# Zero function implies zero polynomial.

I'm trying to help someone with a problem in Apostol's book (Chapter 1 BTW, so before basically any calculus concepts are covered) at the moment and I'm stumped on a question.

I'm trying to prove that if $p$ is a polynomial of degree $n$, that is where $$p(x) = a_0 + a_1x + \cdots + a_nx^n$$ for some real numbers $a_0, \dots, a_n$, and if $p(x) = 0$ for all $x\in \Bbb R$, then $a_k = 0$ for all $k$.

Looking through the site, I find this question, but the solution given uses the derivative. But this before the definition of the derivative in Apostol's book, so I can't use that to prove this. I also know that we can use linear algebra to solve this, but pretend I don't understand the concept of linear independence either as Apostol's book doesn't presuppose that. Then what can we do to prove this? It feels like there should be a proof by induction possible, but I'm not seeing how to do the induction step.

My Attempt: Proving that $a_0 = 0$ is trivial by evaluating $p(0)$. But then I'm left with $$p(x) = x(a_1 + \cdots +a_nx^{n-1})$$ Here I see that for all $x\ne 0$, $a_1 + \cdots a_nx^{n-1}=0$. But because of that $x\ne 0$ part, that means I can't use the same trick to show that $a_1 = 0$.

Any ideas?

• Do you know Lagrange interpolation? – GNUSupporter 8964民主女神 地下教會 Jan 19 '18 at 23:33
• Are you allowed to use the polynomial division and the uniqueness of decomposition into a product of irreducible polynomials? Because, it can be proven that, if $p(a)=0$ then $x-a\mid p$ - so we would end up with $x-a\mid p$ for every $a\in\mathbb R$ which is impossible as $x-a$ and $x-b$ are coprime for $a\ne b$... Or something similar... – user491874 Jan 19 '18 at 23:34
• You can prove (using the root-factor theorem, for example) that a real polynomial of degree $n\ge 0$ has at most $n$ roots. But your argument works fine — A polynomial is continuous, so if it's $0$ for all $x\ne 0$, it's $0$ at $x=0$ as well. – Ted Shifrin Jan 19 '18 at 23:37
• @GNUSupporter I'm not actually familiar with that particular numerical analysis technique, but that seems like it would be over the head of a beginning calculus student. – Dylan Jan 19 '18 at 23:37
• Just using Euclid's algorithm: You already got $a_0=0$. Now, write the polynomial $a_1+a_2x+...+a_nx^{n-1}$, instead of powers of $x$, in powers of $x-1$. You get $b_1+b_2(x-1)+...+b_n(x-1)^{n-1}$. Evaluate at $x=1$. This time the factor $x$ that you had extracted, doesn't vanish, but the polynomial does. Therefore $b_1=0$. Extract the factor $(x-1)$. Continue expanding the remaining factor at other points and evaluating at them. In particular you see that we only needed to know that the polynomial was zero at $n+1$ different points. – orole Jan 19 '18 at 23:48

The polynomial $a_1 + a_2x + \cdots + a_nx^{n-1}$ that you found has root $x=1$ and so, by the factor theorem, has a factor of $x-1$. Thus $$p(x) = x(x-1)(b_0 + b_1x + \cdots + b_{n-2}x^{n-2})$$ for some $b_0, \dots , b_{n-2}$. But because $$a_1 + \cdots + a_nx^{n-1} = (x-1)(b_0 + b_1x + \cdots + b_{n-2}x^{n-2}) = 0$$ for all $x\ne 0$, and $x-1=0$ only for $x=1$, this shows us that $b_0 + \cdots + b_{n-2}x^{n-2}$ is zero for all $x\ne 0, 1$. In particular, it is zero at $x=2$. Hence $$p(x) = x(x-1)(x-2)(c_0 + \cdots + c_{n-3}x^{n-3})$$ Continuing on in this way you'll find that $$p(x) = x(x-1)(x-2)\cdots (x-n)(d_0)$$ for some constant $d_0$. Now we know that $p(n+1) = 0$ and we also know that none of the linear factors $x, x-1, \dots, x-n$ is zero at $x=n+1$. Hence $d_0 = 0$.

