1 = (0.049 382 716)(+0.03) = +0.00 148 .
f (1.5) ≈ 2.59 7 with an error of about
c 1998–2017 Kevin G. TeBeest, Kettering University
Copyright Error ≈ +0.00 148.
file diff.tex
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