Fdtxjr

As I said in the title, in the book where this exercise was proposed, derivatives or limits were not introduced yet. The exercise was to show that given the function $':mathbbC[x]rightarrow mathbbC[x]$ I should show that $(fg)'=f'g+fg'$

$f'(x)$ was defined as $sum_k=1^nka_kx^k-1$ where $f(x)=sum_k=0^na_kx^k$.

I have found out that an equivalent representation for $f'(x)$ is $sum_k=0^n-1(k+1)a_k+1x^k$

I have tried to prove the result myself step by step but there must be an error somewhere in the proof because in the end the result does not fit the bill.

$fg=sum_k=0^m+nc_kx^k$ with $c_k=sum_r+s=ka_rb_s$

$(fg)'=sum_j=0^m+n-1(j+1)c_j+1x^j$

$f=sum_r=0^na_rx^r$

$f'=sum_r=0^n-1(r+1)a_r+1x^r$

$g=sum_s=0^mb_sx^s$

$g'=sum_s=0^m-1(s+1)b_s+1x^s$

$f'g=sum_j=0^m+n-1c'_jx^j$, with $c'_j=sum_r+s=j(r+1)a_r+1b_s$

$fg'=sum_j=0^m+n-1c''_jx^j$, with $c''_j=sum_r+s=ja_r(s+1)b_s+1$

I have tried to prove by induction that for every $j$ we have

$c'_j+c''_j=(j+1)c_j+1$

but I failed to show that the coefficients of both polynomials are equal. Where is the mistake? Can somebody give me a hint please?

The Problem is the induction step; I don't know how I can use the induction hypothesis to my advantage; I have made the induction over $j$ and I could Show it for $j=0$ and $j=1$ (just to be sure):

The claim is

$$(j+1)sum_r+s=j+1a_rb_s=sum_r+s=j(r+1)a_r+1b_s+sum_r+s=ja_r(s+1)b_s+1$$

I also know that $(j+1)sum_r+s=j+1a_rb_s=(j+1)sum_r=0^j+1a_rb_j+1-r$

It is easier to not use coefficients at all - except that we initially check by straightforward computation
$$tag1 (f+g)'=f'+g',$$ $$tag2 (cf)'=cf',$$ $$tag3 (xf)'=f+f'.$$
$$tag 4c'=0$$
Let $Asubseteq Bbb C[x]$ be the set of polynomials $f$ with the property that $(fg)'=f'g+fg'$ for all polynomials $g$.

If $f_1,f_2in A$, then for every polynomial $g$,
$$beginalign((f_1+f_2)g)'&=(f_1g+f_2g)'\&=(f_1g)'+(f_2g)'&textby (1)\&=f_1g'+f_1'g+f_2g'+f_2'g\&=(f_1+f_2)g'+(f_1'+f_2')g\
&=(f_1+f_2)g'+(f_1+f_2)'g&textby (1)endalign $$
Thus,

$A$ is closed under addition.

Assume $f_1,f_2in A$. Then for any polynomial $g$,
$$beginalign
((f_1f_2)g)'&=(f_1(f_2g))'\
&=f_1'f_2g+f_1(f_2g)'\
&=f_1'f_2g+f_1(f_2'g+f_2g')\
&=(f_1'f_2+f_1f_2')g+f_1f_2g'.
endalign$$
We conclude that

$A$ is closed under multiplication.

From $(2)$ and $(4)$, we see that

$A$ contains all constants

and from $(3)$ (after verifying that in particular $x'=1$)

$xin A$.

Now observe that the four highlighted statements imply that $A=Bbb C[x]$.