Lemma (Generalized Rolle's theorem). If a, b = I are distinctand f(a) = f'(a) = f'(a) =···= = f(n) (a) = f(b) = 0, then there existsa point c = (a, b) or (b, a) such that f(n+¹)(c) = 0. (b) Prove that if a, b = I are distinct, then there exists a c strictlybetween a and b such thatf(b)=nf(k) (a)k!(b − a)k +f(n+1) (c)(n + 1)!(b − a)n +¹k=0in the following way. For each t € I, letnTn(t) = f(a) (t-a)*, R₁(t) = f(t)-Tn(t), _g(t) = Tn(t)+k!k=0Apply the generalized Rolle's theorem lemma to h = f - g.Rn (b)(b − a)n + ¹(t-a)n+1

Given a,b∈I are distinct. Without loss of generality, suppose a<b. Let f be a function on I such that its all derivative exists. By Taylor's series formula, the nth degree Taylor polynomial of the function f can be calculated at each t∈I, centered at t=a as follows: Tnt=∑k=0nfkak!t-ak Therefore the Remainder of the Taylor series of the function f can be written as: Rnt=ft-Tnt. Observe that Rnt will contains all the term of Taylor series of function f centered at t=a with degree greater then n. Now consider a function gt defined as: gt=Tnt+Rnbb-an+1t-an+1.Now consider a function h=f-g: ht=ft-gt=Tnt+Rnt-Tnt+Rnbb-an+1t-an+1=Rnt-Rnbb-an+1t-an+1 Since Rnt contains all the term of Taylor series of function f centered at t=a with degree greater then n, therefore all the terms in ht=Rnt-Rnbb-an+1t-an+1 contains a power of t-a of degree at least n+1. Hence h, h', h",…,hn will contains a power of t-a. Therefore ha=h'a=…=hna=0. Now observe that: hb=Rnb-Rnbb-an+1b-an+1=Rnb-Rnb=0 Hence by Generalized Rolle's theorem, there exists a point c∈a,b such that hn+1c=0.Since gt=Tnt+Rnbb-an+1t-an+1 and h=f-g, therefore: ht=ft-gt=ft-Tnt-Rnbb-an+1t-an+1⇒hn+1t=fn+1t-Rnbb-an+1n+1! Since hn+1c=0, therefore: fn+1c-Rnbb-an+1n+1!=0Rnb=fn+1cn+1!b-an+1 Since Rnt=ft-Tnt, therefore: Rnt=ft-Tntft=Tnt+Rntft=∑k=0nfkak!t-ak+Rnt Therefore at t=b, we have: fb=∑k=0nfkak!b-ak+Rnb=∑k=0nfkak!b-ak+fn+1cn+1!b-an+1 Hence fb=∑k=0nfkak!b-ak+fn+1cn+1!b-an+1.