Deriving an Alternate Taylor's Theorem for Functions with a Defined Derivative

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Homework Statement

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 an alternate Taylor’s theorem:

\ f(\beta)=P(\beta)+\frac{Q^{(n-1)}(\alpha)}{(n-1)!}(\beta-\alpha)^{n} (I had to put this here to make the above expression stay on one line)

Homework Equations

The Attempt at a Solution

So first I did the differentiation n-1 times, you notice a pattern and since f(beta) is constant, you get

\ f^{(n}}(t)= nQ^{(n-1)}+(t-\beta)^nQ^{(n)}(t)

Then from Taylor's theorem we know that

\ f(\beta) = P(\beta) + \frac{f^{(n)}(x)}{n!}(\beta - \alpha)^{n}

Just plugging in that expression into Taylor's theorem is real damn close to the result I need. How do I get rid of the extra Q^{(n)} in the numerator? (or is my differentiation wrong and I'm not catching it?)

Thanks a mil guys.

 

[Avodyne]

Both your equations in part (3) are wrong.

 

[Quantumpencil]

The second equation is Rudin 24... I don't see how it could be wrong -_-.

The first is just differentiating the given expression. We'd use the product rule (n-1) (I think there I used it n times), and that would be the result yes?