Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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
show 4 more comments

1 Answer

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).

share|improve this answer
    
I understand now the differenes you have shown me. Could I ask 2 more points of you. –  Tinker Mar 29 '12 at 19:41
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.