Determine the $P(1)$

Question

$P(x)$ is a polynomial function,

$$P(x)-P'(x) = 2x^2-8x$$

Determine the $P(1)$

I've tried to solve the polynomial for $0$

$$P(0) = 2x^2-8x$$

Here we get

$$x = \{4 , 0\}$$

However, I believe that I've gone wrong so far.

Answer 1

A locally nilpotent linear operator on a vector space $V$ (not necessarily finite-dimensional) over a field $\mathbb{K}$ is a linear operator $N:V\to V$ such that, for each $v\in V$, there exists a nonnegative integer $r_v$ depending on $v$ such that $$N^{r_v}(v)=0_V\,.$$ Here, $N^0:=\text{id}_V$. For such a linear operator $N$, the linear operator $\text{id}_V-N:V\to V$ is bijective. To show this, we exhibit the inverse of $\text{id}_V-N$ as $i(N):=\sum\limits_{r=0}^\infty\,N^r$. This linear operator $i(N):V\to V$ is well defined because $N$ is locally nilpotent. It is clear that $$\left(\text{id}_V-N\right)\,i(N)=\text{id}_V=i(N)\,\left(\text{id}_V-N\right)\,.$$

Now, the derivative operator $D$, $Df:=f'$, is a locally nilpotent operator on the vector space $\mathbb{K}[X]$ of polynomials in variable $X$ over an arbitrary field $\mathbb{K}$, namely, for each $f(x)\in\mathbb{K}[x]$, $$D^{\deg(f)+1}f\equiv0\text{ for }f\not\equiv0\,,\text{ and }D^0f\equiv 0\text{ for }f\equiv0\,.$$ By convention, $D^0$ denotes the identity operator $I$.

Now, we go back to the problem. Observe that $$\big((I-D)\,P\big)(X)=P(X)-P'(X)=2X^2-8X\,.$$ By the paragraph above, $$\begin{align}P(X)&=(I-D)^{-1}\,\left(2X^2-8X\right)=\sum_{r=0}^\infty\,D^r\left(2X^2-8X\right)\\ &=(2X^2-8X)+(4X-8)+4=2X^2-4X-4\,.\end{align}$$ That is, $P(1)=-6$.

Answer 2

as mentioned in the comment by @mfl that the polynomial must have degree two because if we take P as degree n polynomial then P-P' also has degree n thus n=2 because the right side polynomial is of degree 2.

you can solve it by taking a degree two polynomial $P(x)=ax^2+bx+c$

I got $b=c$ and $a=2$ and $2a-b=8$

thus $$P(x)=2x^2-4x-4$$ thus $$P(1)=2-4-4=-6$$ $\hspace{20pt}$

Here is method two if want to solve this by differential equations

this is of the form $$\frac{dy}{dx}+P(x)y=Q(x)$$

$$\frac{dy}{dx}-y=8x-2x^2$$

$$P(x)=-1 \space and \space Q(x)=8x-2x^2$$ in this type of differential equations, we multiply the equation by integrating factor

integrating factor $$R(x)=e^{\int P(x) dx}$$

thus $$R(x)=e^{\int -1 dx}=e^{-x}$$

multiply by $R(x)$ $$e^{-x}\frac{dy}{dx}-e^{-x}y=e^{-x}(8x-2x^2)$$ $$\frac{d(e^{-x}y)}{dx}=e^{-x}(8x-2x^2)$$ $$e^{-x}y=\int e^{-x}(8x-2x^2) dx $$

apply integration by part two times on the right side you will get

$$e^{-x}y= e^{-x}(2x^2-4x-4) + C$$

thus $$y= 2x^2-4x-4+ C *e^x$$ since P is a polynomial C must be zero

Answer 3

Let $P(x)$ be any n degree polynomial $$P(x) = a_0 + a_1 x + a_2x^2 + ... + a_nx^n$$

Given that:

$$P(x)-P'(x) = 2x^2-8x$$ $$\implies a_0+a_1x+a_2x^2 + ... a_nx^n - (a_1+2a_2x + ...na_nx^{n-1} )= 2x^2 - 8x$$

$$(a_0-a_1) + (a_1-2a_2)x + (a_2 - 3a_3)x^2 + .... = 2x^2 - 8x$$

Now the r.h.s and l.h.s are two polynomials that are equal for all values of $x$. Therefore they must be equivalent. You should be able to proceed now..

Cheers!