But what does this have to do with the numbers $a_1, \dots, a_n$? Well we can recover the form $a_0 + a_1x + \cdots a_nx^n$ by multiplying through all of the factors of $p$. But it's clear that the highest power of $p$ will be $d_0x^n$. So $d_0 = a_n = 0$. Hence $p$ is in fact the polynomial $a_0 + a_1x + \cdots + a_{n-1}x^{n-1}$.

Applying the exact same argument to this new representation of $p$ shows that $a_{n-1}=0$, and then $a_{n-2}=0$, etc.

Of course I haven't really given the above as a polished proof, but I'm sure you can handle that.

• The question is specifically asking how to show that each coefficient has to equal zero, though. It's asking about polynomials, not the functions they give rise to. – sarahzrf Jan 20 '18 at 2:28
• Alright, fine. I modified my answer. – user137731 Jan 20 '18 at 3:35

Assume that $p$ has degree $n$. If $p(x)=0$ for all $x$, then $$p(1)=0,\ p(2)=0,\ \ldots\ , p(n+1)=0$$ is a linear system of $n+1$ equations on the $n+1$ coefficients $a_0,\ldots,a_n$. This is a Vandermonde matrix, and so its determinant is non-zero. Thus the only possible solution to the system is $$a_0=a_1=\cdots=a_n=0.$$ Of course, this argument shows that already if $p$ has $n+1$ roots, then it has to be zero.

• This looks to me like the best answer, respecting the fact that no calculus should be involved. – Tom-Tom Jan 26 '18 at 11:21

.Suppose you do not want to use the derivative, but are a beginning calculus student as mentioned above. Then you can use the following result:

Lemma : $$\lim_{k \to \infty} \frac{p(k)}{k^n} = a_n$$ (that is, it exists and equals $$a_n$$)

To prove this proposition, expand the polynomial : $$\frac{p(k)}{k^n} = \sum_{i=0}^n a_ik^{i-n}$$. Since $$n \to \infty$$, all the terms in the above expansion go to zero as $$k \to \infty$$, except when $$i=n$$, in which case the limit is just $$a_n$$, since $$k^0 = 1$$.

We now want to prove that if $$p(x) = \sum_{i=0}^n a_ix^i$$ is zero everywhere, then all the coefficients are zero.

Let us perform induction, on the maximum power that of $$x$$ that occurs in the expansion of $$p$$ as a polynomial (This is not the same as the degree). If $$p$$ is (expressed as) a degree $$0$$ polynomial $$a_0$$, then $$a_0$$ is a constant, hence must be zero.

Let $$p(k) = \sum_{i=0}^n a_ix^i$$. By the above argument, $$a_n = \lim_{k \to \infty} \frac{p(k)}{x^k} = 0$$, since $$p$$ is zero everywhere. Now $$p(x)$$ simplifies to $$\sum_{i=0}^{n-1} a_ix^i$$, and by induction, all the $$a_i$$ are zero.

Hence, the proposition follows.

• OP actually says "before basically any calculus concepts are covered", but I think limits like this is fine. – GNUSupporter 8964民主女神 地下教會 Jan 20 '18 at 0:03
• If only the Euclidean algorithm is to be used, then orole's answer above is also very good. – Teresa Lisbon Jan 20 '18 at 0:10
• Of course you don't need limits: $\; \text{|---|}\;$ For any number $a$ and integer $n \gt 0$ there exists an integer $m \gt 0$ with $\frac{a}{m} \lt \frac{1}{n}$. $\; \text{|---|}\;$ Let $\{x_1,x_2,\dots,x_n\}$ satisfy $|x_k| \lt \frac{1}{n}$ for $1 \le k \le n$. Then $x_1 + x_2 + \dots + x_n \lt 1$. – CopyPasteIt Jan 21 '18 at 4:12
• Yes. As is mentioned in another answer below, I could have skipped limits and done something like you did. But I wanted to give a flavor of calculus to the questioner, who is taking baby steps in the subject. Hence the use of such language. But you are correct. – Teresa Lisbon Jan 21 '18 at 4:17

Note that according to the Fundamental Theorem of Algebra a polynomial of degree $n$ has exactly $n$ roots.

