Eli Bendersky's website - Math

Relative velocity and closing speed

2026-08-03T20:01:00-07:00

In Physics simulations or game engines it’s sometimes useful to determine the speed with which two objects are approaching each other. This post will discuss the concept of closing speed, which is the normal component of the relative velocity of two objects.

Relative velocity and its components

Suppose we have objects A and B [1] with velocity vectors \vec{V_A} and \vec{V_B}. The relative velocity of B w.r.t. A is:

\[\vec{V}_{B|A}=\vec{V}_B-\vec{V}_A\]

Put differently, it’s the velocity of B in A’s frame of reference.

This relative velocity is a vector, and we can split it into orthogonal components. Obviously, the nature of such a split depends on the basis we want to use. We could look at the vector’s x an y components (we’ll be using - two dimensional space, but everything here applies to 3D as well), but for this post we’re interested in something slightly different:

We draw a line connecting the two objects. The component of \vec{V}_{B|A} in the direction of this line is called the normal component of relative velocity, while the component perpendicular to this direction is called the tangential component.

Computing the normal component

How do we find the component of a vector in the direction of a specific line? By using a vector projection!. We’ll represent the line by a vector, and find the projection of \vec{V}_{B|A} onto this vector.

The positions of A and B can also be seen as vectors: \vec{P}_A and \vec{P}_B. The line connecting them can then be expressed as the vector \vec{P}_B-\vec{P}_A:

All we need from this position difference vector is its direction, not its magnitude, however [2]. So we’ll use the unit vector of \vec{P}_B-\vec{P}_A, denoted as:

\[\widehat{P}=\frac{\vec{P}_B-\vec{P}_A}{|\vec{P}_B-\vec{P}_A|}\]

Finally, to find the projection of \vec{V}_{B|A} onto \widehat{P}, we compute [3]:

\[S_c=\vec{V}_{B|A}\cdot\widehat{P}=(\vec{V}_B-\vec{V}_A)\cdot\widehat{P}\]

Where the multiplication operator between the vectors is the dot product. Note that the result of the dot product is a scalar; therefore, the quantity S_c is called the closing speed - it expresses the rate at which the relative distance of the two objects is changing. If it’s positive, the objects are drifting farther apart; if it’s negative, the objects are getting closer together. Therefore the term "closing speed" may be slightly confusing; alternatively, this has been called a "signed separation speed", or "normal relative speed" [4].

The signs in these calculations can be tricky to get right, so we have to be very careful. Let’s see a few examples that will help us make these computations more concrete.

Examples

To build up some intuition and get some practice with the equations, we’ll review the following examples:

Example I: We’ll start by computing the relative position unit vector \widehat{P}:

\[\widehat{P}=\frac{\vec{P}_B-\vec{P}_A}{|\vec{P}_B-\vec{P}_A|}=\frac{\langle 4,0\rangle}{|\langle 4,0 \rangle|}=\langle1,0\rangle\]

Then, the closing speed is:

\[S_c=(\vec{V}_B-\vec{V}_A)\cdot\widehat{P}=\langle -3,0\rangle\cdot\langle1,0\rangle=-3\]

Based on our convention, the negative sign of S_c means that the objects are approaching each other. Due to the simple nature of the example, this result is easy to verify, as it can be immediately guessed just by looking at the diagram.

Example II: Here \widehat{P} is the same as in the previous example. The closing speed is:

\[S_c=(\vec{V}_B-\vec{V}_A)\cdot\widehat{P}=\langle 3,0\rangle\cdot\langle1,0\rangle=3\]

Same magnitude, but different sign from before, because the objects are moving farther apart.

Example III This example is to demonstrate that we get consistent results even if B is to the left of A. Here the relative position unit vector is:

\[\widehat{P}=\frac{\vec{P}_B-\vec{P}_A}{|\vec{P}_B-\vec{P}_A|}=\frac{\langle -2,0\rangle}{|\langle -2,0 \rangle|}=\langle-1,0\rangle\]

And the closing speed:

\[S_c=(\vec{V}_B-\vec{V}_A)\cdot\widehat{P}=\langle 3,0\rangle\cdot\langle -1,0\rangle=-3\]

Which is the same as in example I, as expected. The direction of \widehat{P} flipped, but so did the direction of the relative velocity vector, so the result has the same sign.

Example IV: Finally, an example showing more arbitrary positions and velocities.

\[\widehat{P}=\frac{\vec{P}_B-\vec{P}_A}{|\vec{P}_B-\vec{P}_A|}=\frac{\langle 3,4\rangle}{|\langle 3,4 \rangle|}=\langle0.6,0.8\rangle\]

Then:

\[S_c=(\vec{V}_B-\vec{V}_A)\cdot\widehat{P}=\langle -3,-6\rangle\cdot\langle 0.6,0.8\rangle=-6.6\]

This example is a good opportunity to demonstrate something important about S_c: it’s time-dependent, because positions change with time. Here, -6.6 is the closing speed at the exact moment when A’s and B’s positions and velocities are as stated in the example. In the next time step, the position of A will be \langle2,5\rangle and the position of B will be \langle2,3\rangle, while their velocities remain the same. The S_c then will be quite different. This is a good segue to the next topic - which is a more physical view of closing speed.

