# Apostol Calculus Vol.1 Exercise 9 , Chapter 1.5 (Prove property of polynomial function)

Ok, so I have a huge problem with this exercise. It is a property of polynomial functions that needs proving. Thing is, I can not even get a clue and I put in some numbers and it does not seem to hold. So please enlighten me

Here goes the problem: $$f(x) = \sum_{k=0}^n c_k x^k$$ $$f(0) = 0$$ $$\text{For }n \ge 1, \text{ where }g(x)\text{ is a polynomial of degree }n-1$$ $$\text{Show that }f(x)=xg(x).$$

I have assumed that sentence "$g(x)$ is polynomial of degree $n-1$ means that $g(x) = \sum_{k=0}^{n-1}c_kx^k$ but when I put in some numbers the equality $f(x)=g(x)$ does not seem to hold.

Please note that I prefer hints over full answers, because that way I get to understand it lot better, but full answers are also acceptable.

Thanks in forward.

Ok so proof goes like this from fact that f(x)=0 we can prove that $c_0$ is zero.Here is how we do that: Since $f(x)=\sum_{k=0}^{n} c_kx^n$ it implies that for some $n \ge 1$ it holds $f(0) = c_0x^0 + c_1x^1 + c_2x^2+...+c_nx^n$ Since we have $x^0$ and $x=0$ then we can conclude that $x^0$ = 1 thus we thus the first term in expansion becomes $c_0 \times 1$ .For now we know that all terms except first term in expansion are zero because 0 to any power is zero,and $c_k \times 0 =0$ always.

Now we are left with $f(0) = c_0 \times 1 = 0$ from which we easily conclude that that $c_0 = 0$

Next since g(x) is polynomial of degree n-1 we can write it in following manner

$g(x) = \sum_{k=0}^{n} c_kx^{k-1} = c_0x^{-1} + c_1 + c_2x + ... + c_nx^{n-1}$

Now we can come to real business which is:

$f(x) = xg(x)$

$\sum_{k=0}^{n} c_kx^k = x\sum_{k=0}^{n} c_kx^{k-1}$

$c_0x^0 + c_1x + c_2x^2 + ... + c_nx^n = x(c_0x^{-1} + c_1x^0 + c^2x^1 + ... + c_nx^{n-1} )$

since we have concluded that c_0 = 0 then we can write:

$c_1x + c_2x^2 + ... + c_nx^n = x(c_1x^0 + c_2x^1 + ... + c_nx^{n-1} )$

$c_1x + c_2x^2 + ... + c_nx^n = c_1x^1 + c_2x^2 + ... + c_nx^n$

and thus it is proven.

• This proof is in the spirit of the correct proof, but is very muddled and not clear. However, you have all the correct ideas. Since you have finally gotten it, I shall amend my answer to have a full proof. Feb 2, 2014 at 20:34
• Thank you for your constructive criticism.I hope you know that it would not be possible without you and DonAntonio Feb 2, 2014 at 20:36
• Not a problem. Just an observation, in several of your other questions people have given fine answers to your questions. Perhaps you should accept one of them as an answer. You can read more here: meta.stackexchange.com/questions/5234/… or here: meta.stackoverflow.com/help/someone-answers Good luck with the rest of Spivak! It is a good book. Feb 2, 2014 at 21:13
• Oh I did not know I could do that,definitely will.Thanks for heads up Feb 3, 2014 at 16:21

I assume Apostol means that if $f(x)$ is as given and $f(0)=0$, he wants you to show that $f(x)=xg(x)$ for some polynomial $g(x)$ of degree $n-1$. My hint is as follows:

Write down a few examples of $f(x)$'s where $f(0)=0$. Can you factor something out in each of these examples to obtain a polynomial of lower degree?

Once you see that, since $f(x)=\sum_{k=0}^nc_k x^k$, how should you define $g(x)$ using the $c_k$'s? Again, go back to your examples! Then look at $f(x)=xg(x)$ to see that they are the same.

EDIT. Since you have finally obtain the answer, I shall write my proof of this here.

