Question

State and prove Tayler's theorem with Cauchy's form of remember.​

Answer 1

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➡️ I was studying the rigorous definition of Taylor's theorem and various kinds of remainders' forms when I got stuck at this question.

If fii(a)=fiii(a)=fiv(a)=.....=fn−1(a)=0 but fn(x) is continuous non zero at x=a, the we need to prove that

limh→0(θn−1)=1/n

where

f(a+h)=f(a)+hf′(a)+.......+hn−1(n−1)!fn−1(x+θn−1h)

I tried to use the Cauchy's remainder form by writing

Rn=hn−1(1−θ)n−1(n−1)!fn(a+θh)

For the sequence to converge now

limn→∞Rn=0

which seems to be perfectly true because of the hn−1(n−1)! and I use limit of h tending to zero, that just makes the whole expression 0 at once. How can I reach the final answer?

P.S. Though my book mentions the limit in the question as h tending to zero but I just can't fathom as to how can θ depend on h′s decreasing value.

Answer 2

Here,

I don't like to insist, but it seems a missing brace. Anyway, it's true that,

limh→0θn−1=1n

Fristly, from the hypothesis with the remainder of order n−1:

f(a+h)=f(a)+hf′(a)+hn−1(n−1)!f(n−1)(a+θn−1h)

hn−1(n−1)!f(n−1)(a+θn−1h)=f(a+h)−f(a)−hf′(a)(1)

And isolating θn+1 in the mean value theorem applied to the n−1 derivative:

f(n−1)(a+θn−1h)−f(n−1)(a)=θn−1hf(n)(ξ), with ξ∈(a,a+θn−1h). So is,

f(n−1)(a+θn−1h)=θn−1hf(n)(ξ)

θn−1=f(n−1)(a+θn−1h)hf(n)(ξ)=hn−1(n−1)!f(n−1)(a+θn−1h)hn(n−1)!f(n)(ξ)

Taking the limit:

limh→0θn−1=(n−1)!limh→0hn−1(n−1)!f(n−1)(a+θn−1h)hnf(n)(ξ)

But f(n) is continuous and f(n)(a)≠0, then limh→0f(n)(ξ)=f(n)(a) and can be taken out from the limit operation.

limh→0θn−1=(n−1)!f(n)(a)limh→0hn−1(n−1)!f(n−1)(a+θn−1h)hn=

=(1)(n−1)!f(n)(a)limh→0f(a+h)−f(a)−hf′(a)hn

By using the L'Hôpital rule n−1 times, and remembering that f(n−1)(a)=0:

limh→0θn−1=(n−1)!f(n)(a)limh→0f(n−1)(a+h)−f(n−1)(a)n!h=

=1nf(n)(a)limh→0f(n−1)(a+h)−f(n−1)(a)h=

=1nf(n)(a)f(n)(a)=1n