Now your function has infinitely many zeros, therefore it can not be a polynomial of degree $n$ for any $n$.

Thus all the coefficients are zero which makes your function to be identically zero.

• The fundamental theorem of algebra says that there is at least one root for $n>0$, over the complex (!). That there cannot be more than $n$ is a far far simpler result. – orole Jan 19 '18 at 23:59
• OP needs the proof to help someone reading Apostol's book chapter one "before basically any calculus concepts are covered". I believe that one learns the proof of the Fundamental Theorem of Algebra is a complex variables course. Anyways, since this Algebra Theorem is not quite "algebraic", no pure algebraic proof can exist. The one using a 2-Sylow group also makes use of a theorem in elementary analysis (Intermediate Value Theorem in my memory). To sum up, that's beyond the level of beginning calculus. – GNUSupporter 8964民主女神 地下教會 Jan 20 '18 at 0:01
• @GNUSupporter I agree with you that we can not solve this problem with just basic algebra. On the other hand Apostol's book does not follow a specific order in presenting concepts. – Mohammad Riazi-Kermani Jan 20 '18 at 0:08
• However, I do appreciate the beauty of algebraic proofs. They are far more elegant than analytic ones using inequalities. – GNUSupporter 8964民主女神 地下教會 Jan 20 '18 at 0:14
• @GNUSupporter In my experience, in most calculus classes and below, many theorems are presented without proof, and the students are expected to use them; for example in my precalculus class we learned the statement of the fundamental theorem of algebra, obviously without even a smell of the proof. I don't know anything about Apostol's Calculus book, but I think there might be a good chance that the author is expecting the readers to use such "precalculus" facts. – Ovi Jan 24 '18 at 4:47

Given a polynomial $P$ define $(\Delta P)(x)=P(x+1)-P(x)$. Given a polynomial $P$ where $P(x)= a_0 + a_1x + \cdots + a_nx^n$, we claim that $$\Delta^{n}(P)(x)=n!a_{n}\tag{0}\label0.$$ To see this write $$P(x) = \sum_{k=0}^{n} a_k x^k= \sum_{k=0}^{n} a_k \sum_{m=0}^{k}S(k,m)x^{\underline{m}}\tag{1}\label1$$ where $S(k,m)$ is the stirling number of the second kind. and $x^{\underline{m}}$ is the falling factorial. Since $\Delta(x^{\underline m})=mx^{\underline{m-1}}$, $\Delta^{n}(x^{\underline {m}})=m^{\underline n}x^{\underline{m-n}}$. If $m<n$, then $\Delta^{n}(x^{\underline {m}})=0$. Hence taking $\Delta ^{n}$ of both sides of (1) yields that $$(\Delta^nP)(x)=n!S(n,n)a_{n}=n!a_{n}\tag{2}\label2.$$ Now onto the problem.

If $p$ is a polynomial of degree $n$, that is where $$p(x) = a_0 + a_1x + \cdots + a_nx^n$$ for some real numbers $a_0, \dots, a_n$, and if $p(x) = 0$ for all $x\in \Bbb R$, then $a_k = 0$ for all $k$

We prove the claim by induction on $n$. The base case where $n=0$ yields that $a_0=0$ as desired. Suppose that the claim holds for all polynomials with degree less than $n$. Let $p$ be a polynomial where $$0=p(x)= a_0 + a_1x + \cdots + a_nx^n\tag{3}\label3$$ for all $x\in\mathbb{R}$. Take $\Delta^{n}$ of both sides of equation \eqref{3} and use equation \eqref{0}, to get that $a_{n}=0$. The induction hypothesis now implies the result.

My solution share the same spirit with another one using limits, though I think that the discrete, algebraic ones are less restrictive and far more elegant than the continuous, analytic ones. Nonetheless, this gives learners a taste of elementary analysis.

The basic idea of my proof is that when $|x|$ is sufficiently large, the highest degree term will dominate over the remaining one. Given a basic entrance level, the bounds are going to be violent.

