A new and elementary proof of Newton's "favorite" quadrature formulae

0.06 MB·English

by  Cezar Lupu

A new and elementary proof of Newton’s
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”favorite” quadrature formulae
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n Cezar Lupu, Tudorel Lupu
a
J

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1 In this note we shall give a new proof to a quadrature formulae due to Newton(see [1.]) which
states

]
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Theorem 0.1 If f : [a,b] → R is four times differentiable on [a,b] with f(4) continuous, then
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there is some ξ ∈[a,b] such that
.
h
at bf(x)dx= b−a f(a)+3f 2a+b +3f 2b+a +f(b) − (b−a)5f(4)(ξ).
m Za 8 (cid:20) (cid:18) 3 (cid:19) (cid:18) 3 (cid:19) (cid:21) 6480
[ In order to prove this theorem we need to show first three lemmas starting with

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v Lemma 1. If f : [a,b] → R is four times differentiable with f(4) continuous, then there is some
9 ξ ∈[a,b] such that
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2 b a+b (b−a)3 a+b (b−a)5
1 f(x)dx=(b−a)f + f(2) + f(4)(ξ1).
0 Za (cid:18) 2 (cid:19) 24 (cid:18) 2 (cid:19) 1920
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Proof. We begin with the following integral
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/
h b a+x 2b+a−x
t (x−a)4 f(4) +f(4) dx.
a Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
m
Integrating by part we get
:
v
i b a+x 2b+a−x a+x 2b+a−x
X (x−a)4 f(4) +f(4) dx=2(x−a)4 f(3) −f(3) dx|b
a
r Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21) (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
a
b
a+x 2b+a−x a+x 2b+a−x
−2 4(x−a)3 f(3) −f(3) =−16(x−a)3 f(2) +f(2) |ba
Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21) (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
b
a+x 2b+a−x a+b
+48 (x−a)2 f(2) +f(2) dx=−32(b−a)3f(2) +
Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21) (cid:18) 2 (cid:19)
b
a+x 2b+a−x a+x 2b+a−x
+96(x−a)2 f′ −f′ |b−192 (x−a) f′ −f′ dx
a
(cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21) Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
a+b a+x 2b+a−x
=−32(b−a)3f(2) −384(x−a) f +f |b
a
(cid:18) 2 (cid:19) (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
b b
a+x 2b+a−x a+b a+b
+384 f +f dx=768 f(x)dx−768(b−a)f −32(b−a)3f(2) .
Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21) Za (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)
On the other hand, by the mean value theorem, there is η ∈[a,b] such that
b b
a+x 2b+a−x a+η 2b+a−η
(x−a)4 f(4) +f dx= f(4) +f(4) (x−a)4dx.
Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21) (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)Za
1
But, f(4) is continuous, so f(4) has intermediate value property. It follows that there is ξ ∈[a,b]
1
such that
a+η 2b+a−η
f(4) +f(4) =2f(4)(ξ ),
1
(cid:18) 2 (cid:19) (cid:18) 2 (cid:19)
so
b a+x 2b+a−x 2(b−a)5
(x−a)4 f(4) +f dx= f(4)(ξ ).
1
Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21) 5
We get
b a+b a+b 2(b−a)5
768 f(x)dx=768(b−a)f +32(b−a)3f(2) + f(4)(ξ )
1
Za (cid:18) 2 (cid:19) (cid:18) 2 (cid:19) 5
which is equivalent to
b a+b (b−a)3 a+b (b−a)5
f(x)dx=(b−a)f + f(2) + f(4)(ξ ),
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Za (cid:18) 2 (cid:19) 24 (cid:18) 2 (cid:19) 1920
where ξ ∈[a,b]. (cid:3)
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Lemma 2.
If f : [a,b]→ R is four times differentiable with f(4) continuous, then there is some ξ ∈ [a,b]
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such that
b (b−a) a+b (b−a)3 a+b (b−a)5
f(x)dx= f(a)+6f +f(b) + f(2) − f(4)(ξ ).
2
Za 8 (cid:20) (cid:18) 2 (cid:19) (cid:21) 96 (cid:18) 2 (cid:19) 7680
Proof. Like in the proof of lemma 1, we will begin with the following integral
b
a+x 2b+a−x
(x−a)3(x−b) f(4) +f(4) dx.
Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
Integrating by part we get
b
a+x 2b+a−x a+x 2b+a−x
(x−a)3(x−b) f(4) +f(4) dx=2(x−a)3(x−b) f(3) −f(3) |b−
a
Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21) (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
b
a+x 2b+a−x
−2 3(x−a)2(x−b)+(x−a)3 f(3) −f(3) dx=
Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
