Exercise 5.18

Question

Exercise 18:
    Suppose $f$ is a real function on $[a,b]$, $n$ is a positive integer, and $f^{(n-1)}$ exists for every $t\in[a,b]$. Let $\alpha$, $\beta$, and $P$ be as in Taylor's theorem
    (Theorem 5.15). Define $$Q(t)=\frac{f(t)-f(\beta)}{t-\beta}$$ for $t\in[a,b]$, $t
eq\beta$, differentiate $$f(t)-f(\beta)=(t-\beta)Q(t)$$ $n-1$ times at $t=\alpha$, and 
    derive the following version of Taylor's theorem: $$f(\beta)=P(\beta)+\frac{Q^{(n-1)}(\alpha)}{(n-1)!}(\beta-\alpha)^n.$$

analambanomenos · Accepted Answer

We have
    \begin{align*}
        f'(t) &= Q(t)+(t-\beta)Q'(t) \
        f''(t) &= 2Q'(t)+(t-\beta)Q''(t) \
        f^{(3)}(t) &= 3Q''(t)+(t-\beta)Q^{(3)}(t)
    \end{align*}
    and so forth, which can be rewritten
    \begin{align*}
	Q(t) &= f'(t)+Q'(t)(\beta-t) \
	Q'(t) &= \frac{1}{2}\bigl(f''(t)+Q''(t)(\beta-t)\bigr) \
	Q''(t) &= \frac{1}{3}\bigl(f^{(3)}(t)+Q^{(3)}(t)(\beta-t)\bigr).
    \end{align*}
    Hence
    \begin{align*}
        f(\beta) &= f(\alpha)+Q(\alpha)(\beta-\alpha) \
        &= f(\alpha)+f'(\alpha)(\beta-\alpha)+Q'(\alpha)(\beta-\alpha)^2 \
        &= f(\alpha)+f'(\alpha)(\beta-\alpha)+\frac{1}{2}f''(\alpha)(\beta-\alpha)^2+\frac{1}{2}Q''(\alpha)(\beta-\alpha)^3 \
        &= f(\alpha)+f'(\alpha)(\beta-\alpha)+\frac{1}{2}f''(\alpha)(\beta-\alpha)^2+\frac{1}{3!}f^{(3)}(\beta-\alpha)^3+\frac{1}{3!}Q^{(3)}(\alpha)(\beta-\alpha)^4
    \end{align*}
    which easily leads to the desired formula.

Exercise 5.18

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Exercise 5.18

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