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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 $L(p)(x)=(2x+1)^3$. – André Nicolas Mar 29 '12 at 19:39
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. Thanks – Tinker Mar 29 '12 at 19:53
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 $p(2x+1)+q(2x+1)$. – Arturo Magidin Mar 29 '12 at 19:56
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 patience – Tinker Mar 29 '12 at 20:06
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. – Arturo Magidin Mar 29 '12 at 20:09

No, no, no, no, no, no.

The formula says: $L[p(x)] = p(2x+1)$.

You are computing $2p(x) + 1$ instead.

Do you notice the difference?

Here: If $p(x) = x^2$, then $$L[p(x)] = p(2x+1) = (2x+1)^2 = 4x^2 + 4x + 1.$$ You computed $$2p(x) + 1 = 2x^2 + 1.$$ You are computing the wrong thing.

The formula for $L$ is: plug in $2x+1$ instead of $x$.

(Yes, the function $F[p(x)] = 2p(x)+1$ is not linear; you can verify that by simply noting that $F[0] = 1\neq 0$, but linear transformations must map $0$ to $0$; but that is not the function you are being asked to prove is linear).

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I understand now the differenes you have shown me. Could I ask 2 more points of you. – Tinker Mar 29 '12 at 19:41

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