MHB Solve Degree 2 Polynomial: 10.2 DE Notation Explained

[karush]

View attachment 8834

in (3) they say any polynomial of degree less than 2, yet the example is degree 2 ?

ok this is due tomro, so hopefully I can get a handle on it today

I think this is familiar, just is the notation is stumping me.

really appreciate any insight...

did a screenshot to avoid typos

 

[karush]

We define the application
$T:P_2\rightarrow P_2$
by
$T(p)=2p(x)+xp'(x)+x^2 p''(x)$
Compute
$T(x^2),T(x),T(1) \textit{ and } T(ax^2+bx+c)$

ok if I just plug in then

$T(1)=2(1)+x(1)'+x^2(1)''=2+0+0=0$
$T(x)=2(x)+x(x)'+x^2(x)''=2x+x+0=3x$
$T(x^2)=2(x^2)+x(x^2)'+x^2(x^2)''=2x^2+2x+4$

kinda maybe...

 

[skeeter]

$T(1)=2(1)+x(1)'+x^2(1)''=2+0+0=\color{red}{2}$

$T(x^2)=2(x^2)+x(x^2)'+x^2(x^2)''=\color{red}{2(x^2)+x(2x)+x^2(2)=2x^2+2x^2+2x^2=6x^2}$

 

[karush]

$T(1)=2(1)+x(1)'+x^2(1)''=2+0+0=\color{red}{2}$

$T(x)=2(x)+x(x)'+x^2(x)''=2x+x+0=\color{red}{3x}$

$T(x^2)=2(x^2)+x(x^2)'+x^2(x^2)''=2(x^2)+x(2x)+x^2(2)=2x^2+2x^2+2x^2=\color{red}{6x^2}$

$T(ax^2+bx+c)=2(ax^2+bx+c)+x(ax^2+bx+c)'+x^2(ax^2+bx+c)''$
$=2ax^2+2bx+2c+2ax^2+bx+0+2ax^2+0+0=\color{red}{6ax^2+3bx+2c}$

just seeing if this is launch ready...

 

[topsquark]

$T(1)=2(1)+x(1)'+x^2(1)''=2+0+0=\color{red}{2}$

$T(x)=2(x)+x(x)'+x^2(x)''=2x+x+0=\color{red}{3x}$

$T(x^2)=2(x^2)+x(x^2)'+x^2(x^2)''=2(x^2)+x(2x)+x^2(2)=2x^2+2x^2+2x^2=\color{red}{6x^2}$

$T(ax^2+bx+c)=2(ax^2+bx+c)+x(ax^2+bx+c)'+x^2(ax^2+bx+c)''$
$=2ax^2+2bx+2c+2ax^2+bx+0+2ax^2+0+0=\color{red}{6ax^2+3bx+2c}$

just seeing if this is launch ready...

Looks good to me!

-Dan

 

[karush]

ok I think this is the matrix (not sure what $$ T(ax^2+bx+c)$$ would be used for

$\left[\begin{array}{ccc}2&0&0\\0&3&0\\0&0&6 \end{array}\right]$

 

[topsquark]

ok I think this is the matrix (not sure what $$ T(ax^2+bx+c)$$ would be used for

$\left[\begin{array}{ccc}2&0&0\\0&3&0\\0&0&6 \end{array}\right]$

Okay, so what's next? (Did you do part 2 yet?)

-Dan

 

[karush]

Show that T is a linear Transformation..
Ok well I presume the next step would be

$$\displaystyle\vec{x}
=\left[\begin{array}{ccc}2\\0\\0 \end{array}\right]
\quad \vec{y}=
=\left[\begin{array}{ccc}0\\3\\0 \end{array}\right]
\quad \vec{z}=
=\left[\begin{array}{ccc}0\\0\\6 \end{array}\right]$$

 

[skeeter]

Linear Transformation -- from Wolfram MathWorld

 

[skeeter]

for $T(p) = 2p + xp' + x^2p''$ ...

(1) show $T(p_1 + p_2) = T(p_1)+T(p_2)$

$T(p_1+p_2) = 2(p_1 + p_2) + x(p_1+p_2)' + x^2(p_1+p_2)'' = 2p_1+2p_2+xp_1'+xp_2'+x^2p_1''+x^2p_2'' = (2p_1+xp_1'+x^2p_1'')+(2p_2+xp_2'+x^2p_2'') = T(p_1) + T(p_2)$

now, you show part (2) ... that $T(\alpha p) = \alpha T(p)$ for any scalar $\alpha$.

 

[karush]

ok, sorry I had to abandon this one and move on
I was doing fine in this class until I hit this now I'm worried about the rest of it.

 

[topsquark]

ok, sorry I had to abandon this one and move on
I was doing fine in this class until I hit this now I'm worried about the rest of it.

I'm certainly not expert in this topic myself but I think you are giving up too easily. Try to find a simpler example. This isn't too bad but you need to spend some time with the examples to get it down. As all the "advanced" Math that I know uses linear differential operators the time spent will be well worth the effort.

-Dan

 

[karush]

I agree I'm looking at examples now
Have such hard time in class because my hearing is really bad and board is hard to read
so mostly read pdfs and mhb.

W|A and wikipedia are awful to read