Complex periodic functions in a vector space

[jendrix]

Homework Statement

Consider the set V + {all periodic *complex* functions of time t with period 1} Draw two example functions that belong to V.

Show that if f(t) and g(t) are members of V then so is f(t) + g(t)

Homework Equations

The Attempt at a Solution

f(t) = e^(i*w₀*t))

g(t) =e^{(i*w₀*t +φ)})

Where W₀ = 2*πUsing 'Euler's' I can write these as;

f(t) =cos(w₀*t) + i*sin(w₀*t)

g(t) =cos(w₀*t +φ) + i*sin(w₀*t +φ)So for part a) I would plot these functions separating the real and imaginary parts and choosing a value for φ to illustrate the phase shift?Partb) f(t) + g(t) = e^(i*w₀*t)) + e^{(i*w₀*t +φ)})

= (1 + e^φ) * e^i*w₀*t

The signal remains a periodic complex function of t with a period of 1 and is therefore a member of V.Thanks

 

[PeroK]

What makes you think the functions are complex valued? Is that in the question? I ask because that make them hard to draw.

Are all complex functions of the form $e^{iwt}$?

 

[pasmith]

Part (2) is asking you to prove the result for all possible choices of f and g, not just your two example functions.

 

[jendrix]

What makes you think the functions are complex valued? Is that in the question? I ask because that make them hard to draw.

Are all complex functions of the form $e^{iwt}$?

Sorry, I missed the complex part out of the question, I have edited my post now

 

[PeroK]

Sorry, I missed the complex part out of the question, I have edited my post now

Okay. What's the definition of a periodic function?

 

[jendrix]

What makes you think the functions are complex valued? Is that in the question? I ask because that make them hard to draw.

Are all complex functions of the form $e^{iwt}$?

I thought all periodic ones were, or could be represented by the exponential equivalent?

 

[jendrix]

Okay. What's the definition of a periodic function?

A function that repeats itself over a set period

 

[PeroK]

I thought all periodic ones were, or could be represented by the exponential equivalent?

Not at all!

A function that repeats itself over a set period

Can you express that mathematically? In this case for a function of period $1$.

 

[jendrix]

Not at all!
Can you express that mathematically? In this case for a function of period $1$.

Sin(wt) ≡ Sin(wt+t) ?

 

[PeroK]

Sin(wt) ≡ Sin(wt+t) ?

Come on, that's not a definition of anything! A definition would have to be something like:

$f$ is periodic with period $1$ if ...

 

[jendrix]

f is periodic with a period of 1 if ... Sin(2π*t) ≡Sin(2πt+T)

 

[PeroK]

f is periodic with a period of 1 if ... Sin(2π*t) ≡Sin(2πt+T)

To be honest, it's difficult to know what to say to that. It suggests that you lack some basic understanding of mathematics. $\sin$ is an example of a periodic function, but in no way is it the only periodic function or the definition of a periodic function.

https://en.wikipedia.org/wiki/Periodic_function

 

[jendrix]

To be honest, it's difficult to know what to say to that. It suggests that you lack some basic understanding of mathematics. $\sin$ is an example of a periodic function, but in no way is it the only periodic function or the definition of a periodic function.

https://en.wikipedia.org/wiki/Periodic_function

Sorry, I was using Sin as an example, I appreciate it is not the only periodic function. Therefore, a function f(x) is said to be periodic if f(x) = f(x+P)

 

[jendrix]

h(t+1) = f(t+1) + g(t+1) =f(t)+g(t)=h(t)

Would I use a general form for a complex periodic function? And prove the above using that?Thanks

 

[PeroK]

h(t+1) = f(t+1) + g(t+1) =f(t)+g(t)=h(t)

Would I use a general form for a complex periodic function? And prove the above using that?Thanks

Yes, exactly.

 

[jendrix]

Yes, exactly.