Closing speed as a function of time

The computation shown so far represents a static view of the world; perhaps the right word to use is instantaneous. Given the positions and velocities at a given moment, what is the closing speed between the objects at that exact moment?

But there’s no reason to not generalize this using a more standard physical interpretation of velocity.

First, let’s state the position vectors of A and B as a function of time: \vec{P}_A(t) and \vec{P}_B(t). The relative position vector between the objects is also a function of time:

\[\vec{R}(t)=\vec{P}_B(t)-\vec{P}_A(t)\]

Now we’ll define the scalar distance as the magnitude of this vector:

\[r(t)=|\vec{R}(t)|=|\vec{P}_B(t)-\vec{P}_A(t)|\]

We’re interested in \frac{dr(t)}{dt} - the change in this distance over time. Let’s start by breaking R(t) down to its constituents:

\[\begin{aligned} \vec{R}(t)&=\langle x(t),y(t)\rangle \\ r(t)&=\sqrt{x(t)^2+y(t)^2} \end{aligned}\]

By the chain rule [5]:

\[\begin{aligned} \frac{dr(t)}{dt}&=\frac{\frac{d}{dt}(x(t)^2+y(t)^2)}{2\sqrt{x(t)^2+y(t)^2}}\\ &=\frac{2x(t)x'(t)+2y(t)y'(t)}{2\sqrt{x(t)^2+y(t)^2}} =\frac{x(t)x'(t)+y(t)y'(t)}{\sqrt{x(t)^2+y(t)^2}} \end{aligned}\]

Switching back to the vector representation: since \vec{R}(t)=\langle x(t),y(t)\rangle, the numerator of the fraction above is then a dot product between \vec{R}(t) and \vec{R}'(t), we can write this as:

\[\frac{dr(t)}{dt}=\frac{\vec{R}(t)\cdot\vec{R}'(t)}{|\vec{R}(t)|}=\hat{R}(t)\cdot\vec{R}'(t)\]

But \hat{R}(t) is precisely \hat{P} from the earlier section, as a function of time. Moreover:

\[\vec{R}'(t)=\vec{P}'_B(t)-\vec{P}'_A(t)=\vec{V}_B(t)-\vec{V}_A(t)\]

Because velocity is the time derivative of position. Therefore, we end up with the same equation, just as a function of time:

\[\frac{dr(t)}{dt}=\hat{P}(t)\cdot(\vec{V}_B(t)-\vec{V}_A(t))\]

This formulation is more precise because it makes it very obvious that all the quantities we’re dealing with are time dependent.

[1]	Throughout this post, we assume our objects are sufficiently far apart that they can be treated as points (or particles) without any shape, area or volume.

[2]	The velocity components should be the same regardless of how far apart the two objects are.

[3]	In the standard projection formula we’d also divide by the magnitude of \widehat{P}, but in our case it’s a unit vector anyway.

[4]	In some online sources you may see the closing speed described as closing velocity because it has a direction. IMHO this is the wrong framing. Scalars can be signed! The closing speed is certainly a scalar, not a vector.

[5]	The notation x'(t) means \frac{dx(t)}{dt}

Notes on the Fourier Transform

2026-07-14T20:04:00-07:00

The Fourier series is a great tool for analyzing periodic functions. But what about functions that don’t repeat? We’ve seen that we can compute Fourier series for a non-periodic function defined on a finite interval, as long as we don’t care about its behavior beyond that interval.

Let’s extend this idea to functions that never repeat; that is, non-periodic functions defined on the interval (-\infty,\infty).

Visualizing Fourier series for non-repeating functions

To motivate the subject ahead, let’s look back at the example used in the earlier post about Fourier series:

\[t(x)= \begin{cases} x & 0 \leq x \leq 1 \\ 2-x & 1 < x \leq 2 \\ \end{cases}\]

With an odd extension into [-2,0]. In that post, to make the Fourier series work, we assumed t(x) keeps repeating with a period 2L=4 on the entire x axis. Here, let’s face the reality that it does not - in fact - repeat, and observe how our Fourier series work out.

Recall that the Fourier series approximating t(x) are the sine series (since it’s an odd function):

\[t(x)=\frac{8}{\pi^2}\bigg[ sin\frac{\pi x}{2}-\frac{1}{3^2} sin\frac{3\pi x}{2}+\frac{1}{5^2}sin\frac{5\pi x}{2}-\cdots\bigg]\]

The following visualization is interactive. By default, it shows t(x) (with its odd extension) and no Fourier series approximation. We’ll proceed by a series of steps and observe the outcome:

n (terms in the Fourier series)

x min

x max

Step 1: set to some non-zero number; already at 3, the approximation is very good.

