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Fourier Series: part 1


Taylor Series

You are probably already familiar with Taylor series. Taylor series is an approximate expansion of a function around x0x_0 using polynomials. It was usually recognized as:

f(x)=n=0an(xx0)n(E1)\begin{align*} \tag{E1} f(x) = \sum_{n=0}^\infty a_n (x-x_0)^n \end{align*}

The property of Taylor series is that it is a local approximation. So the function ff can be replaced by the series as long as it converges and the function has derivatives. Each of the coefficient of the series ana_n, corresponds to the value of the n-th derivative at point x0x_0.

f(n)(x0)n!=an(E2)\begin{align*} \tag{E2} \frac{f^{(n)}(x_0)}{n!} = a_n \end{align*}

In other words, the Taylor series expansion can be understood as a set of the derivative’s value that defines the function.

But what if the n-th derivative can be inferred from the function? For example, suppose that there exists a function whose its own derivatives?

A function whose its own derivatives

Making a statement: “function whose its own derivatives” is equivalent of defining function f(x)f(x) that is invariant under the derivative operator.

f(x)=f(x)=f(x)=f(n)(x)f(x) = f'(x) = f''(x) = f^{(n)}(x)

Suppose that it has a Taylor expansion. Since derivative operator is linear, we can operate on it term by term. The derivative of the n-th term:

Dx[an(xx0)n]=ann(xx0)n1=an1(xx0)n1D_x \left[ a_n\, (x-x_0)^n \right]= a_n\, n\, (x-x_0)^{n-1} = a_{n-1} \, (x-x_0)^{n-1}

We now have a recurrent relation:

ann=an1a_n \, n = a_{n-1}

However, at n=0n=0, then a0=1a_0=1. This is because the value of the derivative at xx0=0x-x_0=0 must be the same for any n-th derivative. It is the smallest integer possible. Consequently:

an=an1n=1n!a_n = \frac{a_{n-1}}{n} = \frac{1}{n!}

Thus we found our special function:

exp(x)=n=0xnn!\exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}

This function is so special, we called it the “exponential” function. Since it also exhibits properties of exponential form:


We will elaborate this on a separate article. But for now, we realized that there exists a function like this.

Now, what if we replaced xx with exp(x)\exp(x). What is this function?

f(x)=n=0(ex)nn!=n=0enxn!f(x) = \sum_{n=0}^\infty \frac{(e^x)^n}{n!} = \sum_{n=0}^\infty \frac{e^{nx}}{n!}

This would of course blow up to infinity since exe^x would be a very big number if xx is positive. But what about this function?

f(x)=n=0(ex)nn!=n=0enxn!f(x) = \sum_{n=0}^\infty \frac{(e^{-x})^n}{n!} = \sum_{n=0}^\infty \frac{e^{-nx}}{n!}

It will converge really fast.

A function with alternating derivatives

Note that if we swap xx with x-x. We will have this function f(x)=exf(x)=e^{-x}.

The Taylor series will have its terms alternating between positive and negative. The n-th derivative will be:

f(n)(x)=Dxnex=(1)nf(x)Dx1ex=f(x)Dx2ex=f(x)Dx3ex=f(x)Dx4ex=f(x)\begin{align*} f^{(n)}(x)=D_x^n\, e^{-x} &= (-1)^n \, f(x) \\ D_x^1\, e^{-x} &= -f(x) \\ D_x^2\, e^{-x} &= f(x) \\ D_x^3\, e^{-x} &= -f(x) \\ D_x^4\, e^{-x} &= f(x) \\ \end{align*}

As you can see, the sign alternates for every multiple of 2.

