# Prob. 18, Chap. 5, in Baby Rudin: Another Form of Taylor's Theorem

Here is Prob. 18, Chap. 5, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

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 (5.15). Define $$Q(t) = \frac{ f(t) - f(\beta) }{ t- \beta}$$ for $$t \in [a, b]$$, $$t \neq \beta$$, differentiate $$f(t) - f(\beta) = (t-\beta) Q(t)$$ $$n-1$$ times at $$t = \alpha$$, and derive the following version of Taylor's theorem: $$f(\beta) = P(\beta) + \frac{Q^{(n-1)}(\alpha)}{(n-1)!} (\beta - \alpha)^n.$$

And, here is Theorem 5.15 in Baby Rudin, 3rd edition:

Suppose $$f$$ is a real function on $$[a, b]$$, $$n$$ is a positive integer, $$f^{(n-1)}$$ is continuous on $$[a, b]$$, and $$f^{(n)}(t)$$ exists for every $$t \in (a, b)$$. Let $$\alpha$$, $$\beta$$ be distinct points of $$[a, b]$$, and define $$P(t) = \sum_{k=0}^{n-1} \frac{f^{(k)}(\alpha)}{k!} \left( t-\alpha \right)^k.$$ Then there exists a point $$x$$ between $$\alpha$$ and $$\beta$$ such that $$f(\beta) = P(\beta) + \frac{f^{(n)}(x)}{n!} (\beta - \alpha )^n.$$

An Attempt:

For all $$t \in [a, b]$$, we have \begin{align} f(t) - f(\beta) &= ( t-\beta) Q(t), \tag{1} \\ f^\prime(t) &= Q(t) + (t-\beta) Q^\prime(t), \tag{2} \\ f^{\prime\prime}(t) &= 2Q^\prime(t) + (t-\beta) Q^{\prime\prime}(t), \tag{3} \\ f^{(3)}(t) &= 3 Q^{\prime\prime}(t) + (t-\beta)Q^{(3)}(t), \tag{4} \\ f^{(4)}(t) &= 4 Q^{(3)}(t) + (t-\beta) Q^{(4)}(t), \tag{5} \\ \cdots &= \cdots \\ f^{(n-1)}(t) &= (n-1) Q^{(n-2)}(t) + (t-\beta) Q^{(n-1)}(t). \tag{*} \end{align} So, for $$t = \alpha$$, the above chain of equations yields \begin{align} & \qquad f(\beta) \\ &= f(\alpha) + Q(\alpha) (\beta - \alpha ) \qquad \mbox{ [ using (1) ] } \\ &= f(\alpha) + \left[ f^\prime(\alpha) + (\beta - \alpha) Q^\prime(\alpha) \right] (\beta - \alpha ) \\ & \qquad \qquad \mbox{ [ using (2) ] } \\ &= f(\alpha) + f^\prime(\alpha) (\beta - \alpha) + Q^\prime(\alpha) (\beta - \alpha)^2 \\ &= f(\alpha) + f^\prime(\alpha) (\beta - \alpha) \\ & \qquad + \left[ \frac{1}{2} \left( f^{\prime\prime}(\alpha) + (\beta - \alpha) Q^{\prime\prime}(\alpha) \right) \right] (\beta - \alpha)^2 \\ & \qquad \qquad \mbox{ [ using (3) ] } \\ &= f(\alpha) + \frac{f^\prime(\alpha)}{1!} (\beta - \alpha) + \frac{f^{\prime\prime}(\alpha)}{2!} (\beta - \alpha)^2 + \frac{Q^{\prime\prime}(\alpha)}{2!} (\beta - \alpha)^3 \\ &= f(\alpha) + f^\prime(\alpha) (\beta - \alpha) + \frac{f^{\prime\prime}(\alpha)}{2} (\beta - \alpha)^2 + \frac{\frac{1}{3} \left( f^{(3)}(\alpha) + (\beta - \alpha) Q^{(3)}(\alpha) \right) }{2} (\beta - \alpha)^3 \qquad \mbox{ [ using (4) ] } \\ &= f(\alpha) + \frac{f^\prime(\alpha)}{1!} (\beta - \alpha) + \frac{f^{\prime\prime}(\alpha)}{2!} (\beta - \alpha)^2 + \frac{f^{(3)}(\alpha)}{3!}(\beta-\alpha)^3 + \frac{Q^{(3)}(\alpha)}{3!} (\beta-\alpha)^4 \\ &= f(\alpha) + \frac{f^\prime(\alpha)}{1!} (\beta - \alpha) + \frac{f^{\prime\prime}(\alpha)}{2!} (\beta - \alpha)^2 + \frac{f^{(3)}(\alpha)}{3!}(\beta-\alpha)^3 + \frac{ \frac{1}{4} \left( f^{(4)}(\alpha) + (\beta - \alpha) Q^{(4)}(\alpha) \right) }{3!} (\beta-\alpha)^4 \\ & \qquad \qquad \mbox{ [ using (5) ] } \\ &= f(\alpha) + \frac{f^\prime(\alpha)}{1!} (\beta - \alpha) + \frac{f^{\prime\prime}(\alpha)}{2!} (\beta - \alpha)^2 + \frac{f^{(3)}(\alpha)}{3!}(\beta-\alpha)^3 + \frac{f^{(4)}(\alpha)}{4!} (\beta - \alpha)^4 + \frac{Q^{(4)}(\alpha)}{4!} (\beta-\alpha)^5 \\ &= \cdots \\ &= f(\alpha) + \frac{f^\prime(\alpha)}{1!} (\beta - \alpha) + \frac{f^{\prime\prime}(\alpha)}{2!} (\beta - \alpha)^2 \\ & \qquad + \frac{f^{(3)}(\alpha)}{3!}(\beta-\alpha)^3 + \frac{f^{(4)}(\alpha)}{4!} (\beta - \alpha)^4 + \cdots + \frac{f^{(n-1)}(\alpha)}{(n-1)!} (\beta-\alpha)^{n-1} + \frac{Q^{(n-1)}(\alpha)}{(n-1)!} (\beta - \alpha)^n \\ &= P(\beta) + \frac{Q^{(n-1)}(\alpha)}{(n-1)!} (\beta - \alpha)^n, \end{align} as required.

Is this proof correct? If so, then is it rigorous enough for Rudin as well?

I don't see any problems with this proof, I think it's valid. If you want to make it more rigorous, you can prove by induction that $$Q^{(k)}(t) = \frac{f^{(k)}(t) + (\beta - t)Q^{(k+1)}(t)}{k+1},$$ for integer $$k$$, where $$0\leq k \leq n-2$$.