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I am given transformation: $L:\mathbb{R}[x]\rightarrow \mathbb{R}[x], (Lp)(x)=xp'(x)+p(1) $

$R[x]$ is (i believe) polynomial of 1 degree. $ax+b$

I do few transformations: $p(x)=ax+b$

$(Lp)(x)=x\cdot b+a+b$

I am wrinting two polynomial of 1 degree :

$u(x)=a_{1}x+b_{1}$

$v(x)=a_{2}x+b_{2}$

I am taking $\alpha ,\beta \in \mathbb{R}$

$\alpha \cdot u(x) + \beta \cdot v(x) = x(a_{1}\cdot\alpha+a_{2}\cdot\beta)+\alpha\cdot b_{1}+\beta\cdot b_{2}$

$L(\alpha \cdot u(x) + \beta \cdot v(x))=L(x(a_{1}\cdot\alpha+a_{2}\cdot\beta)+\alpha\cdot b_{1}+\beta\cdot b_{2})=x\cdot(\alpha\cdot b_{1}+\beta\cdot b_{2} )+a_{1}\cdot\alpha+a_{2}\cdot\beta+\alpha\cdot b_{1}+\beta\cdot b_{2}=\alpha(x\cdot b_{1} + a_{1} + b_{1})+\beta(x\cdot b_{2}+a_{2}+b_{2})=\alpha(Lu)(x) + \beta(Lv)(x) \implies \text It \ is \ linear \ \ transfomration$

Am i doing this right?

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2  
try know $\alpha L(u)+\beta L(v)$ and compare –  janmarqz Jan 8 at 21:12
    
When I do this the proof is complete and correct? –  Marcin Majewski Jan 8 at 21:14
1  
when you verified αL(u)+βL(v)= L(αu+βv) you are thru –  janmarqz Jan 8 at 21:22

1 Answer 1

You don't mention explicitly what space $p(x)$ is in, but it does seem that you assume $p(x)\in\mathcal{P}_1$. I will take $p$ to be a differentiable function with $p(1)$ defined, since I am not sure what your $\mathbb{R}[x]$ notation is meant to signify.

For linearity of the operator $L$, we need (as mentioned in the comments) $L(\alpha p+\beta q)=\alpha L(p)+\beta L(q)$, for appropriate $p$ and $q$ and scalars $\alpha$ and $\beta$.

Now, $$ \begin{align} L(\alpha p(x)+\beta q(x))&=x\frac{d}{dx}(\alpha p(x)+\beta q(x))+\left.(\alpha p(x)+\beta q(x))\right|_{x=1}&[\text{Operator}]\\ &=x\left(\alpha p'(x)+\beta q'(x)\right)+\left(\alpha p(1)+\beta q(1)\right)&[\text{By definitions}]\\ &=\alpha\left(xp'(x)+p(1)\right)+\beta\left(xq'(x)+q(1)\right)&[\text{Rearrangement}]\\ &=\alpha L(p)+\beta L(q). \end{align} $$

Thus, since $L(\alpha p+\beta q)=\alpha L(p)+\beta L(q)$, $L$ is a linear operator.

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