Let $p(x)=a_0+a_1 x+…+a_nx^n$ be a non-constant polynomial $q(x)=\int_0^xp(t)dt,r(x)=\frac{d}{dx}p(x).$ Let $V$ denote the real vector space of all polynomials in $x.$ Then which one of the following is True?
A.$q$ and $r$ are linearly independent in $V$.
B.$q$ and $r$ are linearly dependent in $V$.
C.$x^n$ belongs to the linear span of $q$ and $r$.
D.$x^{n+1}$ belongs to the linear span of $q$ and $r$.
My Try
$$p(x)=a_0+a_1 x+…+a_nx^n$$
$$q(x)=a_0x+a_1x^2/2+…+a_nx^{n+1}/n+1$$
$$r(x)=a_1+2a_2x+…+na_nx^{n-1}$$
Since $q$ and $r$ are having different degrees. So, It is linearly independent. So, $A$ is True and $B$ is false.
For others. I tried to solve the equation $c_1q+c_2r=x^n$ and $c_1q+c_2r=x^{n+1}.$ I couldn't deduce anything. Please help me.
Best Answer
If you just want to solve the exercise, consider $x+1$ as counterexample. If you want a more "rigorous" proof, read on...
Let's start with your equation $$ c_1q(x)+c_2r(x)=x^n, $$
or equivalently $$ c_1a_0x+\ldots+c_1\frac{a_n}{n+1}x^{n+1}+c_2a_1+\ldots+c_2na_nx^{n-1}=x^n, $$ or in other words $$ c_1\frac{a_n}{n+1}x^{n+1}+\{\text{polynomial with degree}\leq n\}=x^n. $$ Now you can deduce that $c_1=0$, and then your equation is $c_2r(x)=x^n$, and of course you can't solve it because the left side has degree $n-1$ while the right side has $n$. In the same way you can show there aren't $c_1$ and $c_2$ such that $c_1q(x)+c_2r(x)=x^{n+1}$ if $p(x)\neq x^n$ (otherwise $(n+1)q(x)=x^{n+1}$), so C and D are false in general.