The frequency spacing is \frac{\pi}{L} (this is the coefficient of x in the sines). Note that the Fourier series repeats every 2L, as expected.

Step 2: increase L to 6. This means our series are constructed assuming t(x) has a period of 12, not 4. Note how the Fourier series look now - they repeat every 12, and they don’t match t(x) as well as before. We can increase to a higher number to make the match better. As L grows, the spacing between adjacent frequencies decreases.

Step 3: increase L to 10. We no longer see the repetitions, so feel free to increase the values of x min and x max until you do. Note again that we need to add more and more coefficients to match t(x) better with this larger L, and the spacing adjacent frequencies grows smaller.

Increasing L means our function repeats at larger and larger intervals. The logical conclusion of this progression is to ask - what happens if the function never repeats, meaning L\rightarrow\infty? While not mathematically rigorous, the visual experiment here lets us make some conjectures: we’ll likely need an infinite number of coefficients for a good approximation, and moreover, the spacing between these coefficients will tend to zero.

In other words, instead of a discrete set of coefficients, we’ll end up with a continuous line, or function. The function produced by this process is the Fourier transform of t(x), and the next section shows its mathematical derivation.

Fourier series with L\rightarrow\infty leading to Fourier transform

In these notes, we’ll be using the complex exponential formulation of Fourier series:

\[f(x)=\sum_{n=-\infty}^{\infty}C_n\cdot e^{in\pi x/L}\]

With:

\[C_n=\frac{1}{2L}\int_{-L}^{L}f(x)e^{-in\pi x/L}dx\]

We’re interested in a non-periodic defined on the interval (-\infty,\infty). So we’ll be exploring the above equations for L\rightarrow\infty.

First, let’s make a slight change of notation. Instead of writing formulae in terms of the period (2L), we’ll be using the n-th harmonic angular frequency w_n:

\[w_n=\frac{n\pi}{L}\]

So we can slightly rewrite our series as:

\[f(x)=\sum_{n=-\infty}^{\infty}C_n\cdot e^{i w_n x}=\sum_{n=-\infty}^{\infty}C_n\cdot e^{i\cdot n \Delta w x}\]

Using \Delta w as the difference between two consecutive frequencies:

\[\Delta w=w_n-w_{n-1}=\frac{n\pi}{L}-\frac{(n-1)\pi}{L}=\frac{\pi}{L}\]

Using this notation, C_n is expressed as:

\[C_n=\frac{\Delta w}{2\pi}\int_{-\pi/\Delta w}^{\pi/\Delta w}f(x)e^{-i\cdot n \Delta w x}dx\]

So far there are no new insights here, just some new notation. Now we’re going to use it to facilitate the next step.

Since L\rightarrow \infty, then \Delta w\rightarrow 0. Let’s calculate the limit of the Fourier series representation of when \Delta w\rightarrow 0:

\[f(x)=\lim_{\Delta w\rightarrow 0}\sum_{n=-\infty}^{\infty}C_n\cdot e^{i\cdot n \Delta w x}\]

And substitute the latest C_n into this equation, changing its dummy integration variable from x to t to avoid confusion [1]

\[f(x)=\lim_{\Delta w\rightarrow 0}\sum_{n=-\infty}^{\infty}\left[\frac{\Delta w}{2\pi}\int_{-\pi/\Delta w}^{\pi/\Delta w}f(t)e^{-i\cdot n \Delta w t}dt\right]\cdot e^{i\cdot n \Delta w x}\]

Reordering slightly, and also replacing n\Delta w by w_n in the complex exponents:

\[f(x)=\frac{1}{2\pi}\lim_{\Delta w\rightarrow 0}\sum_{n=-\infty}^{\infty}\left[\int_{-\pi/\Delta w}^{\pi/\Delta w}f(t)e^{-i\cdot w_n t}dt\right]\cdot e^{i\cdot w_n x}\Delta w\]

Looking at the limit with the sum carefully, this is a Riemann sum (see Appendix A)! w_n is the "sampled" version of , and \Delta w\rightarrow 0. We can therefore replace it by an integral, changing w_n to and \Delta w to dw [2]:

\[f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\left[\int_{-\infty}^{\infty}f(t)e^{-i\cdot wt}dt\right]\cdot e^{i\cdot w x}dw\]

The inner integral is called the Fourier transform of and denoted [3]:

\[\boxed{\hat{f}(w)=\mathcal{F}\left[f(x)\right]=\int_{-\infty}^{\infty}f(x)e^{-i\cdot wx}dx}\]

And the full equation for is then the inverse Fourier transform:

\[\boxed{f(x)=\mathcal{F}^{-1}\left[\hat{f}(w)\right]=\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{f}(w)e^{i\cdot w x}dw}\]

Example calculation of Fourier transform

Let’s take our favorite odd triangular pulse example and calculate its Fourier transform. The function’s mathematical definition and plot are shown earlier in this post. Note that we’re not extending this function periodically - it’s zero beyond the range [-2,2]; this is exactly why we need the Fourier transform here - as we’ve seen, Fourier series won’t do because the function they reconstruct eventually starts repeating.