If we swap xx with ixix or ix-ix. With ii is the imaginary number unit. We will have alternating derivatives with every multiple of 4.

f(n)(x)=Dxneix=(i)nf(x)Dx1eix=if(x)Dx2eix=f(x)Dx3eix=if(x)Dx4eix=f(x)\begin{align*} f^{(n)}(x)=D_x^n\, e^{ix} &= (i)^n \, f(x) \\ D_x^1\, e^{ix} &= if(x) \\ D_x^2\, e^{ix} &= -f(x) \\ D_x^3\, e^{ix} &= -if(x) \\ D_x^4\, e^{ix} &= f(x) \\ \end{align*}

Fourier Series

We will now began to extend our ideas into something that is known as “Fourier Series”. Instead of swapping xx with eixe^{ix} and confuse ourselves, we renamed the parameter to perform substitution x=reitx=r\,e^{i t}

Suppose that we have a well-behaved Taylor Series of a function f(x)f(x) around x0=0x_0=0. let us swap x=reitx=r\,e^{it} and tries to see what happened. We will have:

f(x)=f(reit)=n=0anrneintn!f(x)=f(re^{i t}) = \sum_{n=0}^\infty a_n\frac{r^n e^{i\,n t}}{n!}

If we use 1x\frac{1}{x}, conveniently we have:

f(1x)=f(r1eit)=n=0anrneintn!f(\frac{1}{x})=f(r^{-1} e^{-i t})=\sum_{n=0}^\infty a_n \frac{r^{-n} e^{-i\,n t}}{n!}

Now, we might be tempted to perform substitution n=k-n=k. This series is similar with f(x)f(x), except the indices comes from the negative integers. The only reason we won’t be able to do that is because (k)!(-k)! doesn’t make sense.

So let’s use the property of derivative to see the Taylor expansion. One nice property of function eite^{-it} is that it is differentiable everywhere. Keeping in mind that the derivative will be around f(0)f(0), we tried to simplify our derivative expression like below. If we use the full chain, it will be very long.

Dtnf(r1eit)=f(n)(0)(i)nr1eitD_t^n \, f(r^{-1} e^{-it}) = f^{(n)}(0) \, (-i)^n\, r^{-1} e^{-it}

However Dtnf(r1eit)D_t^n \, f(r^{-1} e^{-it}) around x0=0x_0=0 is the coefficient of that Taylor series. If we directly match the coefficient, we will have:

f(n)(0)(i)n=ann!f^{(n)}(0) \, (-i)^n = \frac{a_n}{n!}

Now, if we want to make some kind of replacement for negative factorial (k)!(-k)! by doing substitution n=k-n=k, then the left hand side must also makes sense.

The term (i)k(-i)^{-k} can still make sense and defined.

(i)k=1(i)k=ik(ii)k=ik(-i)^{-k} = \frac{1}{(-i)^{k}} = \frac{i^k}{(-i \cdot i)^{k}} = i^k

But the interesting part is f(k)(0)f^{(-k)}(0). What does it mean to have a negative n-th derivative? You might guess intuitively that it is an anti-derivative. Suppose that f(n)(0)f^{(n)}(0) is defined when n=1n=-1. If F(x)F(x) is the anti-derivative of f(x)f(x), then F(x)F'(x) (the first derivative of F(x)F(x)), has to be f(x)f(x) by definition. So, in order for a function to have n-th derivative with negative nn, then it must have an anti-derivative. You can get the anti-derivative using integral.

In summary, the notion of this negative factorial can be completely replaced, if the function itself has anti-derivative, infinitely. Or in other words, you can integrate the function infinitely, and its values on x0=0x_0=0 is defined.

Here comes the good part. If a function have alternating derivative, like eite^{it}, consequently we can integrate it infinitely. The function eite^{it} and eite^{-it}, is a little bit more special because both of them have cycle of multiple of 4. This relation is different from exe^x (cycle of 1) and exe^{-x} (cycle of 2).

Now back to our previous series. We have justification to write ann!\frac{a_n}{n!} as AnA_n, another constants. Even though nn is a negative number.

