Proving the Taylor Series Expansion of a Polynomial

When dealing with polynomials and their Taylor series expansions, one of the essential concepts to understand is the behavior of their derivatives. This article explores how the Taylor series expansion of a polynomial of degree n can be proven to be finite, focusing on the derivatives of such polynomials.

The Taylor Series Formula and Polynomial

The Taylor series expansion of a function p(x) is given by:

Quaddisplaystyle p(x) sum_{i0}^{n}a_ix^i

In this expansion, the coefficients (a_i) are the derivatives of p(x) of different orders divided by some factorials. Specifically, for a polynomial of the form:

Quaddisplaystyle p(x) sum_{i0}^{n}a_ix^i

The nth derivative of a polynomial of degree n is a constant, and all higher-order derivatives are zero. This property significantly simplifies the Taylor series expansion, making it finite. This article will prove this concept by showing that the kth derivative of (x^i) for (0 le k le n) results in a specific form.

Derivatives of (x^i)

To understand the Taylor series expansion, we first need to analyze the derivatives of (x^i), where i is a non-negative integer.

Quaddisplaystyle frac{d^k}{dx^k}x^i begin{cases}frac{i!}{(i-k)!}x^{i-k} text{if} kle i 0 text{if} k gt i end{cases}

We will prove this statement by induction. For the base case, when k 0, both sides of the equation simplify to x^i, which is clearly true. For the induction step, we assume that the statement holds for k m and prove it for k m 1.

Suppose the statement holds for k m:

Quaddisplaystyle frac{d^m}{dx^m}x^i begin{cases}frac{i!}{(i-m)!}x^{i-m} text{if} m le i 0 text{if} m gt i end{cases}

Now, we need to show that the statement holds for k m 1:

Quaddisplaystyle frac{d^{m 1}}{dx^{m 1}}x^i frac{d}{dx}(frac{i!}{(i-m)!}x^{i-m})

By the product rule and chain rule of differentiation, we get:

Quaddisplaystyle frac{d}{dx}(frac{i!}{(i-m)!}x^{i-m}) frac{i!}{(i-m)!}cdot(i-m)x^{i-m-1}

If m 1 le i, then frac{i!}{(i-(m 1))!} frac{i!}{(i-m-1)!} frac{i!}{(i-m)!}cdot(i-m). Thus, the statement holds for k m 1. For k m 1 > i, both sides are zero, validating the statement.

Taylor Series Expansion and Polynomial Derivatives

Given a polynomial (p(x)) and its kth derivative at (x 0), we need to show that:

Quaddisplaystyle p^{(k)}(0) k!a_k

From the Taylor series expansion, we know that the terms of the expansion are of the form (frac{p^{(i)}(x)}{i!}x^i). When we take the kth derivative of (p(x)) and evaluate it at (x 0), only the term where (i k) remains, as all other terms involve (x) and thus evaluate to zero. Therefore, we have:

Quaddisplaystyle p^{(k)}(0) k!a_k

Using the result from the previous step, we know that:

Quaddisplaystyle p^{(k)}(0) frac{d^k}{dx^k}sum_{i0}^{n}a_ix^i |_{x0} k!a_k

This confirms that the kth derivative of a polynomial of degree n at x 0 is indeed (k!a_k).

Conclusion

By understanding the derivatives of (x^i) and applying the induction method, we have proven the finite nature of the Taylor series expansion for a polynomial. This understanding not only simplifies the concept but also provides a useful tool for analyzing and manipulating polynomials and their series expansions.