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Let $L\colon\mathbf{P_2}\to \mathbf{P_2}$ be given by $L[p(x)] = p(2x+1)$. We want to prove it is a linear transformation.

Trying to prove that $L[u+v] = L[u]+L[v]$

$$\begin{align*} L[p(x)+q(x)] &= 2[p(x)+q(x)]+1\\ &= 2p(x)+2q(x)+1 \end{align*}$$

From the other size of $L[u]+L[v]$ $$\begin{align*} L[q(x)]+L[q(x)] &= 2(p(x)+1)+2(q(x)+1)\\ &=2p(x)+2q(x)+4 \end{align*}$$

These do result do no match, and so $L[p(x)]=p(2x+1)$ cannot not be a linear transformation?

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    Evaluated $L(p+q)$ incorrectly. Also $L(p)$ and $L(q)$ are not calculated correctly. Please note, for example, that if $p(x)=x^3$, then $L(p)(x)=(2x+1)^3$.2012-03-29
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    Could I ask 2 more points of you, what would be the result if P(x)=1 (instead of p(x)=x^2 in your example. And could you explain the processing of showing how L(p+q)=L(p)+L(q) in this instance. Thanks2012-03-29
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    If you plug in $2x+1$ into the constant function $1$, you get $1$. So $L(1) = 1$. To show $L(p+q)=L(p)+L(q)$, you need to verify that $(p+q)$, evaluated at $2x+1$, is the same thing as $p(2x+1)+q(2x+1)$.2012-03-29
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    Please help me out a little, I'm really trying to get my head around this conceptually. Can you explain by 'verify the (p+q), evaluated at 2x+1' means, or how I go about it. Thanks in advance for your patience2012-03-29
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    I *am* trying to help you out; I think you will greatly benefit from working this out yourself instead of asking me to do your homework for you.2012-03-29
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    It isn't actually homework - it's related to one of the exercises in my textbook. I think I'm having a conceptual problem of how the (2x+1) applies to the original p(x) (or the p(x)+q(x)), which is why I'm asking for an example.2012-03-29
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    @Tinker: And I've given you examples. **And** you haven't even bothered to correct the problem to specify *what* $\mathbf{V}$ and $\mathbf{W}$ are. If $p$ is a polynomial, $$p(x) = a_0 + a_1x + a_2x^2+\cdots+a_nx^n,$$ then $p(2x+1)$ means$$a_0+a_1(2x+1)+a_2(2x+1)^2+\cdots+a_n(2x+1)^n.$$ This is called "evaluating $p$ at $2x+1$." If $p$ and $q$ are polynomials, then you can add them to get a new polynomial which is called "$p+q$", and you can then evaluate this new polynomial at $2x+1$. You can also evaluate $p$ at $2x+1$, evaluate $q$ at $2x+1$, and add the results. Compare the end results.2012-03-29
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    @ArturoMagidin: Thank you - that is exactly the hint/tip I was looking for. I've only just noticed that I can edit my original question, which I will do so now.2012-03-29
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    @Tinker: If $P)x)$ is the constant polynomial $1$, then $L(p)(x)$ is also identically $1$.2012-03-29

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