We will then have this pair of series:

f(x)=f(reit)=n=0Anrneint(E3)\begin{align*} \tag{E3} f(x)=f(re^{i t}) = \sum_{n=0}^\infty A_n r^n e^{i\,n t} \end{align*}
f(1x)=f(r1eit)=k=0Bkrkeikt(E4)\begin{align*} \tag{E4} f(\frac{1}{x})=f(r^{-1} e^{-i t}) = \sum_{k=-\infty}^0 B_k r^{k} e^{i\,k t} \end{align*}

The above rearrangement is possible by swapping n=k-n=k and then renaming AnA_n to BkB_k. We got a series with the same term representation, but the index counts from negative infinity.

Adding those two together, we have a series that spans from -\infty to \infty, a much larger spans than the Taylor series!

f(x)+f(1x)=n=0Anrneint+k=0Bkrkeiktf(x)+f(1x)=n=(An+Bn)eint\begin{align*} f(x)+f(\frac{1}{x}) &= \sum_{n=0}^\infty A_n r^n e^{i\,n t} + \sum_{k=-\infty}^0 B_k r^{k} e^{i\,k t} \\ f(x)+f(\frac{1}{x}) &= \sum_{n=-\infty}^\infty (A_n + B_n) e^{i\,n t} \end{align*}

Now, let’s ponder a bit. If the right side converges, then any function that is invariant under its input reciprocals can be represented like this. In a much more general sense, even the coefficient can just be combined and renamed as CnC_n.

Let’s define a new function g(t)g(t), based on this series.

g(t)=n=Cneintg(t) = \sum_{n=-\infty}^\infty C_n \, e^{i\,n t}

We will call the right hand side, the Fourier Series with complex coefficient. This is because there are other equivalent representation of Fourier Series, but for now let’s just use this.

Example of Fourier Series

There is a very straightforward example of a function that is invariant under its input reciprocals.

Suppose that f(x)=f(1x)=x+x12f(x)=f(\frac{1}{x})=\frac{x+x^{-1}}{2}. Now suppose x=eitx=e^{it}. We have a function


This is actually a representation of the cos(t)\cos(t) function as complex exponent. If we express it using Fourier Series/Sums, we will have coefficient set of C1=12C_1=\frac{1}{2}, C1=12C_{-1}=\frac{1}{2}, and any other Cn=0C_n=0.

From this example, we can intuitively understand that not all function can be locally expressed as Taylor series, because by definition the series doesn’t have negative nn terms.

However, a Fourier series has index nn that spans from negative infinity to infinity. This made us possible to express function as linear combination of circular complex function.

Fourier Series as linear terms

From the form of the Fourier Series (and Taylor series as well), we can see that a function g(t)g(t) might be represented as infinite sums with each term from something like this CneintC_n \, e^{i\,n t}.

The term conveniently use the same index nn, as if it was a part of vector.

Let’s say that we can represent einte^{i\,n t} as a basis vector. As a notational convenience let’s write it as:

eintene^{i\,n t} \equiv \mathbf{e_n}

In this case, the right hand side ene_n was meant to be the “basis” e\mathbf{e} vector of index nn. Rather than ee as in the exponential function.

Then CnC_n is also a component of vector C\mathbf{C}.

So, it “might” behaves like a vector, but with one important detail. Usually a vector have finite components, or finite index nn. However, if we thought of Fourier Series as vector, then it has infinite components nn.

Interestingly, when we see term CneintC_n \, e^{i\,n t}, there is no real preference whether if we choose C\mathbf{C} as the basis, or if we choose e\mathbf{e} as the basis. If we choose complex number CnC_n as the components of the basis, then it is a basis that is fixed. It doesn’t evolve over parameter tt (which conveniently can be thought of as time). If we choose complex number einte^{i\,n t} as the components of the basis, then it is a basis that is evolving over input parameter time tt.

All the above notation can also be expressed in terms of Tensor notation. But, this is a much more advanced topic, and need more grounds to cover. However, I just think it was worth to mention that if we use Einstein summation convention, we can write tensor g(t)g(t) as CνeνC_\nu \, e^\nu or CνeνC^\nu \, e_\nu

Inner product within Fourier Series basis

From the analogy of using vectors, given terms such as CνeνC^\nu \, e_\nu, or CnenC_n\, \mathbf{e_n}. It is possible to recover CnC_n using dot product (or inner product, in a more abstract term).

