# Properties of Fourier series

Here we will look at the usual properties we ask from series expansion (how it behaves with respect to the usual operations and how it reacts to transformations of functions). Then we look at alternative ways to write Fourier series, namely the amplitude - phase angle form and the complex form.

We start with a result that is not related to the main topic of this section, but it is sometimes useful, so we quickly go through it.

Theorem.
Let T > 0, denote ω = 2π/T.
For any T-periodic function f that is integrable on [0,T ], the Fourier coefficients ak and bk tend to zero.

Now we turn to investigation of operations. First one fairly obvious result that follows from the fact that coefficients of Fourier series are given by integrals, that is, by a linear process.

Theorem.
Let T > 0, denote ω = 2π/T.
Let c be any real number, let f and g be T-periodic functions that are integrable on [0,T ]. Assume that

Then

Now we would like to determine similar formulas for derivative and integral. With derivative it is simple. Assume that f is T-periodic (and integrable) and has Fourier series with coefficients ak and bk. Denote the coefficients of the Fourier series of the derivative f ′ by Ak and Bk. Just by using periodicity it follows that

The integrals for the other coefficients can be changed into integrals featuring f using integration by parts.

Similarly we evaluate bk and obtain the following statement.

Theorem.
Let T > 0, denote ω = 2π/T.
Let f be a T-periodic function integrable on [0,T ]. Assume that

Then

So we see that we can differentiate both sides of the "tilde relation" (the Fourier series we differentiate term by term) and the relation stays valid. How about a true equality? If we use the usual Jordan conditions (but applied to f ′), we get the following implication:

Assume that f is piecewise continuous and has the first and second derivative that are piecewise continuous. Then

That is, we can differentiate a convergent Fourier series term by term. Now we turn to integration. We again start with formal assignment of Fourier series, that is, assume that f is T-periodic (and integrable) and has Fourier series with coefficients ak and bk. Assume that it has an antiderivative F (which need not always exists even for a piecewise continuous function, see for instance this example), denote coefficients of its Fourier series of by Ak and Bk. Here we have a serious problem, this antiderivative need not be T-periodic. To see this, consider one special antiderivative, namely (see The Fundamental Theorem of Calculus)

Now consider some t from the interval [0,T ) and an integer k. Using the fact that f is T-periodic we get the following.

We see that this antiderivative F is T-periodic exactly if the integral of f over the basic interval is 0; this actually means that a0 = 0. Since other antiderivatives are just shifts of this one, we get the following observation.

Antiderivatives of f are T-periodic exactly if a0 = 0.

Under this assumption we may start asking about the Fourier series assigned to F. This time coefficients are given as integrals with F and we use integration by parts to pass to integrals with f, then we use periodicity of f and also the fact that our special F has F(T ) = F(0) = 0.

Similarly we evaluate the integral for Bk. Thus we get the following statement.

Theorem.
Let T > 0, denote ω = 2π/T.
Let f be a T-periodic function continuous on [0,T ]. Assume that

If a0 = 0, then for the antiderivative given by

we have

This tells us about one particular antiderivative. Other antiderivatives differ by a constant, so if we consider the set of all of them (the indefinite integral), we can write the conclusion in the following way:

How about true equality instead of formal assignment? Again, we will use Jordan's conditions. We need to guarantee that an antiderivative exists, for which the natural assumption is that f is continuous. The antiderivative F in the above theorem is then also continuous, therefore F and its derivative f satisfy the "better" version of assumptions in Jordan's theorem and we actually get uniform convergence.

Instead of expressing this formally we will try something else, we look at definite integral (which is in a sense more general). Since we will not work with antiderivatives, we do not have to worry about their periodicity and thus we need not require that a0 = 0. Then we have the following statement.

Theorem.
Let T > 0, denote ω = 2π/T.
Let f be a T-periodic function that is piecewise continuous and integrable on [0,T ]. Assume that

Then for any a < b we have

Now we will look at how Fourier series reacts to transformations of f. Note that if f is a periodic function, then the usual transformations again yield periodic functions, and apart from scaling the variable they even preserve the original period.

Theorem (transformations).
Let T > 0, denote ω = 2π/T.
Let f be a T-periodic function that is integrable on [0,T ]. Assume that

Then for any non-zero real number c we have

The first statement we already saw above, it follows from linearity. Note that a0 is actually the average of f (multiplied by 2), so by scaling the variable (shrinking or expanding the function along the x-axis) or shifting the function in horizontal direction we do not change this average. However, if we shift the function up or down, the average changes accordingly. This explains what is happening to a0 in those formulas above. The other coefficients show how important individual frequencies are in f, which is something that does not change when we shift the function vertically or scale its variable, but by scaling the values of the function we obviously change importance of all frequencies. The last expression is more complicated, since we shift "waves in f" left or right, but the cosines and sines on the right do not get shifted, which makes things rather difficult. However, note that if c is a multiple of the basic period T, then in fact the above formulas show that f (t − c) has the same Fourier series as f. This is to be expected, since shifting a periodic function by a multiple of its period does not change it at all and if the above formula did not yield this result, it would have been wrong.

Note that this last formula looks much better in complex form, see below.

## Amplitude - phase angle form of Fourier series

The basic Fourier series can be rearranged to better suit one's needs. The first rearrangement that we will show here uses a trick that is fairly popular when working with waves (signals, electrical circuits etc.). Given numbers a and b, there is a certain angle φ and a number A such that for any x we have

The number A is called the amplitude and the angle φ is called the phase angle, they are given by the following formulas.

If we exchange sine and cosine in this definition of φ, we get a similar reduction, but this time with cosine on the right (the phase angle will be now different). If we apply this to all terms in the Fourier series, we get the following formulas.

## Complex form of Fourier series

Here we use a different trick. If we replace sines and cosines in a Fourier series by their equivalent expressions with exponentials, we get

Thus if we denote

we get the complex form of Fourier series

This form is actually the natural form of this series, since many formulas become much nicer. For instance, we do not have to find ck by doing several cases, there is a common formula for them all:

Indeed, for instance, for a positive integer k we have

Similarly, there are more convenient forms for the rules for transformations above.

In the last formula, n is obviously some integer.

Going a bit further in this direction would get us to the notion of Fourier transform, which is another story, so we'd better stop.