Vector space of functions defined by a condition

[Hall]

Homework Statement

Main body is to be referred.

Relevant Equations

Main body should be referred.

$f : [0,2] \to R$. $f$ is continuous and is defined as follows:
$$
f = ax^2 + bx ~~~~\text{ if x belongs to [0,1]}$$
$$
f(x)= Ax^3 + Bx^2 + Cx +D ~~~~\text{if x belongs to [1,2]}$$

$V = \text{space of all such f}$

What would the basis for V? Well, for $x \in [0,1]$ the basis for $V$ (or those f’s can be constructed) is $\{x, x^2\}$, and for $x\in [1,2]$ the basis for $V$ is $\{x^3, x^2, x, 1\}$. I wonder how to combine it, and can I find the dimension of V without finding the basis for V?

 

[Orodruin]

To construct a basis you need to find a set of functions that belong to your vector space from which any other function in the vector space can be constructed. What have you thought of doing so far?

 

[Delta2]

I think the basis of this V is some functions that also change formula depending on where x belongs. I got some idea but I ll explain further if this hint is not enough.

 

[pasmith]

Homework Statement:: Main body is to be referred.
Relevant Equations:: Main body should be referred.

$f : [0,2] \to R$. $f$ is continuous and is defined as follows:
$$
f = ax^2 + bx ~~~~\text{ if x belongs to [0,1]}$$
$$
f(x)= Ax^3 + Bx^2 + Cx +D ~~~~\text{if x belongs to [1,2]}$$

$V = \text{space of all such f}$

What would the basis for V? Well, for $x \in [0,1]$ the basis for $V$ (or those f’s can be constructed) is $\{x, x^2\}$, and for $x\in [1,2]$ the basis for $V$ is $\{x^3, x^2, x, 1\}$. I wonder how to combine it, and can I find the dimension of V without finding the basis for V?

On the face of it, six real numbers are required to specify f: (a,b,A,B,C,D). But f must be continuous at 1. What condition does that impose on (a,b,A,B,C,D)?

 

[Hall]

On the face of it, six real numbers are required to specify f: (a,b,A,B,C,D). But f must be continuous at 1. What condition does that impose on (a,b,A,B,C,D)?

$f(1)=f(1)$
$a+ b = A + B + C +D$
$a = A +B +C +D -b$
Only 5 of them are independent.

 

[Hall]

I think the basis of this V is some functions that also change formula depending on where x belongs. I got some idea but I ll explain further if this hint is not enough.

How about this:
$$
\text{Basis} = \{x , x^2\} ~~~~~\text{if x belongs to [0,1]}$$

$$
\text{Basis} = \{1, x, x^2, x^3\} ~~~~\text{if x belongs to [1,2]}$$

 

[Delta2]

How about this:
$$
\text{Basis} = \{x , x^2\} ~~~~~\text{if x belongs to [0,1]}$$

$$
\text{Basis} = \{1, x, x^2, x^3\} ~~~~\text{if x belongs to [1,2]}$$

No I don't mean that unfortunately.
I mean to define some functions like for example $f_3:[0,2]\to R$ with $$f_3(x)=0 ,x\in [0,1]$$$$f_3(x)=x^3,x\in [1,2] $$and then it is obvious at least to me that this $f_3$ is one of the basis vectors of V.

 

[Hall]

No I don't mean that unfortunately.
I mean to define some functions like for example $f_3:[0,2]\to R$ with $$f_3(x)=0 ,x\in [0,1]$$$$f_3(x)=x^3,x\in [1,2] $$and then it is obvious at least to me that this $f_3$ is one of the basis vectors of V.