For any degree $n$ polynomial $p(x) = a_0 + a_1x + \cdots + a_nx^n$, we assume $a_n \ne 0$. Define $A = \max\{a_0,\dots,a_{n-1}\}$. Bound the non-dominating terms.

$$|a_0 + a_1x + \cdots + a_{n-1}x^{n-1}| \le nA \frac{|x|^n - 1}{|x| - 1}$$

\begin{align} |p(x)| &\ge |a_n| |x|^n - nA \frac{|x|^n - 1}{|x| - 1} \\ &= |a_n||x|^n - nA \frac{|x|^n}{|x| - 1} + nA \frac{1}{|x|-1} \\ &> |a_n||x|^n - nA \cdot \frac12 |x|^{n-1} + 0 \\ &= |x|^{n-1} (|a_n||x| - nA/2) \end{align}

To ensure that the lower bound is positive, take $|x| > \dfrac{nA}{2|a_n|}$. As a result, $p(x)$ explodes as $x$ get arbitrarily large. This clearly contradicts the assumption that $p$ is zero on $\Bbb{R}$. Q.E.D.

Remark: $\dfrac{1}{|x|-1}$ is more difficult to manipulate than $\dfrac{1}{|x|}$ To throw away $-1$ in the denominator, choose $|x|$ sufficiently large, so that the difference between $\dfrac{1}{|x|-1}$ and $\dfrac{1}{|x|}$ is arbitrarily small. (Denoted by $\dfrac{1}{|x|-1} \sim \dfrac{1}{|x|}$) Logically, we set a violent upper bound $$\frac{1}{|x|-1} < \frac{2}{|x|} \iff |x| < 2|x|-2 \iff |x| > 1.$$

• Full credit for the answer. – Teresa Lisbon Jan 20 '18 at 13:24

Here's a different approach. Suppose there are some non-zero coefficients, and that $k$ is the largest integer such that $a_k\neq 0$. Also, let $M$ be the maximum of $|a_0|, |a_1|,\dots,|a_k|$, i.e. the maximum absolute value of the coefficients.

All you have to do is note that for $x > k\times \frac{M}{|a_k|}$ the magnitude of the last term in the polynomial, $a_k x^k$, is more than $k$ times bigger than any other term in the polynomial$^1$. Hence, the sum of all other terms besides $a_k x^k$ (there are $k$ of them) can't be big enough to cancel $a_k x^k$. So the polynomial can't be $0$ for such big values of $x$.

Footnotes:

1. Here's a detailed proof:

• $|a_k x|>kM\geq k |a_j|$ for any $j$.
• $\frac{M}{|a_k|} \geq 1$, so $x > 1$. Thus $x^{a}\geq x^b$ for $a\geq b$.
• Hence for any $j<k$, $|a_k x^k| = |a_k x| x^{k-1} > k|a_j| x^j = k|a_j x^j|$.
2. As a brief editorial, I would contend that this approach is philosophically more sound than using the Fundamental Theorem of Algebra, as the latter relies on calculus for its proof.

Lemma

If $$p(x)=a_0 + a_1x + \cdots + a_n{x}^n$$ and $$p(x)=0, \;\forall x\ne 0$$ then $$a_0=0.$$

So $p(x)=0$ for all $x.$ Therefore $$p(x)=x(a_1+a_2x+\cdots a_nx^{n-1})$$ and for $x\ne0$ we have $$a_1+a_2x+\cdots a_nx^{n-1}=0$$ Again we can apply the lemma and conclude that $a_1$ is $0$. We can repeat this process again and again and so show that all $a_k$ are $0.$

We proof the lemma by showing that the absolute value of $$a_1x+\cdots+a_nx^n$$ can be made arbitrary small if we choose an $x=x_0$ that is sufficient near to $0$. So $$a_0+(a_1x+\cdots+a_nx^n)$$ is very near to $a_0$ and therefor not $0$ if $a_0\ne0.$

I assume you can use that $$|x y|=|x||y|$$ the triangle inequality

$$|x+y|\le|x|+|y|$$

and the inequality $$|x-y|\le \;|x|-|y|\;|$$

which follows from the triangle inequality.

