Taylor's theorem $\lim_{x \to a} \frac{f(x)-P_{n,a}(x)}{(x-a)^n}=0$

Question

First, I wanted to ask whether this theorem concerns the remainder of Polynomial at point $a$ and order n-th? I also wanted to know in general why this is important that the difference between function and it's Taylor's polynomial is higher order small function than $(x-a)^n$?

Then I have questions about theorem itself:

I don't understand why $g^{(k)}=\frac{n!(x-a)^{n-k}}{(n-k)!}$? Then the second expression that is underlined with red is not clear to me. $Q(x)=f^{n-1}(a)$. Is it because the $(x-a)$ is derived $(n-1)th$ times and the derivative is $x'=1$ and all the orders $i$ of the derivative cancels out with the factorial on denominator? But then, for example, $Q(x)=f^{(n-k+1)}(x-a)^{(n-k)}$, because the derivative and factorial cancels out?

Answer 1

Since $g(x) = (x - a)^n$, one can compute $g^{(1)}(x) = g'(x) = n \cdot (x-a)^{n-1}$, $g^{(2)}(x) = g''(x) = n(n-1) \cdot (x-a)^{n-2}$, etcetera. By induction this shows the identity in your first question. If the factorial bit is what troubles you, just note that $n(n-1)(n-2) \cdots (n-k + 1) = n! / (n-k)!$.

For the second part of your question, note that $f^{(i)}(a)$ is a constant, so $Q(x)$ really is just a degree $n-1$ polynomial. This means that taking the $(n-1)$-th derivative is equal to the $(n-1)$-th derivative of $x^{n-1}$ multiplied with the coefficient of this term. The factorial $(n-1)!$ in the denominator cancels the factorial $(n-1)!$ obtained from taking the $(n-1)$-th derivative of $x^{n-1}$, so we are just left with $f^{(n-1)}(a)$

Answer 2

Yes, this theorem concerns the remainder of the Taylor polynomial at $a$, since, by definition, this remainder is $f(x)-P_{n,a}(x)$.

This theorem tells us that, as $x$ approaches $a$, $f(x)$ approaches $P_{n,a}(x)$, and that this approach is faster as $n$ gets bigger.

Since $g(x)=(x-a)^n$, you have:

• $g'(x)=n(x-a)^{n-1}=\frac{n!}{(n-1)!}(x-a)^{n-1}$;
• $g''(x)=n(n-1)(x-a)^{n-2}=\frac{n!}{(n-2)!}(x-a)^{n-2}$

and so on.

Finally, since $Q(x)$ is a polynomial function with degree $n-1$, $Q^{(n-1)}(x)$ is constant. So, compute $Q^{(n-1)}(a)$ to know what that constant is. You will get $f^{(n-1)}(a)$.