State and prove Taylor’s formula with the
Lagrange form of the remainder.
Answers
Answer:
The Taylor series of a function is extremely useful in all sorts of applications and, at the same time, it is fundamental in pure mathematics, specifically in (complex) function theory. Recall that, if f(x)f(x) is infinitely differentiable at x=ax=a, the Taylor series of f(x)f(x) at x=ax=a is by definition
\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n = f(a) + f'(a) (x-a) + \frac{f''(a)}{2}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots.
n=0
∑
∞
n!
f
(n)
(a)
(x−a)
n
=f(a)+f
′
(a)(x−a)+
2
f
′′
(a)
(x−a)
2
+
3!
f
′′′
(a)
(x−a)
3
+⋯.
The expression (and its tremendous utility) comes from its being the best polynomial approximation (up to any given degree) near the point x=ax=a. For f(x)=\sin xf(x)=sinx and a=0a=0, it's easy to compute all the f^{(n)}(0)f
(n)
(0) and to see that the Taylor series converges for all x\in\mathbb Rx∈R (by ratio test), but it's by no means obvious that it should converge to \sin xsinx. After all, the derivatives at x=0x=0 only depend on the values of the function very close to x=0x=0. Why should you expect that somehow it "knows" the values of the function far away?
That the Taylor series does converge to the function itself must be a non-trivial fact. Most calculus textbooks would invoke a Taylor's theorem (with Lagrange remainder), and would probably mention that it is a generalization of the mean value theorem. The proof of Taylor's theorem in its full generality may be short but is not very illuminating. Fortunately, a very natural derivation based only on the fundamental theorem of calculus (and a little bit of multi-variable perspective) is all one would need for most functions.
Answer:
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