We’re looking to find:

\[\hat{t}(w)=\int_{-\infty}^{\infty}t(x)e^{-iwx}dx\]

To calculate the integral, let’s decompose the complex exponent using Euler’s formula:

\[\hat{t}(w)=\int_{-\infty}^{\infty}t(x)cos(wx)dx-i\int_{-\infty}^{\infty}t(x)sin(wx)dx\]

Since our t(x) is odd, the first integral is zero. Also t(x)sin(wx) is even, so we can write:

\[\hat{t}(w)=-2i\int_{0}^{\infty}t(x)sin(wx)dx\]

We’ve already calculated a very similar integral in the post on Fourier series, so let’s just skip to the result:

\[\hat{t}(w)=-2i\cdot\frac{2\cdot sin(w)-sin(2w)}{w^2}\]

The only remaining difficulty is its value at 0, which seems undefined at first (division by zero). However, note that as w\rightarrow 0, the numerator also tends to 0, so we can use L’Hopital’s rule (twice!) to find that:

\[\lim_{w\rightarrow 0} \hat{t}(w)=0\]

Therefore:

\[\hat{t}(w)= \begin{cases} -2i\cdot\frac{2\cdot sin(w)-sin(2w)}{w^2} & w\neq 0 \\ 0 & w=0 \\ \end{cases}\]

This function is complex-valued; in fact, it’s purely imaginary. How do we visualize it? A common way to visualize complex-valued functions is by plotting their magnitude and phase separately.

The magnitude of \hat{t}(w) is:

\[|\hat{t}(w)|=\sqrt{\hat{t}(w)\cdot\hat{t}(w)^*}=2\left|\frac{2\cdot sin(w)-sin(2w)}{w^2} \right|\]

Since \hat{t}(w) is purely imaginary, there are only two options for the phase:

When the numerator is positive, we get a negative imaginary number with phase -\pi/2, and when the numerator is negative, we get a positive imaginary number with phase \pi/2. Finally, when \hat{t}(w)=0 (which happens at w=0, by our earlier analysis, but also whenever is a whole multiple of \pi), the phase is undefined.

Here’s the magnitude and phase of \hat{t}(w) plotted against :

It is common to talk about \hat{t}(w) as the frequency domain representation of t(x).

The frequency domain representation of functions

When the functions we’re working with have time as their domain (e.g. the x in t(x) represents time), which is often the case in the study of signals and systems, the Fourier transform can be seen as computing the frequency domain representation of the function.

Here’s the Fourier transform formula again:

\[\hat{f}(w)=\mathcal{F}\left[f(x)\right]=\int_{-\infty}^{\infty}f(x)e^{-i\cdot wx}dx\]

It takes - the time domain representation of a function, and converts it to \hat{f}(w) - a frequency domain representation. For well-behaved functions, these two representations are dual - each one describes the function completely, just in a different way.

To convert back from a frequency domain representation to the time domain, we use the inverse Fourier transform:

\[\mathcal{F}^{-1}\left[\hat{f}(w)\right]=\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{f}(w)e^{i\cdot w x}dw\]

While a time-domain plot (t(x)) shows how a signal changes over time, a frequency-domain plot (\hat{t}(w)) shows how the signal is distributed across all possible frequencies. Moreover, as we’ve seen, \hat{t}(w) is complex valued. Each frequency therefore has both a magnitude and a phase: the magnitude tells us how strongly that frequency contributes, while the phase tells us how that component is shifted.

The frequency domain is extremely useful in signal analysis; for example, when designing filters.

The Fourier transform also has a number of properties that are very useful in signal analysis and processing. But first, let’s discuss what a "well-behaved function" means for the purpose of applying Fourier transforms.

Existence condition for the Fourier transform

The simplest existence condition for Fourier transforms is absolute integrability (also known as Lebesgue integrable):

\[\int_{-\infty}^{\infty}|f(x)|dx<\infty\]

With this condition, \hat{f}(w) exists on the entire domain, is continuous and vanishes (tends to 0) as |w|\rightarrow\infty [4].

While this condition is sufficient, it’s not necessary; there are less well-behaved functions that also have Fourier transforms defined with some limitations. In these notes, we’re mostly interested in well-behaved functions that are used in real-world engineering, so we won’t discuss the other cases.

Another assumption commonly made for real-world functions is that they vanish (tend to 0) as |x|\rightarrow\infty. While this is not a direct outcome of absolute integrability [5], it’s a reasonable assumption in engineering. After all, real-world signals have finite energies.

Intuitively, when we also assume is uniformly continuous, the assumption of vanishing at |x|\rightarrow\infty is a logical conclusion, because otherwise how can the total area for |f(x)| be finite?

