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


*

*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 $j<k$, $|a_k x^k| = |a_k x| x^{k-1} > k|a_j| x^j = k|a_j x^j|$.


*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.
A: 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 $x_0\ne0.$
A: 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.
A: 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. 
A: 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 
$$p\left(\frac 12\delta\right)
   = \left(\frac 12\delta\right)^k 
     \left(a_k + R \left(\frac 12\delta\right) \right)
   \ne 0
$$
A: 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.
A: I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.
In this book, the author proved the following proposition on p.120:

4.7:
Suppose $a_0,\dots,a_m\in\mathbb{F}$. If $$a_0+a_1z+\cdots+a_mz^m=0$$ for every $z\in\mathbb{F}$, then $a_0=\cdots=a_m=0.$

In the above proposition, $\mathbb{F}=\mathbb{C}$ or $\mathbb{F}=\mathbb{R}$.
I felt the author's proof difficult.
So, I tried to prove this proposition by myself.

Let $f(x)=a_0+a_1z+\cdots+a_mz^m$.
Since $f$ is the zero function, the $i$th derivative $f^{(i)}$ is also the zero function for $i=1,2,\dots$.
Since $f(0)=0$ and $f(0)=a_0$, $a_0=0$.
Since $f'(0)=0$ and $f'(0)=1\cdot a_1$, $a_1=0$.
Since $f^{(2)}(0)=0$ and $f^{(2)}(0)=2!\cdot a_2=0$, $a_2=0$.
$\dots$
Since $f^{(m)}(0)=0$ and $f^{(m)}(0)=m!\cdot a_m$, $a_m=0$.

