Determine if function forms a vector space

[Carson]

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)

 

[fresh_42]

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.

 

[Carson]

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)

 

[Mark44]

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.