Vector space of functions defined by a condition

In summary: 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.
  • #1
Hall
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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?
 
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  • #2
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?
 
  • #3
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.
 
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  • #4
Hall said:
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 [itex]f[/itex]: [itex](a,b,A,B,C,D)[/itex]. But [itex]f[/itex] must be continuous at 1. What condition does that impose on [itex](a,b,A,B,C,D)[/itex]?
 
  • #5
pasmith said:
On the face of it, six real numbers are required to specify [itex]f[/itex]: [itex](a,b,A,B,C,D)[/itex]. But [itex]f[/itex] must be continuous at 1. What condition does that impose on [itex](a,b,A,B,C,D)[/itex]?
##f(1)=f(1)##
##a+ b = A + B + C +D##
##a = A +B +C +D -b##
Only 5 of them are independent.
 
  • #6
Delta2 said:
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:
$$
\text{Basis} = \{x , x^2\} ~~~~~\text{if x belongs to [0,1]}$$

$$
\text{Basis} = \{1, x, x^2, x^3\} ~~~~\text{if x belongs to [1,2]}$$
 
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  • #7
Hall said:
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.
 
  • #8
Delta2 said:
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:
$$
f_3(x) = x^2 +x ~~~~x \in [0,1]$$
$$
f_3(x) = x^3 +x^2 +x +1 ~~~~ x\in [1,2]$$
 
  • #9
Hall said:
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...
 
  • #10
Well it seems that ##f_3## of yours generates a subspace of V.
 
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  • #11
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)
 
  • #12
Office_Shredder said:
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.
 
  • #13
Delta2 said:
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.
 
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  • #14
Delta2 said:
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.
 
  • #15
Hall said:
##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?
 
  • #16
Hall said:
##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.
 
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  • #17
Office_Shredder said:
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##?
 
  • #18
Hall said:
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.

Hall said:
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.
 
  • #19
Orodruin said:
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.
 
  • #20
Hall said:
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.
 
  • #21
Orodruin said:
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##
 
  • #22
Check that ##V## is a subspace.

By continuity at ##x=1## we have
[tex]
a+b = A+B+C+D.
[/tex]
The solution space for the above is of dimension ##5##. More explicitly, put
[tex]
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)
[/tex]
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##.
 
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  • #23
Consider [tex]
\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}
[/tex] Then every degree 1 polynomial in [itex]V[/itex] is a linear combination of [itex]f_1[/itex] and [itex]f_2[/itex].
 
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  • #24
Hall said:
##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.
 
  • #25
Orodruin said:
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##.
 
  • #26
Elements of ##V## are defined piecewise. So too should your basic elements be defined as such.
 
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  • #27
Hall said:
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)
 
  • #28
pasmith said:
Consider [tex]
\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}
[/tex] Then every degree 1 polynomial in [itex]V[/itex] is a linear combination of [itex]f_1[/itex] and [itex]f_2[/itex].
$$
f_3(x)=
\begin{cases}
x^2 & x \in [0,1]\\
1 & x \in(1,2] \\
\end{cases}$$
$$
f_4(x) =
\begin{cases}
0 & x \in[0,1] \\
x^2 -1 & \in (1,2]\\
\end{cases}$$
All the second order polynomials in V are linear combinations of ##f_1, f_2, f_3## and ##f_4##?
 
  • #29
Orodruin said:
Start with this: What is the function for which a=A=1 and B=C=D=0? (The value of b should follow)
$$
f_1 =
\begin{cases}
x^2 & x \in[0,1]\\
x^3 & x \in[1,2]\\
\end{cases}$$
 
  • #30
Hall said:
$$
f_1 =
\begin{cases}
x^2 & x \in[0,1]\\
x^3 & x \in[1,2]\\
\end{cases}$$
Ok, so now, what if you pick ##A=1## and ##B=C=D=a=0##?
 
