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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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    @Tinker: If $P)x)$ is the constant polynomial $1$, then $L(p)(x)$ is also identically $1$.2012-03-29

1 Answers 1

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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.2012-03-29