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

Question

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.

*EDIT(ANSWER): **

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.

Answer 1

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).

Answer 2

Hints:

With your notation:

$$0=f(0)=c_0\implies f(x)=c_1x+c_2x^2+\ldots+c_nx^n=x(c_1+c_2x+\ldots+c_nx^{n-1})\;\;\ldots$$