Let’s take a look at simple example of 3D vectors, with basis e0=i^\mathbf{e_0}=\mathbf{\hat{i}}, e1=j^\mathbf{e_1}=\mathbf{\hat{j}}, e2=k^\mathbf{e_2}=\mathbf{\hat{k}}. A vector A\mathbf{A} can be written as

A=n=02Anen=A0i^+A1j^+A2k^\mathbf{A} = \sum_{n=0}^2 A_n \, \mathbf{e_n} = A_0 \, \mathbf{\hat{i}} + A_1 \, \mathbf{\hat{j}} + A_2 \, \mathbf{\hat{k}}

The result of a dot product of A\mathbf{A} with one of the basis, like i^\mathbf{\hat{i}}, will result in its component. This is because a dot product of orthogonal basis will result in 11 if the index is the same, and 00 otherwise. As an example:

i^i^=1i^j^=0j^k^=0\begin{align*} \mathbf{\hat{i}} \cdot \mathbf{\hat{i}} &= 1 \\ \mathbf{\hat{i}} \cdot \mathbf{\hat{j}} &= 0 \\ \mathbf{\hat{j}} \cdot \mathbf{\hat{k}} &= 0 \end{align*}

So, expanding a dot product of A\mathbf{A} with i^\mathbf{\hat{i}}, will result something like this:

Ai^=(A0i^+A1j^+A2k^)(i^)=A0i^i^+A1j^i^+A2k^i^=A0\begin{align*} \mathbf{A} \cdot \mathbf{\hat{i}} &= ( A_0 \, \mathbf{\hat{i}} + A_1 \, \mathbf{\hat{j}} + A_2 \, \mathbf{\hat{k}}) \cdot (\mathbf{\hat{i}}) \\ &= A_0 \, \mathbf{\hat{i}} \cdot \mathbf{\hat{i}} + A_1 \, \mathbf{\hat{j}} \cdot \mathbf{\hat{i}} + A_2 \, \mathbf{\hat{k}} \cdot \mathbf{\hat{i}} \\ &= A_0 \end{align*}

We got the corresponding component of the basis i^\mathbf{\hat{i}}. Which is A0A_0.

Now we want to know if Fourier Series has similar properties. We have two problem at the moment. First, a Fourier Series has infinite basis like what we understand before. Second, the basis evolve over time. Let’s tackle the problem one by one.

Assume the position and direction of the basis at specific time tt from the function of g(t)g(t). The term CnenC_n \, \mathbf{e_n} has component CnC_n and basis en\mathbf{e_n}. Suppose we want to recover only the component from multiplicative rule. Then, since the basis is a complex number, we must multiply it with the reciprocal of the number, so that we got 11.

einteint=1enen=1\begin{align*} e^{int} \cdot e^{-int} &= 1 \\ \mathbf{e_n} \cdot \mathbf{e_{-n}} &= 1 \end{align*}

From this, we got some general rule that if any Fourier basis en\mathbf{e_n} has its inverse basis en\mathbf{e_{-n}} such that its product were 1.

So, if we want to retrieve the component of C1C_1, we can start by multiplying the function with e1\mathbf{e_{-1}}.

g(t)e1=nCnene1=C1+n1Cnene1\begin{align*} g(t) \cdot \mathbf{e_{-1}} &= \sum_n C_n \, \mathbf{e_n} \cdot \mathbf{e_{-1}} \\ &= C_1 + \sum_{n \ne -1} C_n \, \mathbf{e_n} \cdot \mathbf{e_{-1}} \end{align*}

The problem now, if n1n \ne -1, then the basis still exists for the rests of the component! Let’s say if n=2n=2, then the basis for C2C_2 becomes ei(21)t=eite^{i(2-1)t} = e^{it}. This is not zero like the case in vector above. It is only shifted.

