# Can every polynomial be written as $a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 +\ldots + a_n (x-x_0)^n$

Can every polynomial of degree $n$ be written as $a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 + \ldots + a_n (x-x_0)^n$ while $x_0$ is an arbitrary but given real number and all $a_k$ can be freely chosen? Is there a proof for this representation without using the concept of taylor series?

Explanation to my question: I'm reading about the taylor formula, where one author made the assumption that the polynomial of degree $n$ which approximate a function at a given point $x_0$ shall have the form $a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 +\ldots + a_n (x-x_0)^n$. With the further assumption that the polynomial and the original function shall have the same first $n$ derivatives at $x_0$ the taylor formula is concluded. I can understand the second assumption, also that the approximative function shall be a polynomial. But I ask myself why every polynomial can be represented via $a_0 + a_1 (x-x_0) + a_2 (x-x_0)^2 +\ldots + a_n (x-x_0)^n$. Because I want to proof the concept of taylor expansion, I of course cannot use this concept to proof the answer of my question.

• Hint: let $y = x - x_0$ and use the binomial theorem on $x^n = (y + x_0)^n$. Commented Jun 4, 2012 at 22:33
• Do you know how to prove this when $a_0 = 0$? Do you know how to show that if $p(x)$ is a polynomial then so is $p(x + x_0)$? Commented Jun 4, 2012 at 22:33
• Prove this by induction on the number of terms $n$ using summation. You want to represent this polynomial as a sum: $P(x) = \sum _{k=0} ^{n} c_k x^k$
– A.S
Commented Jun 4, 2012 at 22:36
• $x^k=((x-x_0)+x_0)^k$. Expand. Commented Jun 4, 2012 at 22:40
• of course, there is a more elegant proof if you know about linear algebra, but that seems unlikely... Commented Jun 4, 2012 at 22:42

Here is an easy proof by induction.

Let $g(x)=f(x)-f(x_0)$. Then $g(x_0)=0$ and so $g(x)=(x-x_0)h(x)$. By induction, you get an expansion for $h(x)$ around $x=x_0$ and this gives one for $g(x)$ and so one for $f(x)$.

Hint $\$ Either make a change variables $\rm\:t = x-c\:$ or employ induction. For the latter approach, $\rm\: f = a_0 + (x-c)\, g\:$ by the Factor Theorem or Division algorithm. By induction on degree $\rm\: g = a_1 + a_2\,(x-c) +\cdots + a_n (x-c)^{n}.\:$ Plugging that in for $\rm\:g\:$ above gives the result.

If you don't expand the intermediate results you obtain the nested Horner form that results from iteratively applying the Division Algorithm, which reveals the recursive structure, namely

$$\rm\: a_0 + (x\!-\!c) (a_1 + (x\!-\!c) (a_2 + (x\!-\!c)(\cdots (a_{n-1} + (x\!-\!c)\,a_n )\cdots )))$$

One can replace $\rm\:x-c\:$ by any monic polynomial $\rm\:f(x)\:$ over any ring. Then the "coefficients" (remainders) will be polynomials of degree $\rm\le deg\ f.\:$ This is useful e.g. for partial fraction expansions of rational functions, e.g. for integration.

Suppose we have a polynomial

$$p(x)=a_0+a_1x+a_2x^2+\cdots+a_n x^n$$

We know that we can express $p$ in terms of a polynomial of degree $n-1$ times $x-x_0$ and a remainder $r$. Thus

$$p(x)=r_0+(x-x_0)p_1(x)$$ get

By the same idea we can express $p_1(x)$ in terms of a remainder $r_1$ and a polynomial of degree $n-2$ times $x-x_0$. Thus we get:

$$p(x)=r_0+r_1(x-x_0)+(x-x_0)^2p_2(x)$$

You can carry out this algorithm to get the desired expansion. With this proven, you can actually show what each coefficient is:

We want

$$p(x)=b_0+b_1(x-x_0)+b_2(x-x_0)^2+\cdots+b_n(x-x_0)^n$$

Note that $p(x_0)=b_0$ since all other terms cancel out. We differentiate now since we have retrieved the constant, to get

$$p'(x)=b_1+2b_2(x-x_0)+\cdots+nb_n(x-x_0)^n-1$$

We now plug in $x=x_0$, and we see that all terms cancel except one and get

$$p'(x_0)=2b_2$$

You can carry this out and get that

$$p^{(n)}(x_0)=n!b_n$$ or, more nicely:

$$\frac{p^{(n)}(x_0)}{n!}=b_n$$

so we can always express $p$ as

$$p(x) = p(x_0)+p'(x_0)(x-x_0)+\frac{p''(x_0)}{2!}(x-x_0)^2+\cdots+\frac{p^{(n)}(x_0)}{n!}(x-x_0)^n$$

Thx @Robert Israel and the overs. I have found the solution:

\begin{align} b_0 + b_1 x + b_2 x^2 + \ldots + b_n x^n &= b_0 + b_1 ((x-x_0)+x_0) + \ldots + b_n ((x-x_0)+x_0)^n \\ &= \ldots = a_0 + a_1 (x-x_0) + \ldots + a_n (x-x_0)^n \end{align}

Here's a concrete point of view.

Say you've got $$f(x) = a_n x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2}+ \cdots.$$

Then $$f(x) = \Big(a_n (x-x_0)^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2}+ \cdots\Big)+(\text{something that compensates}).$$ That last term must of course be $$a_n x^n - a_n(x-x_0)^n.$$ Expand that last term via the binomial theorem: $$a_n x^n - a_n\left(x^n -\binom{n}{n-1}x^{n-1} x_0 + \binom{n}{n-2} x^{n-2} x_0^2 - \binom{n}{n-3} x^{n-3} x_0^3 + \cdots\right).$$ Notice that $a_nx^n$ at the beginning cancels the first term inside the parentheses, and you'e left with $$-a_n\left(-\binom{n}{n-1}x^{n-1} x_0 + \text{lower-degree terms}\right).$$ So you've written $$f(x) = a_n(x-x_0)^n + \text{lower-degree terms}.$$ That largest-degree term still has the same coefficient that it had before, namely $a_n$. But $x^{n-1}$ has a different coefficient, $a_{n-1}+ \binom{n}{n-1}x_0$, and similarly with all of the lower-degree terms.

Next, do with the $(n-1)$th-degree term what you just did with the $n$th-degree term, and so on.