Linear Map Input: Solving P'(1-x) | \pi\rangle

[Petrus]

Hello!
I have hard to understand this input for this linear map $$T:P_3(R)->P_2(R)$$
$$T(p(x))=P'(1-x)$$
so they get this value when they put in which I have hard understanding

I don't understand how they get those, I am totally missing something basic...!
The only logical explain is that in $$p'(x)=3x^2$$ where $$x=(1-x)$$ I don't get it.

Regards,
$$|\pi\rangle$$

 

[I like Serena]

Hello!
I have hard to understand this input for this linear map $$T:P_3(R)->P_2(R)$$
$$T(p(x))=P'(1-x)$$
so they get this value when they put in which I have hard understanding

I don't understand how they get those, I am totally missing something basic...!
The only logical explain is that in $$p'(x)=3x^2$$ where $$x=(1-x)$$ I don't get it.

Regards,
$$|\pi\rangle$$

Hey Petrus!

I suspect you have trouble with the substitution step.

Let's take the last one as an example.
I'm assuming that the step from $p(x)=x^3$ to $p'(x)=3x^2$ is clear.

The latter is equivalent to saying:
$$p'(u)=3u^2$$
That is, $p'$ is a function and if you apply it to $u$ you get $3u^2$.

Now, if we substitute $u=1-x$, we replace all $u$'s by $1-x$ and get:
$$p'(1-x)=3(1-x)^2$$

 

[Petrus]

Hey Petrus!

I suspect you have trouble with the substitution step.

Let's take the last one as an example.
I'm assuming that the step from $p(x)=x^3$ to $p'(x)=3x^2$ is clear.

The latter is equivalent to saying:
$$p'(u)=3u^2$$
That is, $p'$ is a function and if you apply it to $u$ you get $3u^2$.

Now, if we substitute $u=1-x$, we replace all $u$'s by $1-x$ and get:
$$p'(1-x)=3(1-x)^2$$

Thanks a lot for the fast respond! I got it now! Have a nice day!:)

Regards,
$$|\pi\rangle$$

 

[Deveno]

Although most people know the derivative from calculus, you don't need calculus to define the derivative of a polynomial.

Given a polynomial in $P_n(\Bbb R)$, say:

$p(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n$

We define the derivative $p'(x)$ (I prefer to write this as $Dp$ for reasons we shall see later) as:

$Dp(x) = a_1 + 2a_2x + \cdots + na_nx^{n-1}$

If one uses the basis: $B = {1,x,x^2,\dots,x^n}$ for $P_n(\Bbb R^n)$ one can identify (such an identification is called a linear isomorphism) $P_n(\Bbb R)$ with $\Bbb R^n$ like so:

$a_0 + a_1x + a_2x^2 + \cdots + a_nx^n \mapsto (a_0,a_1,a_2,\dots,a_n)$

In this basis (actually "two bases" but the basis for $P_{n-1}(\Bbb R)$ is "just like" the basis for $P_n(\Bbb R)$), $D$ has the nx(n+1) matrix:

$[D]_B = \begin{bmatrix}0&1&0&\cdots&0\\0&0&2&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&n \end{bmatrix}$

which makes it clear $D$ is a linear mapping.

We also have the linear mapping: $L: P_n(\Bbb R) \to P_n(\Bbb R)$ given by:

$L(p(x)) = p(a - x)$ for $a \in \Bbb R$.

It may be instructive to see the matrix of $L$ with respect to the basis $B$:

$[L]_B = \begin{bmatrix}1&a&a^2&\cdots&a^n\\0&-1&-2a&\cdots&-na^{n-1}\\0&0&1&\cdots&\frac{n(n-1)}{2}a^{n-2}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&\cdots&(-1)^n \end{bmatrix}$

So the mapping you are given is just $L \circ D$ (with $a = 1$):

$(L \circ D)(p(x)) = L(D(p(x)) = L(p'(x)) = p'(1 - x)$.

For the case $n = 3$, we can see $D$ as having the 3x4 matrix:

$\begin{bmatrix}0&1&0&0\\0&0&2&0\\0&0&0&3 \end{bmatrix}$

and $L$ as being the 3x3 matrix (with $a = 1$):

$\begin{bmatrix}1&1&1\\0&-1&-2\\0&0&1\\ \end{bmatrix}$

Then multiplying these two matrices together, we get the 3x4 matrix:

$\begin{bmatrix}0&1&2&3\\0&0&-2&-6\\0&0&0&3 \end{bmatrix}$ which is the matrix for $T$ in the basis $B$.

Now the vector representing the constant polynomial $p(x) = 1$ in the basis $B$ is just: (1,0,0,0), while the vector representing $x$ is (0,1,0,0), the vector representing $x^2$ is (0,0,1,0), and the vector representing $x^3$ is (0,0,0,1). So applying the matrix of $T$ to these vectors, we get

$T(1) \to [T(1,0,0,0)]_B = [0,0,0]_B \to 0$
$T(x) \to [T(0,1,0,0)]_B = [1,0,0]_B \to 1$
$T(x^2) \to [T(0,0,1,0)]_B = [2,-2,0]_B \to 2-x = 2(1-x)$
$T(x^3) \to [T(0,0,0,1)]_B = [3,-6,3]_B \to 3 - 6x + 3x^2 = 3(1 - x^2)$

If, instead, your book had defined:

$T(p(x)) = (p(x))'$

we would have taken $D \circ L$, giving a different 3x4 matrix (using a 4x4 "L").

I find that the use of derivative notation in linear algebra problems involving polynomial spaces is sort of confusing, and obscures the underlying vector algebra: we just have some linear map $D$ on a vector space $V$, to understand what $D$ does, we need only examine what it does to a basis (and once having CHOSEN a basis, we can do everything in matrices relative to that basis).

Or, perhaps more simply put: everything you want to know about a polynomial vector space is given by the coefficients of the polynomials (the $x$ is just "excess baggage" in terms of the "vector-space-ness", although it does come into play when considering the RING structure of this vector space...but that is a topic beyond the scope of most linear algebra courses).