Now, since the basis evolve over time, we might want to evaluate the basis in the span of time TT. We want to see if we can do any specific transform in this time interval/period that causes the basis term to become zero and make the constant term CnC_n vanish for other n1n \ne -1.

If we transform the equation using derivative, the component C1C_1 that we figured out before, will vanish because it is a constant term. So, rather than using derivative/differential, we will try to use integral transform.

As a proposition, let’s say there exists interval TT, in which the terms we are trying to integrate becomes zero. Let’s take a look at the basis term, adjacent to n=1n=1, which is n=2n=2. The integral when we are trying to evaluate the component of CkC_k with k=1k=1 is:

Cneint=C2ei2tTCneinteiktdt=TC2ei2teitdt=TC2eitdt0=1iC2eitT=1iC2[eiTfinaleiTinitial]\begin{align*} C_n e^{int} &= C_2 e^{i2t} \\ \int_T C_n e^{int}\, e^{-ikt} dt &= \int_T C_2 e^{i2t} \, e^{-it} dt \\ &= \int_T C_2 e^{it} dt \\ 0 &= \frac{1}{i} C_2 e^{it} |_T = \frac{1}{i} C_2 \left[ e^{i T_{final}} - e^{i T_{initial}} \right] \\ \end{align*}

We can be sure from the definition of Fourier Series we established before that C2C_2, or rather in general CnC_n, were all coefficient without tt parameter. So the only way possible for the left hand side to become 0, is if the right hand side uses interval where eiTfinaleiTinitial=0e^{i T_{final}} - e^{i T_{initial}} = 0. This interval can be π-\pi to π\pi or 00 to 2π2\pi. To attain symmetry around 00, let’s use interval Tinitial=πT_{initial}=-\pi to Tfinal=πT_{final}=\pi.

Now note that the integral above can be successfully evaluated to zero because ei(nk)t=eite^{i(n-k)t}=e^{it}. But every possible (nk)(n-k) will be an integer. Also, it just really convenient that ei(nk)Tfinalei(nk)Tinitial=0e^{i (n-k) T_{final}} - e^{i (n-k) T_{initial}} = 0 for any integer (nk)(n-k). Except when nk=0n-k=0, something special with the integral happens.

For our target component which is nk=0n-k=0, the integral becomes:

Cneint=C1eitTCneinteiktdt=TC1eiteitdt=TC1dt=C1T\begin{align*} C_n e^{int} &= C_1 e^{it} \\ \int_T C_n e^{int}\, e^{-ikt} dt &= \int_T C_1 e^{it} \, e^{-it} dt \\ &= \int_T C_1 dt \\ &= C_1 \, T \\ \end{align*}

With this integral transform, the result were scaled by TT. So, if we want to have the result C1C_1, then the integral itself must be normalized by the span/interval TT.

From this result, we conclude that there exists specific integral transform such that we can extract the component CkC_k from the function using operation, just like an inner product. To summarize:

Ck=1TnNCnTei(nk)tdt=1TT(nNCneint)eiktdt=1TTg(t)eiktdt(E5)\begin{align*} \tag{E5} C_k &= \frac{1}{T} \, \sum_n^N C_n \int_T e^{i(n-k)t} \, dt \\ &= \frac{1}{T} \, \int_T \left( \sum_n^N C_n \, e^{int} \right) \, e^{-ikt} \, dt \\ &= \frac{1}{T} \, \int_T g(t) \, e^{-ikt} \, dt \\ \end{align*}

With some rearrangement, (and some more fundamental proof needed, regarding how we can swap sums and integral around), we obtain the kind of transform needed to extract the component/coefficient CkC_k.

As a test, let’s use it against cos(t)\cos(t) function that we happened to know has components for n=1n=-1 and n=1n=1.

