Determine if function forms a vector space

In summary: The thing you need to focus on is that a vector space must be closed under addition and multiplication by a scalar. That is what you need to show for this problem.
  • #1
Carson
2
0

Homework Statement


Problem- Determine if the set of all function y(t) which have period 2pi forms a vector space under operations of function addition and multiplication of a function by a constant.

What I know- So I know this involves sin, cos, sec, and csc. Also I know that a vector space must be closed under addition and multiplication by a scalar. I'm aware of how to prove polynomials are vector spaces, but not any other function.

Homework Equations


u+v ∈ V
c*u ∈ V

The Attempt at a Solution


[/B]
u=a1*sin(t)+a2*cos(t)+a3*sec(t)+a4*csc(t)
v=b1*sin(t)+b2*cos(t)+a3*sec(t)+a4*csc(t)
 
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  • #2
Carson said:

Homework Statement


Problem- Determine if the set of all function y(t) which have period 2pi forms a vector space under operations of function addition and multiplication of a function by a constant.

What I know- So I know this involves sin, cos, sec, and csc. Also I know that a vector space must be closed under addition and multiplication by a scalar. I'm aware of how to prove polynomials are vector spaces, but not any other function.

Homework Equations


u+v ∈ V
c*u ∈ V

The Attempt at a Solution


[/B]
u=a1*sin(t)+a2*cos(t)+a3*sec(t)+a4*csc(t)
v=b1*sin(t)+b2*cos(t)+a3*sec(t)+a4*csc(t)
You should forget the periodic functions you know for the moment. All you have here is a general periodic function. That is a function ##y(t)## for which ##y(t+2\pi)=y(t)## for all ##t\in \mathbb{R}##. Now you have to show that the two properties mentioned under 2. hold.
 
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  • #3
So does it make sense to say the following in order to prove this set is a vector space:

f(t+2π) + g(t+2π)= (f+g)(t+2π) = f(t) + g(t) = (f+g)(t)

(cf)(t+2π) = c [ f(t+2π) ] = c [f(t)] = (cf) (t)
 
  • #4
Carson said:
So does it make sense to say the following in order to prove this set is a vector space:

f(t+2π) + g(t+2π)= (f+g)(t+2π) = f(t) + g(t) = (f+g)(t)

(cf)(t+2π) = c [ f(t+2π) ] = c [f(t)] = (cf) (t)
This is close, but should be better organized in the first part of your proof.
Let's define U as ##U = \{f | f(t + 2\pi) = f(t)\}##, and assume that ##f, g \in U##.
Start with ##(f + g)(t + 2\pi)## and show that this equals ##(f + g)(t)##. This shows that when f and g are in U, then f + g is also in U; IOW, it shows that set U is closed under vector addition.
The other part looks fine, and shows that for a function f in U, then cf is also in U; i.e, that U is closed under scalar multiplication.

I agree with what @fresh_42 said about forgetting about the periodic functions that you already know. There are lots more periodic functions than the ones you listed. For example, there are sawtooth functions that are periodic with period ##2\pi## and lots more.
 
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Related to Determine if function forms a vector space

1. What is a vector space?

A vector space is a mathematical structure that consists of a set of elements called vectors, along with operations such as addition and scalar multiplication. It is a fundamental concept in linear algebra and is used to describe real-world phenomena such as forces, velocities, and electrical fields.

2. How do you determine if a function forms a vector space?

To determine if a function forms a vector space, we need to check if it satisfies the 10 axioms of vector space. These axioms include closure under addition and scalar multiplication, associativity, commutativity, distributivity, and the existence of an identity element and inverse elements.

3. Can a single function form a vector space?

Yes, a single function can form a vector space as long as it satisfies all the necessary axioms. For example, the set of all linear functions is a vector space under the operations of function addition and scalar multiplication.

4. What happens if a function fails to satisfy one of the vector space axioms?

If a function fails to satisfy even one of the vector space axioms, then it cannot be considered a vector space. This means that it would not have all the necessary properties to be classified as a vector space, and therefore, it cannot be used to describe real-world phenomena in a consistent and accurate manner.

5. Are there different types of vector spaces?

Yes, there are different types of vector spaces, such as finite and infinite-dimensional vector spaces, real and complex vector spaces, and inner product spaces. These different types have specific properties and structures that make them useful for different applications in mathematics and other fields.

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