Would the general form be related to the complex Fourier series?Thanks

 

[PeroK]

Would the general form be related to the complex Fourier series?Thanks

Absolutely nothing to do with Fourier series. This is an algebraic result and requires no analytical structure whatsoever.

 

[jendrix]

Absolutely nothing to do with Fourier series. This is an algebraic result and requires no analytical structure whatsoever.

Ah, I though as Fourier series was for representing a periodic signal I could use the C_ke^(-iwt) as a model for a complex periodic signalWould I instead set

f(t) =A*e^{i(w_ot +θ₁})

g(t) =B*e^{i(w_ot +θ₂})And set about proving h(t+1) = f(t+1) + g(t+1) =f(t)+g(t)=h(t) ?
Thanks

 

[PeroK]

Ah, I though as Fourier series was for representing a periodic signal I could use the C_ke^(-iwt) as a model for a complex periodic signalWould I instead set

f(t) =A*e^{i(w_ot +θ₁})

g(t) =B*e^{i(w_ot +θ₂})And set about proving h(t+1) = f(t+1) + g(t+1) =f(t)+g(t)=h(t) ?
Thanks

You still have a fundamental misunderstandinhg of what is a periodic function. It has absolutely nothing to do with sines, cosines, exponentials, Fourier Series or whatever else. Periodicity is a simple algebraic property:

A function, $f$ is periodic with period $P$ if $\forall x \ f(x+P) = f(x)$. That is it. And that is all you can use.

Here's an example:

$f(x) = 1$ when $x$ is an integer, and $f(x) = 0$ otherwise.

$f$ is periodic with period $1$, but is clearly not a sine, cosine or exponential.

So, if you tried to prove that all periodic functions are of the form $exp(iwt)$ then you would be wrong, as the function above demonstrates.

I misunderstood your post #14, which I thought was a proof for general periodic functions, $f$ and $g$. Post #14 is essentially a valid proof of the result. So, your misunderstanding extends to not recognising a proof even when you've done it!

Can you see why post #14 is a proof? And why there is nothing more to do, other to to say more formally what you doing?

 

[jendrix]

Can you see why post #14 is a proof? And why there is nothing more to do, other to to say more formally what you doing?

It's definitely becoming clear now, the complex part was confusing me though, is it possible that while h(t+1) = f(t+1) + g(t+1) =f(t)+g(t)=h(t) holds true, is it possible that h(t) could no longer be a complex function and thus not a part of V?

Say if f(t) =-g(t) then h(t) would still = h(t+T) but would it still be considered a complex function?Thanks

 

[PeroK]

It's definitely becoming clear now, the complex part was confusing me though, is it possible that while h(t+1) = f(t+1) + g(t+1) =f(t)+g(t)=h(t) holds true, is it possible that h(t) could no longer be a complex function and thus not a part of V?

Say if f(t) =-g(t) then h(t) would still = h(t+T) but would it still be considered a complex function?Thanks

Real numbers are complex numbers too. You can't disqualify a complex function because it only takes real values.

 

[PeroK]

h(t+1) = f(t+1) + g(t+1) =f(t)+g(t)=h(t)

Would I use a general form for a complex periodic function? And prove the above using that?Thanks

Let me show you how to take what you have and make it more formal:

Let $f$ and $g$ be periodic functions of period $1$, and $h = f + g$. Now:

$\forall t \ h(t+1) = f(t+1) + g(t+1) = f(t) + g(t) = h(t)$

Hence, $h$ is periodic with period $1$. QED

Note that complex numbers don't come into it. It doesn't matter what the range of the functions is, as long as adding functions is well defined.

 

[jendrix]

Real numbers are complex numbers too. You can't disqualify a complex function because it only takes real values.

So all functions can be considered complex but only those that satisfy x(t) = x(t+T) can be considered periodic. So for part a) I can draw any periodic function with a period of T=1?