C1=1TTcos(t)eitdt=1TTcos(t)eictdt\begin{align*} C_1 &= \frac{1}{T} \, \int_T \cos(t) \, e^{-it} \, dt \\ &= \frac{1}{T} \, \int_T \cos(t) \, e^{-ict} \, dt \\ \end{align*}

First, we must solve this indefinite integral. If we use integration by parts, the parts will cycle, so we can’t represent it as a closed form. We will slightly modify the integral by adding a parameter cc.

cos(t)eictdt=sin(t)eict+icsin(t)eictdt=sin(t)eicticcos(t)eict+c2cos(t)eictdt(1c2)cos(t)eictdt=sin(t)eicticcos(t)eictcos(t)eictdt=sin(t)eicticcos(t)eict1c2Tcos(t)eictdt=[sin(t)eicticcos(t)eict1c2]ππ\begin{align*} \int \cos(t) \, e^{-ict} \, dt &= \sin(t)\, e^{-ict} + i c\int \sin(t) \, e^{-ict} \, dt \\ &= \sin(t)\, e^{-ict} - i c \cos(t) \, e^{-ict} + c^2 \int \cos(t) \, e^{-ict} \, dt \\ (1-c^2) \int \cos(t) \, e^{-ict} \, dt &= \sin(t)\, e^{-ict} - i c \cos(t) \, e^{-ict} \\ \int \cos(t) \, e^{-ict} \, dt &= \frac{\sin(t)\, e^{-ict} - i c \cos(t) \, e^{-ict}}{1-c^2} \\ \int_T \cos(t) \, e^{-ict} \, dt &= \left[ \frac{\sin(t)\, e^{-ict} - i c \cos(t) \, e^{-ict}}{1-c^2} \right]_{-\pi}^{\pi} \\ \end{align*}

If we evaluate above expression by substituting c=1c=1, we got indeterminate form 00\frac{0}{0}. We will use L’Hôpital’s rule. We took the derivative of numerator and denumerator with respect to cc. Then substitute c=1c=1.

Tcos(t)eictdt=[itsin(t)eicticos(t)eictctcos(t)eict2c]ππ=[t2cos(t)eit+ieit2(tsin(t)+cos(t))]ππ=π\begin{align*} \int_T \cos(t) \, e^{-ict} \, dt &= \left[ \frac{-it\sin(t)\, e^{-ict} - i \cos(t) \, e^{-ict} - ct \cos(t) \, e^{-ict}}{-2c} \right]_{-\pi}^{\pi} \\ &= \left[ \frac{t}{2} \cos(t) e^{-it} + \frac{ie^{-it}}{2} \left( t \sin(t) + \cos(t) \right) \right]_{-\pi}^{\pi}\\ &= \pi \end{align*}

We use it back to the previous expression:

C1=1TTcos(t)eitdt=12ππ=12\begin{align*} C_1 &= \frac{1}{T} \, \int_T \cos(t) \, e^{-it} \, dt \\ &= \frac{1}{2\pi} \, \pi = \frac{1}{2} \\ \end{align*}

The result is the same with what we have in the previous example.


This article concluded the following section about Fourier Series. Stepping up from the ideas of Taylor Series/Expansion, we want to find similar Series by substituting x=reitx=re^{it}. Although it doesn’t guarantee (yet) that such substitution preserves the function, we rediscovered the concept of Fourier Series.

For now, the construction is rather from bottom to top. By combining multiple periodic function with basis einte^{i\,n\,t}, we can represent it as function with a nice property.

The construction seems to be in opposite direction of Taylor Series.

For a Taylor Series/Expansion to work, we evaluate a function locally at point x0x_0, then we can take the coefficient from its subsequent derivative at that point.

The Fourier Series on the other hand, kind of like the opposite. Given a periodic function in the span of TT, if we integrate it globally over each value of tt within TT, we can get the coefficient needed to construct the series.

Rizky Maulana Nugraha

Written by Rizky Maulana Nugraha
Software Developer. Currently remotely working from Indonesia.
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