Let $f(x)=\sum_{k=0}^n c_k x^k$ be a polynomial of degree $n\geq 1$ such that $f(0)=0$ (as given in the problem). We know that $$f(0)=c_n\cdot 0^n+c_{n-1}\cdot 0^{n-1}+\cdots+c_1 \cdot 0+c_0=c_0$$ But since $f(0)=0$, we conclude that $c_0=0$. So we have $$f(x)=\sum_{k=0}^n c_k x^k=\sum_{k=1}^n c_k x^k$$

Now I shall create a function $g(x)$. Given a function $f(x)$ as above, define $$g(x)=\sum_{k=1}^{n} c_{k}x^{k-1}$$ where the $c_k$ are the same as those given by the function $f(x)$. First, notice that the degree of $g(x)$ is $n-1$. Finally, notice that $$xg(x)=x \sum_{k=1}^n c_kx^{k-1}=\sum_{k=1}^n c_k x^k=f(x)$$ Since $f(x)$ was arbitrary polynomial of degree $n\geq 1$ such that $f(0)=0$, we know that there must be a polynomial of degree $n-1$, $g(x)$, such that $f(x)=xg(x)$ (i.e., the one we constructed in the proof).

• I have a very hard time understanding this.If you could give me a full answer that would be a blessing.Thanks in advance Feb 2, 2014 at 19:03
• I won't do the problem for you. However, I can get you started with possible examples. Look at $f(x)=2x^2+3x-x$ and $f(x)=x^3-2x^2+x$. Notice $f(0)=0$ and what do all the terms have in common? Does this 'get you' a polynomial of lower degree? Feb 2, 2014 at 19:05
• yes this does yield a polynomial of lower degree but if I write it more "general" then $\sum_{k=0}^{n} c_kx^k=c_0 + c_1x + ... + c_nx^n$ and you can see that first term does not have an x along with it so it would be $c_0 + x(c_1 + c_2x + ... + c_nx^_{n-1}$ .And this does not equal to g(x).It does equal to g(x) when x=0. :/ Feb 2, 2014 at 19:14
• Take your time. You noticed that $f(x)=x(2x+3-1)$ and $f(x)=x(x^2-2x+1)$. So why not let $g(x)=2x+3-1$ and $g(x)=x^2-2x+1$ respectively? Then notice that $xg(x)=f(x)$. So if $f(x)=c_k x^k+\cdots+ c_1 x=x(c_k x^{k-1}+\cdots +c_1)$, what should $g(x)$ be? Then does $xg(x)=f(x)$? One you show that, you only need to explain why $f(0)=0$ forces $f(x)$ to always have the form I gave; that is, why $f(x)$ never has a nonzero constant term-which is a simple thing to explain. Feb 2, 2014 at 19:17
• What if $f(x) = 4x^2 + 4x + 4$ .It seems that you are somehow implying that $c_0$ is always zero.If $c_0 = 0$ always then this becomes easy ,but it does not seem to be(I am so sorry for bothering you this much though,but I am kind of answer junkie) Feb 2, 2014 at 19:26

Hints:

$$0=f(0)=c_0\implies f(x)=c_1x+c_2x^2+\ldots+c_nx^n=x(c_1+c_2x+\ldots+c_nx^{n-1})\;\;\ldots$$
• I think this is false,correct me if I am wrong,but we are passing x to the function thus $f(0) = c_0 \times 0$ and by that logic $c_0$ can be any number it is due to x=0 we get that function is 0. And thus $f(x) =c_0x^0 + c_1x + c_2x^2 +...+ c_nx^n and we can not really factor it out Feb 2, 2014 at 18:59 • What is false, @VanioBegic? I've no idea what you mean by "we are passing x to the function" and etc. In any polynomial, the value the polynomial function takes on zero is the free coefficient. This is trivial and you can check it at once. What else isn'tclear here? Feb 2, 2014 at 19:18 • I think he missed the fact that the first term is$c_0$( does not depend on$x$), even though it formally appears as$c_kx^k$with$k=0$. It is traditional to take that term as constant even though it really can't be defined for$x=0$. Otherwise you'd always have to write that term separately from the rest of the series, which is a pain. – MPW Feb 2, 2014 at 19:25 • @MPW, I think you're right, but I thought the OP would know that as he wrote that expression for the polynomial... Feb 2, 2014 at 19:28 • I am sorry for that passing of x to function(I do lot of programming,so it is term from there).I just think that$c_0\$ is not equal to 0 ,and you did not write it when you expanded f(x). Feb 2, 2014 at 19:33