 

[PeroK]

So all functions can be considered complex but only those that satisfy x(t) = x(t+T) can be considered periodic. So for part a) I can draw any periodic function with a period of T=1?

Yes, you could make it simple by keeping the functions real (a subset of complex) or perhaps imaginary; otherwise they are hard to draw. It's a strange question to ask you to draw a complex function.

 

[jendrix]

Hello, I have a further two questions on this topic and I was wondering if you could check my answers.

Qc) Suppose f(t) belongs to V. For a complex number c, show that h(t) =cf(t) also belongs to V.

Answer

Let f be a periodic function of period 1 and h =cf where c is a complex number.

∀t h(t+1) =cf(t+1) =cf(t) = h(t)

Hence h is periodic with period 1 QED.

Qd)
Is V with the above addition and scaling operations a vector space?

Would I need to go through the axioms to prove that V was definitely a vector space?Thanks

 

[PeroK]

Qc) Suppose f(t) belongs to V. For a complex number c, show that h(t) =cf(t) also belongs to V.

Answer

Let f be a periodic function of period 1 and h =cf where c is a complex number.

∀t h(t+1) =cf(t+1) =cf(t) = h(t)

Hence h is periodic with period 1 QED.

Yes, that's it.

Qd)
Is V with the above addition and scaling operations a vector space?

Would I need to go through the axioms to prove that V was definitely a vector space?

It depends. If you can assume that the set of all complex functions is a vector space, then you need only show that addition and scalar multiplication are closed on the subset you are dealing with (in this case functions with period 1).

This is true for vector spaces in general:

If $U$ is a subset of a vector space $V$ and:

$u, v \in U$ and $c \in \mathbb{C} \ \Rightarrow u + v \in U$ and $cu \in U$

then $U$ is a vector subspace. Most axioms are met directly because the axioms hold for $V$, but you might like to think about why the above ensures that the $0$ vector is in $U$ and also why it ensures that $u \in U \Rightarrow -u \in U$. And, you might like to look at the other axioms as well and convince yourself why they must be met for any subset of a vector space.

With that assumption, you have shown that the set of functions of period 1 is a vector space.

 

[jendrix]

Yes, that's it.
It depends. If you can assume that the set of all complex functions is a vector space, then you need only show that addition and scalar multiplication are closed on the subset you are dealing with (in this case functions with period 1).

This is true for vector spaces in general:

If $U$ is a subset of a vector space $V$ and:

$u, v \in U$ and $c \in \mathbb{C} \ \Rightarrow u + v \in U$ and $cu \in U$

then $U$ is a vector subspace. Most axioms are met directly because the axioms hold for $V$, but you might like to think about why the above ensures that the $0$ vector is in $U$ and also why it ensures that $u \in U \Rightarrow -u \in U$. And, you might like to look at the other axioms as well and convince yourself why they must be met for any subset of a vector space.

With that assumption, you have shown that the set of functions of period 1 is a vector space.

Is it the case that as the addition and scaling axioms are present in U , these can then be used to find to -u by scaling by a factor of (-1) and the 0 vector can found by summing u + (-u) = 0 ?

 

[PeroK]

Is it the case that as the addition and scaling axioms are present in U , these can then be used to find to -u by scaling by a factor of (-1) and the 0 vector can found by summing u + (-u) = 0 ?

Not quite. The zero vector is in $U$ because $0u = 0 \in U$ for $u \in U$.

If you were a pure maths student, at this point you might jump up and point out that means you need at least one $u \in U$. And, therefore, $U$ must be a non-empty subset of $V$ in order to be a vector space.

 

[jendrix]

Not quite. The zero vector is in $U$ because $0u = 0 \in U$ for $u \in U$.

If you were a pure maths student, at this point you might jump up and point out that means you need at least one $u \in U$. And, therefore, $U$ must be a non-empty subset of $V$ in order to be a vector space.

Is it a fair assumption to make to say that the set of all complex functions is a vector space?