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I am curious why the following is true. The text I am reading is "An Introduction to Numerical Analysis" by Atkinson, 2nd edition, page 133, line 4.

$p(x)$ is a polynomial of the form:

$ p(x) = b_0 + b_1 x + \cdots + b_n x^n$

If $p(x) = 0$ for all $x$, then $b_i = 0$ for $i=0,1,\ldots,n$.

Why is this true? For example, for $n=2$, I can first prove $b_0=0$, then set $x=2$ to get a linear system of two equations. Then I can prove $b_1=b_2 = 0$. Similarly, for $n=3$, I first prove $b_0=0$, then I calculate the rank of the resulting linear system of equations. That shows that $b_1=b_2=b_3=0$. But if $n$ is very large, I cannot keep manually solving systems of equations. Is there some other argument to show all the coefficients must be zero when the polynomial is always zero for all $x$?

Thanks.

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    As you are interested in numerical methods the following is probably not of interest to you, but for the sake of completeness I feel compelled to point out that the conclusion is false, if your domain is finite. A non-zero polynomial can easily vanish at all points of, e.g. $\mathbf{Z}_m$ without being the zero polynomial. The easiest example is the polynomial $p(x)=x^2-x\in \mathbf{Z}_2[x]$ that vanishes at all the elements of $\mathbf{Z}_2$. If you don't study algebra, you can safely ignore this comment for now.2011-08-28

5 Answers 5

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HINT $\ $ A nonzero polynomial over a field (or domain) has no more roots than its degree, as is easily proved by induction and the Factor Theorem. In fact if every natural was a root then the polynomial would be divisible by $\rm\:(x-1)\:(x-2)\:\cdots\:(x-n)\:$ for all $\rm\:n\in \mathbb N\:,\:$ which yields a contradiction for $\rm\:n\:$ greater than the degree of the polynomial.

Note that the proof of said statement depends crucially on the hypothesis that coefficient ring is an integral domain, i.e. a ring satisfying $\rm\:ab = 0\iff a=0\ \ or\ \ b=0\:.\:$ Over non-domains such as the integers modulo $\rm\:m\:$ not prime, polynomials can have more roots than their degree. In fact if this is true then one can use such roots to factor $\rm\:m\:,\:$ see here.

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    @jrand: The degree of the polynomial zero i$s$ not zero.2011-08-28
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Note that $p(x)=0$ implies derivatives of all orders are also $0$. Let $x=0$ into $p(x) = b_0 + b_1 x + \cdots + b_n x^n$ to obtain $b_0 = 0$.

Now differentiate both sides: p'(x) = b_1 + 2b_2 x + 3b_3 x^2 + \cdots + nb_n x^{n-1} . Let $x=0$ again, and we get that $b_1=0$.

If we continue to differentiate and substitution $x=0$ we will get that $b_k=0$ for $k=0,1,2,\cdots, n$. This idea can easily be made into a rigorous induction proof.

A nice corollary of this result is that two polynomials are equal if and only if they have the same degree and coefficients.

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If you are willing to accept $\displaystyle \lim_{x \rightarrow \infty}\frac{1}{x^{m}}=0$ and $\displaystyle \lim_{x \rightarrow \infty}x^m=\infty$ for all $m>0$, then you can argue as follows. Suppose you have a polynomial $f(x)=b_nx^n+ \ldots b_0$ with $b_n \neq 0$ and $n\geq 1$, then $\lim_{x\rightarrow \infty}f(x)=\lim_{x \rightarrow \infty} x^n(b_n+ \ldots \frac{a_0}{x^n})=\lim_{x \rightarrow \infty} x^nb_n=\infty.$

Since the constant function $0$ does not have this property, it cannot be equal to a polynomial with degree greater or equal to $1$.

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Let $n\in \mathbb{N}$, and $p(x)=b_0+b_1x+\ldots +b_nx^n$ a polynomial of degree at most $n$. If $p(x)=0$ for every $x$, then $b_i=0$ for $i=0,\, 1,\ldots\, n$.

Proof:

By induction on $n$:

If $\deg(p)\leq 1$, then $p(x)=b_0+b_1x$. Since $p(0)=p(1)=0$ you get $b_0=b_1=0$.

Suppose that for any $r(x)=c_0+c_1x+\ldots +c_nx^n$ of degree at most $n$, if $r(x)=0$ for every $x$, then $c_i=0$ for $i=0,\, 1,\ldots\, n$. Let $p(x)=b_0+b_1x+\ldots +b_{n+1}x^{n+1}.$ Suppose that $p\equiv 0$. Since $p(0)=0$, you get that $b_0=0$. From here, I have two possible arguments. The first, that was my original one, is as follows:

We have $p(x)=b_1x+b_2x^2+\ldots+b_{n+1}x^{n+1}=x(b_1+b_2x+\ldots+b_{n+1}x^{n}).$ Then for all $x$, $\begin{align*} 0&= p(x)\\ 0&= x(b_1+b_2x+\ldots+b_{n+1}x^{n}).\end{align*}$ Then for any $x\neq 0$, $q(x):=b_1+b_2x+\ldots+b_{n+1}x^{n}=0$ If $q\not\equiv 0$, then we have a polynomial of degree at most $n$ with infinitely many roots. This can't be, therefore $q\equiv 0$. Now, $\deg(q)\leq n$ and $q\equiv 0$, therefore by the induction hypothesis, $b_i=0$ for $i=1,\,\ldots ,\, n+1$.

The second argument was inspired by @Ragib Zaman's answer. Just differentiate both sides of $p(x)=0$, then $b_1+2b_2x+\ldots+(n+1)b_{n+1}x^n=0$ for all $x$. By the induction hypothesis, $kb_k=0$ for $k=1,\,\ldots,\,n+1$, and this implies that $b_k=0$ for $k=1,\,\ldots,\,n+1$

Since $b_0=0$, this proves that $b_i=0$ for $i=0,\, 1,\ldots,\, n+1$.

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    Why the -1 to this?2011-08-28
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A polynomial can be uniquely fitted by knowing its value at $d+1$ points, where $d$ is the degree of the polynomial. If it's 0 at all points, that's clearly more than $d+1$ points.

But we also know that a polynomial can be written as $(x-r_0)(x-r_1)...(x-r_k)$, where the $r_i$ are the roots of polynomial (possibly complex). Again, if there are an infinite number of roots...