An important outcome of this discussion is that the Fourier transform is unsuitable for periodic functions. Functions that repeat at intervals are not absolute integrable. For periodic functions, we use Fourier series.

Some useful properties of Fourier transforms

Linearity

The Fourier transform is a linear operator, because the integral is linear:

\[\begin{aligned} \mathcal{F}\left[\alpha f(x)+\beta g(x)\right]&=\int_{-\infty}^{\infty}\alpha f(x)e^{-i\cdot wx}dx+\int_{-\infty}^{\infty}\beta g(x)e^{-i\cdot wx}dx\\ &=\alpha\int_{-\infty}^{\infty}f(x)e^{-i\cdot wx}dx+\beta\int_{-\infty}^{\infty}g(x)e^{-i\cdot wx}dx\\ &=\alpha\mathcal{F}\left[f(x)\right]+\beta\mathcal{F}\left[g(x)\right] \end{aligned}\]

So is the inverse Fourier transform; it’s similarly easy to show that:

\[\mathcal{F}^{-1}\left[\alpha\hat{f}(w)+\beta\hat{g}(w)\right]= \alpha\mathcal{F}^{-1}\left[\hat{f}(w)\right]+\beta\mathcal{F}^{-1}\left[\hat{g}(w)\right]\]

Scaling

If we scale the domain of a function by a constant, its transform changes only slightly:

\[\mathcal{F}\left[f(ax)\right]=\int_{-\infty}^{\infty}f(ax)e^{-i\cdot wx}dx\]

Let’s do the variable substitution u=ax:

\[\mathcal{F}\left[f(ax)\right]=\frac{1}{a}\int_{-\infty}^{\infty}f(u)e^{-i\cdot \frac{wu}{a}}du\]

This is the Fourier transform evaluated at \frac{w}{a}, so:

\[\mathcal{F}\left[f(ax)\right]=\frac{1}{a}\hat{f}\left(\frac{w}{a}\right)\]

There’s one small caveat here; when a is negative, the integral bounds should be flipped, causing a minus sign in front of the transform. So we can write:

\[\mathcal{F}\left[f(ax)\right]=\frac{1}{|a|}\hat{f}\left(\frac{w}{a}\right)\]

Which works for any a\ne 0.

This property is intuitive when thinking about signals: suppose a>0, then f(ax) means the signal is compressed in the time domain by a factor a. The scaling property says that the frequency domain is expanded using the same factor; in other words, the higher frequencies become more prominent because we need sharper transitions to represent the compressed signal.

Time shifting

What happens to the Fourier transform if we time-shift the input signal by some constant: f(x-x_0). By definition:

\[\mathcal{F}\left[f(x-x_0)\right]=\int_{-\infty}^{\infty}f(x-x_0)e^{-i\cdot wx}dx\]

Substituting u=x-x_0, we get du=dx, so:

\[\begin{aligned} \mathcal{F}\left[f(x-x_0)\right]&=\int_{-\infty}^{\infty}f(u)e^{-i\cdot w(u+x_0)}du\\ &=e^{-iwx_0}\int_{-\infty}^{\infty}f(u)e^{-i\cdot wu}du\\ &=e^{-iwx_0}\mathcal{F}\left[f(x)\right] \end{aligned}\]

Transform of a derivative

An extremely useful property that’s often employed in the solution of partial differential equations; let’s calculate the Fourier transform of the derivative of :

\[\mathcal{F}\left[f'(x)\right]=\int_{-\infty}^{\infty}f'(x)e^{-i\cdot wx}dx\]

We’ll use integration by parts, where dv=f'(x) and u=e^{-i\cdot wx}. Therefore, v=f(x) and du=-iw\cdot e^{-i\cdot wx}:

\[\mathcal{F}\left[f'(x)\right]=\left[f(x)e^{-i\cdot wx}\right]^{\infty}_{-\infty}-\int_{-\infty}^{\infty}f(x)(-iw\cdot e^{-i\cdot wx})dx\]

Recall the assumption made in the "Existence condition..." section about vanishing at infinities. So the first part of the equation above is zero, and we’re left with:

\[\begin{aligned} \mathcal{F}\left[f'(x)\right]&=-\int_{-\infty}^{\infty}f(x)(-iw\cdot e^{-i\cdot wx})dx\\ &=iw\int_{-\infty}^{\infty}f(x)e^{-i\cdot wx}dx\\ &=iw\cdot\mathcal{F}\left[f(x)\right] \end{aligned}\]

Transform of convolution

The convolution between two continuous functions and g(x) is defined as:

\[(f\ast g)(x)=\int_{-\infty}^{\infty}f(\xi)g(x-\xi)d\xi\]

Let’s calculate the Fourier transform of this function:

\[\begin{aligned} \mathcal{F}\left[(f\ast g)(x)\right]&=\int_{-\infty}^{\infty}e^{-i\cdot wx}\left[\int_{-\infty}^{\infty}f(\xi)g(x-\xi)d\xi\right]dx\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i\cdot wx}f(\xi)g(x-\xi)d\xi\ dx \end{aligned}\]

This step of combining the integrals into a double integral, as well as the next step (changing the order of integration) is possible due to Fubini’s theorem and our assumption that and g(x) are Lebesgue integrable.