  • #31
Orodruin said:
Ok, so now, what if you pick ##A=1## and ##B=C=D=a=0##?
$$
f_2 =
\begin{cases}
x & x \in [0,1]\\
x^3 & x \in [1,2]\\
\end{cases}$$
 
  • #32
Hall said:
$$
f_2 =
\begin{cases}
x & x \in [0,1]\\
x^3 & x \in [1,2]\\
\end{cases}$$
So based on those examples, can you pick 5 independent assignments of ##a##, ##A##, ##B##, ##C## and ##D## and quote the corresponding functions?
 
  • #33
Orodruin said:
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}

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

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

$$
f_5 =
\begin{cases}
x^2 + x& x \in[0,1] \\
0& x \in [1,2] \\
\end{cases}
$$
 
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  • #34
Is ##f_5## continuous at x=1?
 
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  • #35
Orodruin said:
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
$$
f_5 =
\begin{cases}
x^2 & x \in [0,1]\\
x^3 & x \in [1,2]\\
\end{cases}$$
 
<h2>1. What is a vector space of functions defined by a condition?</h2><p>A vector space of functions defined by a condition is a mathematical concept where a set of functions is defined by a specific condition or constraint. This condition can be a mathematical equation, inequality, or any other rule that the functions must satisfy. The set of functions that satisfy the condition form a vector space, which means they can be added, subtracted, and multiplied by scalars.</p><h2>2. How is a vector space of functions defined by a condition different from a regular vector space?</h2><p>The main difference between a vector space of functions defined by a condition and a regular vector space is that the former has a specific constraint or condition that the functions must satisfy. This means that not all functions can be a part of the vector space, unlike a regular vector space where any function can be a member as long as it follows the properties of a vector space.</p><h2>3. What are some examples of vector spaces of functions defined by a condition?</h2><p>Some examples of vector spaces of functions defined by a condition include the space of continuous functions, the space of differentiable functions, and the space of polynomial functions. In each of these examples, the condition is different, but the set of functions that satisfy the condition forms a vector space.</p><h2>4. How are vector spaces of functions defined by a condition used in real life?</h2><p>Vector spaces of functions defined by a condition have various applications in fields such as physics, engineering, and economics. For example, in physics, the space of differentiable functions is used to describe the motion of objects, while in economics, vector spaces of functions are used to model consumer preferences and production functions.</p><h2>5. What are the advantages of using vector spaces of functions defined by a condition?</h2><p>One of the main advantages of using vector spaces of functions defined by a condition is that they provide a more specific and structured framework for studying functions. This allows for a deeper understanding of the functions and their properties. Additionally, vector spaces of functions can also simplify complex problems and make them more manageable to solve.</p>

1. What is a vector space of functions defined by a condition?

A vector space of functions defined by a condition is a mathematical concept where a set of functions is defined by a specific condition or constraint. This condition can be a mathematical equation, inequality, or any other rule that the functions must satisfy. The set of functions that satisfy the condition form a vector space, which means they can be added, subtracted, and multiplied by scalars.

2. How is a vector space of functions defined by a condition different from a regular vector space?

The main difference between a vector space of functions defined by a condition and a regular vector space is that the former has a specific constraint or condition that the functions must satisfy. This means that not all functions can be a part of the vector space, unlike a regular vector space where any function can be a member as long as it follows the properties of a vector space.

3. What are some examples of vector spaces of functions defined by a condition?

Some examples of vector spaces of functions defined by a condition include the space of continuous functions, the space of differentiable functions, and the space of polynomial functions. In each of these examples, the condition is different, but the set of functions that satisfy the condition forms a vector space.

4. How are vector spaces of functions defined by a condition used in real life?

Vector spaces of functions defined by a condition have various applications in fields such as physics, engineering, and economics. For example, in physics, the space of differentiable functions is used to describe the motion of objects, while in economics, vector spaces of functions are used to model consumer preferences and production functions.

5. What are the advantages of using vector spaces of functions defined by a condition?

One of the main advantages of using vector spaces of functions defined by a condition is that they provide a more specific and structured framework for studying functions. This allows for a deeper understanding of the functions and their properties. Additionally, vector spaces of functions can also simplify complex problems and make them more manageable to solve.

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