MHB 11.1 Determine if the polynominal.... is the span of (....,....)

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Span
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
Determine if the polynomial
$3x^2+2x-1$
is the $\textbf{span}\{x^2+x-1,x^2-x+2,1\}$ok from examples it looks like we see if there are scalars such that
$c_1(x^2+x-1)+c_2(x^2-x+2,1)=3x^2+2x-1$so far not sure how this is turned into a simultaneous eq

I did notice that it is common to get over 100 views on these DE problems
so thot it would be good to show sufficient steps
 
Last edited:
Physics news on Phys.org
karush said:
$c_1(x^2+x-1)+c_2(x^2-x+2,1)=3x^2+2x-1$

It should be
$$c_1(x^2+x-1)+c_2(x^2-x+2)+c_3\ \equiv\ 3x^2+2x-1$$
Expand out the LHS and compare coefficients. This will give you a system of equations in $c_1,c_2,c_3$. If the system is consistent (i.e. has at least one solution) then the given polynomial is in the given span, otherwise not.
 
https://www.physicsforums.com/attachments/9043

ok here is an example but I don't see how they got the numbers"
 
That solution is for the problem of determining if this polynomial
$$2x^2+x+1$$
is in $\mathrm{Span}\{x^2+x,x^2-1,x+1\}$.

In your OP, you have this:
$$c_1(x^2+x-1)+c_2(x^2-x+2)+c_3\ \equiv\ 3x^2+2x-1$$
so comparing coefficients of $x^2$, $x$ and the constant terms should give you
$$\begin{array}{rcrcrcr}c_1 & +&c_2 && &=& 3 \\ c_1 &-&c_2 && &=& 2 \\ c_1 &+&2c_2 &+&c_3 &=& -1\end{array}.$$
 
Last edited:
$c_1(x^2+x-1)+c_2(x^2-x+2)+c_3\ \equiv\ 3x^2+2x-1$ok I got ...\begin{array}{rcrcrcr}c_1 & +&c_2 && &=& 3 \\ c_1 &-&c_2 && &=& 2 \\ -c_1 &+&2c_2 &+&c_3 &=& -1\end{array}

could we multiply thru the $R_3$ with -1

well anyway got this ...

$\left[ \begin{array}{ccc|c} 1 & 0 & 0 & \frac{5}{2} \\ 0 & 1 & 0 & \frac{1}{2} \\ 0 & 0 & 1 & - \frac{9}{2} \end{array} \right]$
 
Last edited:
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
9
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
2
Views
2K
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K