Switch order of integration:

\[\mathcal{F}\left[(f\ast g)(x)\right]=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-i\cdot wx}f(\xi)g(x-\xi)dx\ d\xi\]

Now, f(\xi) in the inner integral doesn’t depend on x, so we can pull it out:

\[\mathcal{F}\left[(f\ast g)(x)\right]=\int_{-\infty}^{\infty}f(\xi)\int_{-\infty}^{\infty}e^{-i\cdot wx}g(x-\xi)dx\ d\xi\]

The inner integral is just the Fourier transform of a time-shifted g(x-\xi), so we can write:

\[\mathcal{F}\left[(f\ast g)(x)\right]=\int_{-\infty}^{\infty}f(\xi)e^{-i\cdot w\xi}\mathcal{F}\left[g(x)\right]d\xi=\mathcal{F}\left[g(x)\right]\int_{-\infty}^{\infty}e^{-i\cdot w\xi}f(\xi)d\xi\]

And the remaining integral is the Fourier transform of , so:

\[\mathcal{F}\left[(f\ast g)(x)\right]=\mathcal{F}\left[f\right]\cdot\mathcal{F}\left[g\right]\]

Convolution in the time domain translates to multiplication in the frequency domain! This result is so important in signal processing that it’s called the convolution theorem.

Appendix A: Riemann sum and the definite integral

Suppose we have some function and we want to know the area bounded between this function’s graph and the x axis in a certain interval [a,b]. One way to do this is to take a partition of the interval:

\[a=x_0<x_1<\cdots<x_{n-1}<x_n=b\]

And calculate the area under for every element of the partition. We can then approximate such sub-areas by rectangles, as follows:

We’ll denote the area of each rectangle as f(x^*_i)\cdot\Delta x:

\Delta x=(b-a)/n is the width of one interval (assuming a uniform partition, but the math works just as well for non-uniform ones).
x^*_i is some value in the interval [x_{i-1},x_i].

There are many ways to choose which point of the interval [x_{i-1},x_i] to denote as x^*_i: left point (x_{i-1}), right point (), mid-point between the two (which is what our plot shows) or anything in between. The distinction doesn’t really matter for our purpose, as we will soon see.

We can approximate the area under the curve of in the interval [a,b] with the Riemann sum, using a uniform partition:

\[S=\sum_{i=1}^{n}f(x^*_i)\Delta x\]

If is continuous on [a,b], then as n\rightarrow \infty:

\[S=\lim_{n\rightarrow \infty}\sum_{i=1}^{n}f(x^*_i)\Delta x=\int_{a}^{b}f(x)dx\]

This is known as the Riemann integral, or just the definite integral. The limit is why the exact choice of x^*_i doesn’t matter: as n\rightarrow\infty we have \Delta x\rightarrow 0, and all points within [x_{i-1}, x_i] are equally good.

[1]	Note that C_n is not a function of x; in its definition, x only serves as a dummy integration variable and can be called anything we choose. When we substitute C_n into the equation for , which is a function of x, we have to be careful. Thus the renaming.

[2]	Note we apply the limit; therefore, the bounds of the inner integral (in the square brackets) are now also between -\infty and \infty

[3]	We change the dummy integration variable back to x here, for consistency. Once again, since x is just the integration variable and the integral is definite, the final result doesn’t depend on x. It’s a function of .

[4]	The vanishing at infinity part is the Riemann-Lebesgue lemma; you can find a proof on Wikipedia

[5]	A pathological absolute-integrable function can have spikes at infinity but still have a finite total area.

Dot product: Component vs. Geometric definition

2026-07-10T18:42:00-07:00

The goal of this post is to answer a simple question: why are the following two definitions of the vector dot product in Euclidean space [1] equivalent for vectors \vec{a} and \vec{b}:

Component definition: \vec{a}\cdot\vec{b}=\sum_{i=1}^{n}a_i b_i
Geometric definition: \vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|cos(\theta), where |\vec{a}| is the magnitude of \vec{a} and is the angle between the vectors’ directions

Here’s a graphical depiction of our vectors (focusing on for clarity, though this applies to any-dimensional vectors). It shows both the components of the vectors and the angle between them. The length of the arrow for \vec{a} is |\vec{a}|.

We’ll show two proofs of the equivalence here, the geometric proof and the projection proof. The Appendix describes some properties of dot products that facilitate these proofs.