I’m sorry I’m unable to grasp the concept, but let me try once again:
$$
f_3(x) = x^2 +x ~~~~x \in [0,1]$$
$$
f_3(x) = x^3 +x^2 +x +1 ~~~~ x\in [1,2]$$

 

[Delta2]

I’m sorry I’m unable to grasp the concept, but let me try once again:
$$
f_3(x) = x^2 +x ~~~~x \in [0,1]$$
$$
f_3(x) = x^3 +x^2 +x +1 ~~~~ x\in [1,2]$$

hmm I am not sure if this $f_3$ qualifies as basis for V let me think...

 

[Delta2]

Well it seems that $f_3$ of yours generates a subspace of V.

 

[Office_Shredder]

Can you try to write down 5 different functions that you think might be linearly independent? Try to keep them as simple as possible (e.g. monomials are better than many terms in a sum)

 

[Hall]

Can you try to write down 5 different functions that you think might be linearly independent? Try to keep them as simple as possible (e.g. monomials are better than many terms in a sum)

$f_1(x)=1$
$f_2(x) =x$
$f_3(x) = x^2$
$f_4(x)= x^3$

Keeping the context of this question in mind, I couldn’t write the fifth one.

 

[Orodruin]

Well it seems that $f_3$ of yours generates a subspace of V.

Not even, that function is not in $V$ because it is not continuous at x=1.

 

[Orodruin]

No I don't mean that unfortunately.
I mean to define some functions like for example $f_3:[0,2]\to R$ with $$f_3(x)=0 ,x\in [0,1]$$$$f_3(x)=x^3,x\in [1,2] $$and then it is obvious at least to me that this $f_3$ is one of the basis vectors of V.

This too is not continuous at x=1.

 

[Orodruin]

$f(1)=f(1)$
$a+ b = A + B + C +D$
$a = A +B +C +D -b$
Only 5 of them are independent.

So can you figure out 5 ways of assigning the independent constants such that the resulting functions are linearly independent?

 

[Office_Shredder]

$f_1(x)=1$
$f_2(x) =x$
$f_3(x) = x^2$
$f_4(x)= x^3$

Keeping the context of this question in mind, I couldn’t write the fifth one.

The middle two are fine, but $f_1$ and $f_3$ aren't elements of your space! You have to define these piecewise.

 

[Hall]

The middle two are fine, but $f_1$ and $f_3$ aren't elements of your space! You have to define these piecewise.

Sorry, but $f_2$ and $f_3$ are the middle ones.

Considering that you meant $f_1$ and $f_4$, I would say
$$
f_1(x) = 1 ~~~~~~~ x \in [1,2]$$
Else,
$$
f_1(x) = 0 ~~~~~~~~ x \in [0,1]$$

Why would $f_4(x) = x^3$ not belong to V? It is the case when $A=1, B=C=D=0$?

 

[Orodruin]

Sorry, but $f_2$ and $f_3$ are the middle ones.

Considering that you meant $f_1$ and $f_4$, I would say
$$
f_1(x) = 1 ~~~~~~~ x \in [1,2]$$
Else,
$$
f_1(x) = 0 ~~~~~~~~ x \in [0,1]$$

Why would $f_4(x) = x^3$ not belong to V? It is the case when $A=1, B=C=D=0$?

Again, that $f_1$ is not in V because it is not continuous at x=1. Your basis functions must satisfy all requirements to be in V.

Why would f4(x)=x3 not belong to V? It is the case when A=1,B=C=D=0?

Because on [0,1] it is not of a correct functional form to be in V.

Please, do not overthink this and consider post #15.

 

[Hall]

So can you figure out 5 ways of assigning the independent constants such that the resulting functions are linearly independent?

Yes, but it is not something I have done before so I’m unable to kinda absorb it.

I can understand, for example, that $x^3 + 3x^2+ 5x$ and $x^3+ x^2 +x$# are independent functions.

 

[Orodruin]

Yes, but it is not something I have done before so I’m unable to kinda absorb it.

Well, start by writing down one such combination.

 

[Hall]

Well, start by writing down one such combination.