Proof of the Lemma

$$|p(x_0)| = |a_0 + a_1x_0 + \cdots + a_n{x_0}^n| \ge \left| \; |a_0| - |a_1x_0 + \cdots + a_n{x_0}^n| \;\right | \tag{1}$$

We choose $$\alpha=\max\{|a_1|, \sqrt{|a_2|},\ldots, \sqrt[n]{|a_n|}\}$$

and $$x_0=\frac{|a_0|}{2\alpha}$$

So we get $$\frac{a_k}{\alpha^k}\le\frac{a_k}{(\sqrt[k]{a_k})^k}=1$$ and we have $$|a_1x_0 + \cdots + a_n{x_0}^n| \\ \le |a_1|\cdot|x_0| + \cdots + |a_n|\cdot|{x_0}|^n \\ \le|a_1|\frac{|a_0|}{2\alpha}+|a_2|\left(\frac{\sqrt{|a_0|}}{2\alpha}\right)^2+\cdots+|a_n|\left(\frac{\sqrt[n]{|a_0|}}{2\alpha}\right)^n\\ \le \frac{|a_0|}{2}\frac{|a_1|}{\alpha}+\frac{|a_0|}{2^2}\frac{|a_2|}{\alpha^2}+\cdots+\frac{|a_0|}{2^n}\frac{|a_n|}{\alpha^n}\\ \le |a_0|\left(\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^n}\right)\\ = |a_0|\left(1-\frac{1}{2^{n+1}}\right)\\<|a_0|$$ if $a_0\ne 0.$

So from this an $(1)$ follows $p(x_0)\ne0.$ This is a contradiction because $x_0\ne0.$

Proof of the contrapositive.

Suppose
$$p(x) = a_k x^k + a_{k+1} x^{k+1} + a_{k+2} x^{k+2} + \cdots + a_nx^n$$ where $a_k \ne 0$. Then

$$p(x) = x^k(a_k + a_{k+1} x + a_{k+2} x^2 + \cdots + a_nx^{n-k}) = x^k(a_k + R(x))$$

where $R(x) = a_{k+1} x + a_{k+2} x^2 + \cdots + a_nx^{n-k}$.

Since $R(x)$ is a continuous function, then $\displaystyle \lim_{x \to 0} R(x) = 0$. So there exists $\delta > 0$ such that $|x| < \delta$ implies $|R(x)| < \frac 12|a_{k}|$, which implies $a_k+R(x) \ne 0$. Hence

$$p\left(\frac 12\delta\right) = \left(\frac 12\delta\right)^k \left(a_k + R \left(\frac 12\delta\right) \right) \ne 0$$

Let $p_d(x)$ denote a polynomial of degree $d$. Then as induction basis, we show that $p_0(x) = 0$ requires $a_0 = 0$:

$$p_0(x) = a_0 = 0$$

As an induction-step, we assume $p_d(x) = 0$ iff $\forall i \leq d: a_i = 0$. Then

\begin{align} p_{d+1}(x) &= a_{d + 1} x^{d+1} + p_d(x) = 0 \\ a_{d+1} x^{d+1} &= - p_d(x) \end{align}

This cannot hold, unless $\forall i \leq d + 1: a_i = 0$.

Proof: Assume $a_{d+1} \neq 0$, then there must exist $x_0$, such that $\forall x' \geq x_0: |a_{d+1} x^{d+1}| > |p_d(x)|$. This can be shown using the fact that a polynomial is always asymptotically bound from above by it's largest exponent:

\begin{align} &|a_{d} x^{d} + a_{d - 1} x^{d - 1} + ... a_0| \leq |x^d (a_d + a_{d + 1} + ... + a_0)| \quad for \,\, x \geq 1\\ \\ &x > max\left( \left| \sum_{i=0}^d a_i \right|, 1\right) / a_{d + 1}\Rightarrow \left| a_{d+1} x^{d+1} \right| \gt |p_d(x)|\\ \end{align}

Now this leaves us with only one option: $a_{d+1} = 0$, from which we can conclude $p_d(x) = 0$ and using the induction hypothesis $\forall i \leq d + 1: a_i = 0$.

Thus $p_d(x) = \sum_{i=0}^d a_i x^i = 0$ implies that $\forall i \leq d: a_i = 0$

Note:
This is my first post here. Please feel free to edit or comment if you find any mistakes or ways to improve this post.