Geometric proof

We’ll be using this diagram of our vectors \vec{a} and \vec{b}, as well as the vector \vec{c}=\vec{a}-\vec{b}:

Using the law of cosines [2] on the triangle formed by the three vectors:

\[|\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}|cos(\theta)\]

Since for any vector \vec{a}, we have \vec{a}\cdot\vec{a}=|\vec{a}|^2 (see Appendix), let’s rewrite this equation as:

\[\vec{c}\cdot\vec{c}=\vec{a}\cdot\vec{a}+\vec{b}\cdot\vec{b}-2|\vec{a}||\vec{b}|cos(\theta)\]

But \vec{c}=\vec{a}-\vec{b} and the dot product obeys the distributive property (see Appendix). Therefore:

\[\begin{aligned} (\vec{a}-\vec{b})\cdot(\vec{a}-\vec{b})&=\vec{a}\cdot\vec{a}+\vec{b}\cdot\vec{b}-2|\vec{a}||\vec{b}|cos(\theta)\\ \vec{a}\cdot\vec{a}-2\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{b}&=\vec{a}\cdot\vec{a}+\vec{b}\cdot\vec{b}-2|\vec{a}||\vec{b}|cos(\theta)\\ -2\vec{a}\cdot\vec{b}&=-2|\vec{a}||\vec{b}|cos(\theta)\\ \vec{a}\cdot\vec{b}&=|\vec{a}||\vec{b}|cos(\theta) \end{aligned}\]

Projection proof

For this proof, we’ll assume the geometric definition is correct and will see how it leads to the component definition. We’ll begin by denoting vectors \vec{e}_1,\vec{e}_2\dots\vec{e}_n as the standard orthonormal basis for . For example, in 2D space, these basis vectors are \vec{e}_1=[1\ 0] and \vec{e}_2=[0\ 1], shown in this diagram:

If we take an arbitrary \vec{a}\in\mathbb{R}^n and calculate its dot product with a basis vector, we can use the geometric definition:

\[\vec{a}\cdot\vec{e}_i=|\vec{a}||\vec{e}_i|cos(\theta_i)=|\vec{a}|cos(\theta_i)=a_i\]

where a_i is the component of \vec{a} in the direction of \vec{e}_i. The diagram makes it easy to see why this is true from basic trigonometry, but in the more general case this is just a vector projection.

Now let’s represent vectors \vec{a} and \vec{b} as linear combinations of the basis vectors:

\[\begin{aligned} \vec{a}&=\sum_{i=1}^{n}a_i\vec{e}_i\\ \vec{b}&=\sum_{i=1}^{n}b_i\vec{e}_i\\ \end{aligned}\]

And calculate the dot product \vec{a}\cdot\vec{b}, beginning by rewriting \vec{b} with its linear combination of basis vectors representation:

\[\vec{a}\cdot\vec{b}=\vec{a}\cdot\sum_{i=1}^{n}b_i\vec{e}_i\]

Using the fact that the dot product distributes over linear combinations:

\[\vec{a}\cdot\vec{b}=\sum_{i=1}^{n}b_i(\vec{a}\cdot\vec{e}_i)\]

But earlier we’ve shown that \vec{a}\cdot\vec{e}_i=a_i. Therefore:

\[\vec{a}\cdot\vec{b}=\sum_{i=1}^{n}b_i a_i=\sum_{i=1}^{n}a_i b_i\]

Which is the component definition \blacksquare.

Appendix A: Inner product space

A generalization of dot products in is the inner product, which is an operation meeting some specific requirements, defined on a vector space.

The inner product is denoted as \langle x,y\rangle:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}, and must satisfy the following requirements for all vectors x,y,z\in\mathbb{R}^n and scalars a,b\in\mathbb{R}:

Symmetry: \langle x,y\rangle=\langle y,x\rangle
Linearity in the first argument: \langle ax+by,z\rangle=a\langle x,z\rangle+b\langle y,z\rangle
Positive-definiteness: if x\ne 0 then \langle x,x\rangle>0

For , we define the inner product operation in its component formulation as:

\[\langle x,y\rangle=\sum_{i=1}^{n}x_i\cdot y_i\]

Let’s prove the requirements listed above for this operation; this is fairly straightforward, given the well-known properties of scalar multiplication and addition on :

Symmetry:

\[\langle x,y\rangle=\sum_{i=1}^{n}x_i\cdot y_i=\sum_{i=1}^{n}y_i\cdot x_i=\langle y,x\rangle\]

Linearity in the first argument:

\[\begin{aligned} \langle ax+by,z\rangle&=\sum_{i=1}^{n}(ax+by)_i\cdot z_i\\ &=\sum_{i=1}^{n}a x_i\cdot z_i+b y_i\cdot z_i\\ &=a\sum_{i=1}^{n}x_i\cdot z_i+b\sum_{i=1}^{n}y_i\cdot z_i=a\langle x,z\rangle+b\langle y,z\rangle \end{aligned}\]

Positive-definiteness:

Consider the components of vector x. Clearly, \forall i\quad x_i\cdot x_i=x_i^2\ge 0. Since the vector x is not the zero vector, at least one of its components is nonzero, and for that component x_i\cdot x_i>0. Therefore:

\[\langle x,x\rangle=\sum_{i=1}^{n}x_i\cdot x_i>0\]

Now that we’ve proved all the inner product requirements on our operation \langle x,y\rangle, we can say that is an inner product space with this operation.