$f_1(x) = 5 x^2 + 5x$
$f_2(x)= x^3 + 2x^2 + 3x + 4$
$f_3(x)= x^2 + 13 x$
$f_4(x)= 2x^3 +3 x^2 +4x+ 5$
$f_5{x} = x^2$

 

[nuuskur]

Check that $V$ is a subspace.

By continuity at $x=1$ we have
<br /> a+b = A+B+C+D.<br />
The solution space for the above is of dimension $5$. More explicitly, put
<br /> V\to \mathbb R^5,\quad \begin{cases} ax^2+bx, &amp;x\in [0,1] \\ Ax^3+Bx^2+Cx+D, &amp;x\in [0,2] \end{cases} \mapsto (A+B+C+D-b,b,A,B,C)<br />
for example and convince yourself it is an injective linear map. Since isomorphisms map bases to bases, pick your favourite basis in $\mathbb R^5$ and map it to a basis in $V$.

 

[pasmith]

Consider <br /> \begin{split}<br /> f_1 &amp;: x \mapsto \begin{cases} x &amp; x \in [0,1] \\ 1 &amp; x \in (1,2] \end{cases} \\<br /> f_2 &amp;: x \mapsto \begin{cases} 0 &amp; x \in [0,1] \\ x - 1 &amp; x \in (1,2]. \end{cases} \end{split}<br /> Then every degree 1 polynomial in V is a linear combination of f_1 and f_2.

 

[Orodruin]

$f_1(x) = 5 x^2 + 5x$
$f_2(x)= x^3 + 2x^2 + 3x + 4$
$f_3(x)= x^2 + 13 x$
$f_4(x)= 2x^3 +3 x^2 +4x+ 5$
$f_5{x} = x^2$

This is just guessing. $f_2$ and $f_4$ are not in V. 1,3 and 5 are linearly dependent.

 

[Hall]

This is just guessing

I simply took $A = 1, B=2, C= 3, D= 4, a= 5$ for $f_1$ and $f_2$, and $a= 1, A=2, B=3, C=4, D=5$ for $f_3$ and $f_4$.

 

[nuuskur]

Elements of $V$ are defined piecewise. So too should your basic elements be defined as such.

 

[Orodruin]

I simply took $A = 1, B=2, C= 3, D= 4, a= 5$ for $f_1$ and $f_2$, and $a= 1, A=2, B=3, C=4, D=5$ for $f_3$ and $f_4$.

Why would you put all of the constants non-zero? Your interpretation of what it means becomes lacking.

Start with this: What is the function for which $a=A=1$ and $B=C=D=0$? (The value of $b$ should follow)

 

[Hall]

Consider <br /> \begin{split}<br /> f_1 &amp;: x \mapsto \begin{cases} x &amp; x \in [0,1] \\ 1 &amp; x \in (1,2] \end{cases} \\<br /> f_2 &amp;: x \mapsto \begin{cases} 0 &amp; x \in [0,1] \\ x - 1 &amp; x \in (1,2]. \end{cases} \end{split}<br /> Then every degree 1 polynomial in V is a linear combination of f_1 and f_2.

$$
f_3(x)=
\begin{cases}
x^2 & x \in [0,1]\\
1 & x \in(1,2] \\
\end{cases}$$
$$
f_4(x) =
\begin{cases}
0 & x \in[0,1] \\
x^2 -1 & \in (1,2]\\
\end{cases}$$
All the second order polynomials in V are linear combinations of $f_1, f_2, f_3$ and $f_4$?

 

[Hall]

Start with this: What is the function for which a=A=1 and B=C=D=0? (The value of b should follow)

$$
f_1 =
\begin{cases}
x^2 & x \in[0,1]\\
x^3 & x \in[1,2]\\
\end{cases}$$

 

[Orodruin]

$$
f_1 =
\begin{cases}
x^2 & x \in[0,1]\\
x^3 & x \in[1,2]\\
\end{cases}$$

Ok, so now, what if you pick $A=1$ and $B=C=D=a=0$?