Functions forming a vector space

In summary, the conversation discussed the properties of functions that vanish at the endpoints x=0 and L=0 and whether they form a vector space. It was determined that they do form a vector space, but only if the functions also have a null vector. The concept of a periodic function was also brought up, with the conclusion that a function with a period of 0 is possible but not very useful. Finally, it was noted that the set of functions that obey f(0)=4 does not form a vector space due to a lack of closure.
  • #1
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Homework Statement


1.1.3
1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
2) How about periodic functions? obeying f(0)=f(L) ?
3) How about functions that obey f(0)=4 ?

If the functions do not qualify, list what go wrong.

Homework Equations


upload_2018-8-12_22-22-57.png


The Attempt at a Solution


1) Considering a set of functions which vanish at the end points x = {0,L}. Let's say f,g,h belong to this set of functions. Then all the properties in the above image are verified. So, this set of functions form a vector space.
2) Similarly, for 2 also all properties get verified except existence of a null vector. Can a function h(x) = 0 for all x, be a periodic function?
3) For 3, this set of functions do not have closure feature. So, it does not form a vector space.

Is this correct?
 

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  • #2
Technically, the zero function is periodic for any given period. In particular, it has the same value at the end points.
 
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  • #3
Pushoam said:

Homework Statement


1.1.3
1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
2) How about periodic functions? obeying f(0)=f(L) ?
3) How about functions that obey f(0)=4 ?

If the functions do not qualify, list what go wrong.

Homework Equations


View attachment 229213

The Attempt at a Solution


1) Considering a set of functions which vanish at the end points x = {0,L}. Let's say f,g,h belong to this set of functions. Then all the properties in the above image are verified. So, this set of functions form a vector space.
2) Similarly, for 2 also all properties get verified except existence of a null vector. Can a function h(x) = 0 for all x, be a periodic function?
3) For 3, this set of functions do not have closure feature. So, it does not form a vector space.

Is this correct?

You need to have confidence in your own reasoning. (i) Have you verified all the defining properties of vector spaces in (1)-(2)? (ii) Have you shown that some vector-space property fails in (3)?
 
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  • #4
Pushoam said:

Homework Statement


1.1.3
1) Do functions that vanish at the endpoints x=0 and L=0 form a vector space?
2) How about periodic functions? obeying f(0)=f(L) ?
3) How about functions that obey f(0)=4 ?

If the functions do not qualify, list what go wrong.

Homework Equations


View attachment 229213

The Attempt at a Solution


1) Considering a set of functions which vanish at the end points x = {0,L}. Let's say f,g,h belong to this set of functions. Then all the properties in the above image are verified. So, this set of functions form a vector space.
2) Similarly, for 2 also all properties get verified except existence of a null vector. Can a function h(x) = 0 for all x, be a periodic function?
What's the definition of a periodic function?
Pushoam said:
3) For 3, this set of functions do not have closure feature. So, it does not form a vector space.

Is this correct?
 
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  • #5
Pushoam said:
1) Considering a set of functions which vanish at the end points x = {0,L}. Let's say f,g,h belong to this set of functions. Then all the properties in the above image are verified. So, this set of functions form a vector space.
The notion of the corresponding equation would have been almost shorter: ##(\alpha f + \beta g)(0)=\alpha f(0)+\beta g(0)=\alpha f(L)+ \beta g(L)=(\alpha f + \beta g)(L)##
2) Similarly, for 2 also all properties get verified except existence of a null vector.
Why this? ##0(0)=0=0(L)##
Can a function h(x) = 0 for all x, be a periodic function?
This is the extreme version of a periodic function, namely with period ##0##. There is no reason to exclude it.
3) For 3, this set of functions do not have closure feature. So, it does not form a vector space.
Yes, and again ##0(0)=0\neq 4##

Is this correct?
Yes, except that I couldn't see your reasoning why the null vector in section 2 shouldn't work.

Edit: Please do the following little exercise: A set which is closed under addition and scalar multiplication automatically contains zero.
 
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  • #6
Ray Vickson said:
You need to have confidence in your own reasoning. (i) Have you verified all the defining properties of vector spaces in (1)-(2)? (ii) Have you shown that some vector-space property fails in (3)?
Yes, I verified it in mind. I didn't write the verification here as it was easy.
 
  • #7
fresh_42 said:
This is the extreme version of a periodic function, namely with period 0.
Or any period.
If f(x) = 0 for all x, then f(x) = f(x + L), for any L.

The vector space properties in the attached image in post 1 is a bit unusual. The usual definition includes statements about closure under vector addition and closure under scalar multiplication.

Minor point: The usual definitions are of the properties these operations. I've never seen them called features.
 
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  • #8
fresh_42 said:
This is the extreme version of a periodic function, namely with period ##0##. There is no reason to exclude it.
Typically the period is required to be positive to exclude possibilities like claiming f(x)=x is periodic since f(x)=f(x+0) for all x.
 
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  • #9
vela said:
Typically the period is required to be positive to exclude possibilities like claiming f(x)=x is periodic since f(x)=f(x+0) for all x.
Makes sense. I thought the zero might be necessary for the set of all periods, or to get vector spaces for periodic functions, but "any" sounds better than "0" although "any" includes it.
 
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  • #10
fresh_42 said:
The notion of the corresponding equation would have been almost shorter: ##(\alpha f + \beta g)(0)=\alpha f(0)+\beta g(0)=\alpha f(L)+ \beta g(L)=(\alpha f + \beta g)(L)##Edit: Please do the following little exercise: A set which is closed under addition and scalar multiplication automatically contains zero.
Let's say that ## \alpha ## and ## \beta ## belongs to this set and a, b belong to the scalar field. Then, the set being closed under addition and scalar multiplication means that ## a \alpha + b \beta ## must belong to the set.
Taking ## b \beta = - a \alpha ## implies that 0 belongs to the set.

Thus, a set which is closed under addition and scalar multiplication automatically contains zero.
 
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1. What is a vector space?

A vector space is a mathematical concept that refers to a set of objects, called vectors, that can be added together and multiplied by a scalar to form new vectors. These operations follow specific rules and properties, such as commutativity and associativity, making vector spaces an important tool in many areas of mathematics and science.

2. How do functions form a vector space?

Functions can form a vector space when they satisfy a set of conditions, such as closure under addition and scalar multiplication, and the existence of an additive identity and inverse. These conditions ensure that the operations on the functions follow the same rules and properties as those of traditional vectors in a vector space.

3. What are the benefits of studying functions as a vector space?

Studying functions as a vector space allows us to apply concepts and tools from linear algebra, such as basis vectors and dimensionality, to functions. This can provide a deeper understanding of the properties and behavior of functions, and can also be useful in solving problems in various fields, such as physics, engineering, and computer science.

4. Can all types of functions form a vector space?

No, not all types of functions can form a vector space. For a set of functions to form a vector space, they must satisfy the conditions mentioned earlier, such as closure under addition and scalar multiplication. Functions that do not meet these conditions, such as discontinuous functions, cannot form a vector space.

5. How is the dimensionality of a vector space of functions determined?

The dimensionality of a vector space of functions is determined by the number of linearly independent functions in the set. This means that the number of functions in the set is the maximum number of functions that can be combined to create any other function in the space. The dimensionality can also be calculated using the rank-nullity theorem, which relates the dimensionality to the number of elements in the basis of the vector space.

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