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If $p (x)= a_0 + a_1x^2+....a_nx^n$ in $\Bbb R [x]$ and $a \in \Bbb R$, then $p (x)$ can be written as $b_0+b_1(x-a)+.....b_n(x-a)^n$ ,Where $b_i \in\Bbb R \forall i\in\{0,1,2\ldots n\}$.

Confusion: IF two polynomials are equal they both must have the same degree and the coefficients of the corresponding unknowns with equal powers are also equal. the polynomial $f(x) = x^3 + 2x^2 + 5$ is basically the structure $( )^3 + 2( )^2 + 5$ and not what we put into those parenthesis. Then how this problem is meaningful . I must be going horribly wrong some where. Please help.

* Why can't I type mathematical symbols? I am following the codes in the link : MathJax basic tutorial and quick reference

Sourav Chakraborty
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2 Answers2

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Your definition is correct in a sense. In algebra one often defines a polynomial to be just the structure you mentioned. (I.e. $x$ is an "indeterminate", not a real variable.)

But this question is not asking about polynomials defined this way. When the question writes "$p(x)$ can be written as $q(x-a)$", it's not referring to $x-a$ as an indeterminate, in the way that you're thinking. It's asking, in the sense of functions on $\mathbb{R}$, for you to show there are coefficients on $q$ such that the functions $p(x)$ and $q(x-a)$ are everywhere equal.

You should be able to do the base case. Suppose it's true for degree $n-1$. Then for degree $n$, we have $$a_nx^n + \dotsb + a_1x+a_0 \\[12pt] =a_n(x-a)^n + (\text{Stuff of degree $<n$})+a_{n-1}x^{n-1}+\dotsb + a_0.$$

Can you find which terms to use induction on?

To type math code, your syntax should look like

$p(x) = a_0 + a_1x + \cdots b + a_nx^n$

gives $p(x) = a_0 + a_1x + \dotsb + a_nx^n$.

Comment: I would prefer to prove this a different way: given $p(x) = a_nx^n + \dotsb + a_0$, let $\zeta + a=x$, and expand $p(\zeta + a)$. Then substitute $\zeta = x-a$.

Eric Auld
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  • No \ldots?${}$ – Tim Ratigan Dec 20 '13 at 14:05
  • Can you help me proving this using induction ? – Sourav Chakraborty Dec 20 '13 at 14:05
  • @amWhy why use \cdots b rather than \dotsb? – Eric Auld Dec 20 '13 at 14:09
  • If they format identically, then feel free to roll-back to your post, and I'm sorry if it turns out there's no difference. I just learned to use \ldots (lower dots)for typing a sequence of entries: x_1, x_2, \ldots, x_n, and \cdots ("center" dots) for typing an iteration of terms being operated on: x_1 + x_2 + \cdots + x_n. But sorry if it all comes out the same anyway! – amWhy Dec 20 '13 at 14:12
  • @amWhy I use \dotsc for the former. I learned that they stood for "dots comma" and "dots binary", with the latter being used in between binary operations. – Eric Auld Dec 20 '13 at 14:15
  • Interesting: I learn something everyday! It's nice that there is some flexibility with MathJax: e.g. \Bbb R and \mathbb R = $\mathbb R$. – amWhy Dec 20 '13 at 14:18
  • But b$_{n-1}$(x-a)$^{n-1}$ is also a stuff of degree < n – Sourav Chakraborty Dec 20 '13 at 14:30
  • Let the statement for n=n-1 is P(n-1)=$b_{n-1}(x-a)^{n-1} + .......b_0$ Now can we assume that P(n)= (n-1)P(n-1)? – Sourav Chakraborty Dec 20 '13 at 14:41
  • @SouravChakraborty I edited my answer a bit. Try it now. – Eric Auld Dec 20 '13 at 14:53
  • Nah, My brain is chocking, don't know why Something is bothering me at the back of my mind.So unable to follow the logical arguments.Stupid of me. – Sourav Chakraborty Dec 20 '13 at 15:16
  • @SouravChakraborty The idea is that you can take care of the stuff of degree $<n$ (i.e. everything but the first term) by the induction assumption, and I showed you how to take care of the degree $n$ term. – Eric Auld Dec 20 '13 at 15:20
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Two polynomials are called equal, if they yield the same value for any x.

Peter
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