# Vector space of functions defined by a condition

• Hall
This too is not continuous at...##f_2(x) =x####f_3(x) = x^2####f_4(x)= x^3##I'm sorry, but no.Those functions are not in the space.f

#### 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?

• Delta2
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?

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.

• 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?

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 $(a,b,A,B,C,D)$?

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.

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.
$$\text{Basis} = \{x , x^2\} ~~~~~\text{if x belongs to [0,1]}$$

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

• PeroK and Delta2
$$\text{Basis} = \{x , x^2\} ~~~~~\text{if x belongs to [0,1]}$$

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

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:
$$f_3(x) = x^2 +x ~~~~x \in [0,1]$$
$$f_3(x) = x^3 +x^2 +x +1 ~~~~ x\in [1,2]$$

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...

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

• 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)

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.

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.

• Delta2
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.

##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?

##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.

• Steve4Physics
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##?

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.

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.

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.

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##

Check that ##V## is a subspace.

By continuity at ##x=1## we have
$$a+b = A+B+C+D.$$
The solution space for the above is of dimension ##5##. More explicitly, put
$$V\to \mathbb R^5,\quad \begin{cases} ax^2+bx, &x\in [0,1] \\ Ax^3+Bx^2+Cx+D, &x\in [0,2] \end{cases} \mapsto (A+B+C+D-b,b,A,B,C)$$
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##.

• Delta2
Consider $$\begin{split} f_1 &: x \mapsto \begin{cases} x & x \in [0,1] \\ 1 & x \in (1,2] \end{cases} \\ f_2 &: x \mapsto \begin{cases} 0 & x \in [0,1] \\ x - 1 & x \in (1,2]. \end{cases} \end{split}$$ Then every degree 1 polynomial in $V$ is a linear combination of $f_1$ and $f_2$.

• Delta2
##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.

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##.

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

• Delta2
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)

Consider $$\begin{split} f_1 &: x \mapsto \begin{cases} x & x \in [0,1] \\ 1 & x \in (1,2] \end{cases} \\ f_2 &: x \mapsto \begin{cases} 0 & x \in [0,1] \\ x - 1 & x \in (1,2]. \end{cases} \end{split}$$ 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##?

Start with this: What is the function for which a=A=1 and B=C=D=0? (The value of b should follow)
$$f_1 = \begin{cases} x^2 & x \in[0,1]\\ x^3 & x \in[1,2]\\ \end{cases}$$

$$f_1 = \begin{cases} x^2 & x \in[0,1]\\ x^3 & x \in[1,2]\\ \end{cases}$$
Ok, so now, what if you pick ##A=1## and ##B=C=D=a=0##?

Ok, so now, what if you pick ##A=1## and ##B=C=D=a=0##?
$$f_2 = \begin{cases} x & x \in [0,1]\\ x^3 & x \in [1,2]\\ \end{cases}$$

$$f_2 = \begin{cases} x & x \in [0,1]\\ x^3 & x \in [1,2]\\ \end{cases}$$
So based on those examples, can you pick 5 independent assignments of ##a##, ##A##, ##B##, ##C## and ##D## and quote the corresponding functions?

So based on those examples, can you pick 5 independent assignments of ##a##, ##A##, ##B##, ##C## and ##D## and quote the corresponding functions?
Consider, 5-tuple vector ##(A, B, C, D, a)##, its basis elements are
##(1,0,0,0,0)##
##(0,1,0,0,0)##
##(0,0,1,0,0)##
##(0,0,0,1,0)##
##(0,0,0,0,1)##
$$f_1 = \begin{cases} x & x \in [0,1]\\ x^3 & x \in [1,2]\\ \end{cases}$$

$$f_2 = \begin{cases} x & x \in[0,1] \\ x^2 & x \in[1,2]\\ \end{cases}$$
f_3 =
\begin{cases}
x & x\in [0,1]\\
x & x \in [1,2]\\
\end{cases}

f_4 =
\begin{cases}
x & x \in [0,1]\\
1& x \in [1,2] \\
\end{cases}

f_5 =
\begin{cases}
x^2 + x& x \in[0,1] \\
0& x \in [1,2] \\
\end{cases}
$$• nuuskur Is ##f_5## continuous at x=1? • SammyS Is ##f_5## continuous at x=1? I can fix it by changing ##(0,0,0,0,1)## to ##(1,0,0,0,1)## thus converting ##f_5## to$$
f_5 =
\begin{cases}
x^2 & x \in [0,1]\\
x^3 & x \in [1,2]\\
\end{cases}