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page20-1
(1)p.19−3 Rolle′s thm ⇒ ∃ ς1∈ ] 0,1 [ st G(1)(ς1)=(1)0
Now G(1)(0) =(2)0 why?
(1) p.19−1: G(1)(t)=(3)e(1)(t)−5t4e(1)
(5)p.18−3: e(t)=A(t)−A2L(t) (4)
e(1)(t)=A(1)−A2L(1)(t) (5)
page20-2
A(t)=∫+t−t−=∫k−t−+∫tk−
A(1)=(1)F(−t)+F(t)
k∈]−t,t[ (kisconstant)
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A2L(1)(t)=(2)13[F(−t)+4F(0)+F(t)]+t3[F(1)(−t)+F(1)(t)
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e(1)(0)=A(1)(0)−A2L(1)(0)=2F(0)−[2F(0)+0]=0 (3)
(3) and (3)p.20−1⇒(2)p.20−1
G(1)(ξ1)=0 (1)p.20−1
G(1)(0)=0 (2)p.20−1
page20-3
Rolle′stheorem⇒ ∀ξ2∈]0,ξ1[st G(2)(ξ2)=0 (1)
Again, G(2)(0)=(2)0 HW*4.3
(1) and (2) Rolle′stheorem ⇒ ∀ξ3∈]0,ξ2[ G(2)(ξ2)=0 st G(3)(ξ3)=(2)0
(1)p.19−1: G(3)(t)e(4)(3)(t)−t2⏟(ξ)(4)(3)e(1)
e(3)(t)=HW*4.4−t3[F(3)(t)−F(3)(−t)] (5)
G(3)(ξ3)=−ξ33[F(3)(ξ3)−F(3)(−ξ3)]⏞Apply DMVT−60(ξ3)2e(1)=(6)from(3)0
=DMVT(f)−ξ33[2ξ3⏟ξ3−(−ξ3)F(4)(ξ4)]−60(ξ3)2e(1) ξ4∈ ]−ξ3,ξ3[
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