(cid:2) (cid:3)
a+x 2b+a−x
=−4 3(x−a)2(x−b)+(x−a)3 f(2) +f(2) |ba+
(cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
(cid:2) (cid:3)
b
a+x 2b+a−x
+24 (x−a)2+(x−a)(x−b) f(2) +f(2) dx=
Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
(cid:2) (cid:3)
a+b a+x 2b+a−x
=−8(b−a)3f(2) +48 (x−a)2+(x−a)(x−b) f′ −f′ |ba−
(cid:18) 2 (cid:19) (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
(cid:2) (cid:3)
b
a+x 2b+a−x
′ ′
−48 [3(x−a)+x−b] f −f dx=
Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
a+b a+x 2b+a−x
=−8(b−a)3f(2) −96[3(x−a)+(x−b)] f +f |b
a
(cid:18) 2 (cid:19) (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
b b
a+x 2b+a−x a+b
+384 f +f dx=768 f(x)dx−96·6(b−a)f
Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21) Za (cid:18) 2 (cid:19)
b
a+b a+b
−96(b−a)[f(a)+f(b)]−8(b−a)3f(2) =768 f(x)dx−96(b−a) f(a)+6f +f(b) −
(cid:18) 2 (cid:19) Za (cid:20) (cid:18) 2 (cid:19) (cid:21)
2
a+b
−8(b−a)3f(2) . On the other hand, by the mean value theorem there is some η′ ∈ [a,b]
(cid:18) 2 (cid:19)
such that
b a+x 2b+a−x a+η′ 2b+a−η′
(x−a)3(x−b) f(4) +f(4) dx= f(4) +f(4)
Za (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21) (cid:20) (cid:18) 2 (cid:19) (cid:18) 2 (cid:19)(cid:21)
b
· (x−a)3(x−b)dx. By the intermediate value property, there is some ξ ∈[a,b] such that
2
Za
a+η′ 2b+a−η′
f(4) +f(4) =2f(4)(ξ ).
2
(cid:18) 2 (cid:19) (cid:18) 2 (cid:19)
b (b−a)5
A simple calculation shows that (x−a)3(x−b)dx = − . Now, combining these two
Za 20
equalities obtained, the lemma follows immediately. (cid:3)
Lemma 3(Simpson-Cavalieri).
If f : [a,b]→ R is four times differentiable with f(4) continuous, then there is some ξ ∈ [a,b]
3
such that
b (b−a) a+b (b−a)5
f(x)dx= f(a)+4f +f(b) − f(4)(ξ ).
3
Za 6 (cid:20) (cid:18) 2 (cid:19) (cid:21) 2880
Proof. This formulae can be proved in many ways, but we prefer to deduce it by eleminating
a+b
f(2) from lemma 1 and lemma 2. (cid:3)
(cid:18) 2 (cid:19)
Now, we shall prove the main result of our paper. First, we apply lemma 1 on the interval
2a+b 2b+a 2a+b 2b+a
′
, . Then, there is some ξ ∈ , such that
(cid:20) 3 3 (cid:21) 1 (cid:20) 3 3 (cid:21)
2b3+a f(x)dx= b−af a+b + (b−a)3f(2) a+b + (b−a)5 f(4)(ξ′).(∗)
Z2a+b 3 (cid:18) 2 (cid:19) 24·27 (cid:18) 2 (cid:19) 1920·35 1
3
a+b
By eliminating f(2) from lemma 2 and (∗) we get
(cid:18) 2 (cid:19)
b 2b3+a b−a a+b (b−a)5
4 f(x)dx=27 f(x)dx+ f(a)−12f +f(b) − f(4)(ξ ),(∗∗)
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Za Z2a+b 2 (cid:20) (cid:18) 2 (cid:19) (cid:21) 9·192
3
with ξ ∈[a,b]. Now, applying Lemma 3(Simpson-Cavalieri), we get
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2b3+a b−a 2a+b a+b 2b+a (b−a)5
f(x)dx= f +4f +f − f(4)(ξ )(∗∗∗)
Z2a+b 18 (cid:20) (cid:18) 3 (cid:19) (cid:18) 2 (cid:19) (cid:18) 3 (cid:19)(cid:21) 35·2880 5
3
2a+b 2b+a
with ξ ∈ , . Now, replacing (∗∗∗) in (∗∗) we have
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(cid:20) 3 3 (cid:21)
b 27(b−a) 2a+b a+b 2b+a (b−a)5
4 f(x)dx= f +4f +f − f(4)(ξ )
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Za 18 (cid:20) (cid:18) 3 (cid:19) (cid:18) 2 (cid:19) (cid:18) 3 (cid:19)(cid:21) 9·2880
(b−a) a+b (b−a)5
+ f(a)−12f +f(b) − f(4)(ξ ),
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2 (cid:20) (cid:18) 2 (cid:19) (cid:21) 9·192
with ξ ,ξ ∈[a,b]. The above formulae is equivalent to
4 5
b b−a 2a+b 2b+a (b−a)5 (b−a)5
4 f(x)dx= f(a)+3f +3f +f(b) − f(4)(ξ )+ f(4)(ξ ) .
5 4
Za 2 (cid:20) (cid:18) 3 (cid:19) (cid:18) 3 (cid:19) (cid:21) (cid:20)9·2880 9·192 (cid:21)
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Now, using intermediate value property, we get a ξ ∈[a,b] such that
(b−a)5 (b−a)5 (b−a)5
f(4)(ξ )+ f(4)(ξ )= f(4)(ξ).
5 4
9·2880 9·192 1620
Now, Newton’s result follows immediately.
References.
[1.] D.V. Ionescu-Restulˆın formula de cuadratura˘ ”preferata˘” a lui Newton, Gaz.Mat., vol 7,
1965,pag. 298-303.
[2.] J.V. Steffensen-Interpolation, Baltimore, 1927
[3.] B.P. Demidovich, I.A Maron-ComputationalMathematics, Moscow: Mir Publishers, 1976.
2nd edition. Translated from Russian.
Cezar Lupu, student Faculty of Mathematics, University of Bucharest, Bucharest,
Romania. e-mail: [email protected]
Tudorel Lupu, teacher, Decebal High school, Constanza, Romania.
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