# How to prove a polynomial can be written as Taylor-style?

I know that by Taylor's theorem, a function $f$ under some assumptions, can be computed by $$f(x)=f(a)+f'(a)(x-a)+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n+\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{(n+1)}$$ If $f$ itself is a polynomial of degree $n$, then $$f(x)=f(a)+f'(a)(x-a)+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n.$$ This can be directly deduced from Taylor's theorem as I mentioned.

However, since this is a much simpler result, can we prove it without using that theorem? And is there intuitive understanding of the above equation?

• If you're the sort (as I am) who likes a constructive proof, you might want to look up Horner's method, which can be used to explicitly generate the (scaled) derivative values of your polynomial. You then need to prove that this algorithm will halt. Oct 11, 2017 at 15:44

This simply is a re-writing of the binomial formula.

First note it suffices to prove it for monomials, since differentiation is a linear operation. So let's set $f(x)=x^n$.

Second, the binomial formula yields $$x^n=\bigl(a+(x-a)\bigr)^n=\sum_{k=1}^n\binom nk a^{n-k}(x-a)^k$$ and observe that $$\binom nk a^{n-k}=\frac{n(n-1)\dots(n-k+1)a^{n-k}}{k!}=\frac{f^{(k)}(a)}{k!}.$$

• Simplest answer +1. The result is purely algebraic (in fact it is common to define derivative of polynomials over a ring in formal manner based on $(x^{n}) '=nx^{n-1}$) and this is the way to prove it using just algebraic manipulation. Oct 11, 2017 at 17:07

Use induction on $n\in\Bbb Z_{\ge 0}$.

If $n=0$ the statement is obvious.

If the statement is true for polynomials of degree $n$ or lower and $p$ is a polynomial of degree $n+1$ then $p'$ is a polynomial of degree $n$ and $$p'(x)=\sum_{k=0}^n \frac{p^{(k+1)}(a)(x-a)^k}{k!}$$ Therefore $$p(x)=C+\sum_{k=0}^n \frac{p^{(k+1)}(a)(x-a)^{k+1}}{(k+1)!}$$ for some constant $C$. By taking $x=a$, we see that $C=p(a)$. Thus, $$p(x)=\sum_{k=0}^{n+1}\frac{p^{(k)}(a)(x-a)^k}{k!}$$

Just observe that: $$D^k(x^n)\big|_{x=0}=\cases{ k! & if n=k,\\ 0 & if n\ne k.\\ }$$ It follows that if $$P(x)=a_nx^n+\ldots+a_1x+a_0$$ then $$P^{(k)}(0)=k!\,a_k, \quad\hbox{that is:}\quad a_k={P^{(k)}(0)\over k!}.$$ An analogous reasoning can be repeated for the case $P(x)=a_n(x-a)^n+\ldots+a_1(x-a)+a_0$, computing derivatives at $x=a$.

• You also have to show that every polynomial can be expanded in powers of $x-a$. That can be shown in several ways, e.g. by noting that $(x-a)^n = x^n + (\text{lower order terms})$ and applying induction. Oct 11, 2017 at 20:10
• What is $D^k(x^n)$, might I ask? Oct 11, 2017 at 22:27
• @ChaseRyanTaylor - it is another notation for $$\frac{d^k}{dx^k}x^n$$. Oct 11, 2017 at 22:55

Polynomials are already in this Taylor form.

If you want to start with a polynomial, then convert it into this Taylor form, you can do that simply by differentiating the polynomial up to $n$ times and then calculating the derivatives at $a$. If you want to prove that a polynomial is the Taylor form of a given function, then this question should help.

You might find the Taylor form more intuitive if you interpret it as a ‘standard form’ for calculus and approximations. It is easy to apply shifts and stretches to, and it can be truncated as desired to estimate a derivative at a point or to estimate the value of the original function at a point.

For me the intuitive understanding is that the information contained in a polynomial of degree $n$ -- which can be seen as its $n+1$ coefficients, for example--, is encoded in its $n+1$ derivatives at any point. Knowing the behaviour of the polynomial at any point in full detail, you can resconstruct it entirely.
Let $$f(x):=\sum_{k=0}^n c_k x^k$$ be a polynomial of degree $\leq n$, and let $a\in{\mathbb R}$ be an arbitrary point on the real axis. Then $$g(y):=f(a+y)=\sum_{k=0}^n c_k(a+y)^k=\sum_{l=0}^n c_l' y^l\qquad(y\in{\mathbb R})$$ for certain constants $c_l'\in{\mathbb R}$, because expanding the $(a+y)^k$ with $k\leq n$ does only create powers $y^l$ with $l\leq n$. It follows that $$f(x)=g(x-a)=\sum_{l=0}^n c_l'(x-a)^l\qquad(x\in{\mathbb R})\ .$$ For those who know higher derivatives it is easy to check that $$f^{(\ell)}(a)=l!\>c'_l\qquad(l\geq 0)\ .$$