20
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ Do you know Lagrange interpolation? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Jan 19 '18 at 23:33
  • $\begingroup$ 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... $\endgroup$ – user491874 Jan 19 '18 at 23:34
  • $\begingroup$ 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. $\endgroup$ – Ted Shifrin Jan 19 '18 at 23:37
  • $\begingroup$ @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. $\endgroup$ – Dylan Jan 19 '18 at 23:37
  • 7
    $\begingroup$ 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. $\endgroup$ – orole Jan 19 '18 at 23:48

10 Answers 10

15
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – sarahzrf Jan 20 '18 at 2:28
  • $\begingroup$ Alright, fine. I modified my answer. $\endgroup$ – user137731 Jan 20 '18 at 3:35
10
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ This looks to me like the best answer, respecting the fact that no calculus should be involved. $\endgroup$ – Tom-Tom Jan 26 '18 at 11:21
9
$\begingroup$

.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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ OP actually says "before basically any calculus concepts are covered", but I think limits like this is fine. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Jan 20 '18 at 0:03
  • $\begingroup$ If only the Euclidean algorithm is to be used, then orole's answer above is also very good. $\endgroup$ – Teresa Lisbon Jan 20 '18 at 0:10
  • $\begingroup$ 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 $x_1 + x_2 + \dots + x_n \lt 1$. $\endgroup$ – CopyPasteIt Jan 21 '18 at 4:12
  • $\begingroup$ 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. $\endgroup$ – Teresa Lisbon Jan 21 '18 at 4:17
8
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 7
    $\begingroup$ 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. $\endgroup$ – orole Jan 19 '18 at 23:59
  • 1
    $\begingroup$ 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. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Jan 20 '18 at 0:01
  • $\begingroup$ @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. $\endgroup$ – Mohammad Riazi-Kermani Jan 20 '18 at 0:08
  • $\begingroup$ However, I do appreciate the beauty of algebraic proofs. They are far more elegant than analytic ones using inequalities. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Jan 20 '18 at 0:14
  • $\begingroup$ @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. $\endgroup$ – Ovi Jan 24 '18 at 4:47
6
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

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 $$\frac{1}{|x|-1} < \frac{2}{|x|} \iff |x| < 2|x|-2 \iff |x| > 1.$$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Full credit for the answer. $\endgroup$ – Teresa Lisbon Jan 20 '18 at 13:24
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
3
$\begingroup$

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 $x_0\ne0.$

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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

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

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.