By meeting these requirements, it can be readily shown that our inner product operation has additional useful properties:

\langle x,0\rangle=\langle 0,x\rangle=0
\langle x,x\rangle=0 if and only if x=0
\langle x,ay+bz\rangle=a\langle x,y\rangle+b\langle x,z\rangle
\langle x+y,x+y\rangle=\langle x,x\rangle+2\langle x,y\rangle+\langle y,y\rangle

The third property is particularly helpful, because it means the inner product is bilinear, and thus is distributive over addition.

Note that these are shown for the component definition of dot product. It’s not too hard to prove distributivity for the geometric definition using the notion of projections and how they add up.

Norm

The norm of a vector x in an inner product space is defined as |x|=\sqrt{\langle x,x\rangle}. Therefore, the square of the norm is |x|^2=\langle x,x\rangle.

The norm is used to express the notion of magnitude, or length of a vector. If you think of a vector x\in\mathbb{R}^n in Cartesian coordinates, the definition of the norm is a generalization of the Pythagorean theorem.

[1]	By this we mean , where each vector is an n-tuple of real numbers, with the usual mathematical operations making this a vector space.

[2]	Which is a very fundamental theorem in geometry; it is (or rather, its non-trigonometric version) is proven from the basic Euclidean axioms in The Elements.

Notes on Fourier series

2026-05-27T19:30:00-07:00

The trigonometric Fourier series is a beautiful mathematical theory that shows how to decompose a periodic function into an infinite sum of sinusoids. These are my notes on the subject, with some examples and the connection to linear algebra in Hilbert space.

Coefficients of Fourier series

Let’s assume that is a well-behaved 2L-periodic [1] function and that we can find coefficients a_n and b_n such that:

\[f(x)=\sum_{n=0}^{\infty}\left(a_n cos\frac{n\pi x}{L}+b_n sin\frac{n\pi x}{L}\right)\]

Then we say that the Fourier series on the right-hand side converges to . We’ll talk more about the assumptions mentioned above and convergence in the next section.

Note that when n=0, the sum becomes just ; therefore it’s customary to write the series starting with n=1, with a separate constant component (which is the function's average over one period). To make computations nicer, this constant is typically called a_0 / 2, so:

\[f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n cos\frac{n\pi x}{L}+b_n sin\frac{n\pi x}{L}\right)\]

Our goal is to find the coefficients a_n and b_n that satisfy this equation. We’ll do this in three steps.

Step 1: Integrate both sides of the equation between -L and L [2].

\[\int_{-L}^{L}f(x)dx=\int_{-L}^{L}\frac{a_0}{2}dx+\sum_{n=1}^{\infty}\bigg (\int_{-L}^{L}a_n cos\frac{n\pi x}{L}dx+\int_{-L}^{L}b_n sin\frac{n\pi x}{L}dx\bigg )\]

Per Appendix A, all integrals within the sum are zero, so we’re left with:

\[\int_{-L}^{L}f(x)dx=\int_{-L}^{L}\frac{a_0}{2}dx=\bigg[\frac{x\cdot a_0}{2}\bigg]_{-L}^{L}=a_0\cdot L\]

And thus we find :

\[a_0=\frac{1}{L}\int_{-L}^{L}f(x)dx\]

Step 2: Multiply both sides by cos\frac{m\pi x}{L} (where m is a positive integer constant) and integrate between -L and L.

\[\begin{aligned} \int_{-L}^{L}f(x)cos\frac{m\pi x}{L}dx&=\int_{-L}^{L}\frac{a_0}{2}cos\frac{m\pi x}{L}dx\\ &+\sum_{n=1}^{\infty}\bigg (\int_{-L}^{L}a_n cos\frac{n\pi x}{L}cos\frac{m\pi x}{L}dx+\int_{-L}^{L}b_n sin\frac{n\pi x}{L}cos\frac{m\pi x}{L}dx\bigg ) \end{aligned}\]

Looking at the right-hand side, the first integral is zero per Appendix A, and the last integral is zero per Appendix B. We’re left with:

\[\int_{-L}^{L}f(x)cos\frac{m\pi x}{L}dx=\sum_{n=1}^{\infty}\int_{-L}^{L}a_n cos\frac{n\pi x}{L}cos\frac{m\pi x}{L}dx\]

Per Appendix B, the integral on the right is zero for all n\neq m, and L for n=m. Therefore, we can write:

\[\int_{-L}^{L}f(x)cos\frac{m\pi x}{L}dx=a_m\cdot L\]

Recall that m is an arbitrary integer, just like ; for consistency, we’ll replace m by and isolate a_n:

\[a_n=\frac{1}{L}\int_{-L}^{L}f(x)cos\frac{n\pi x}{L}dx\]

Step 3: Hopefully it’s clear where this is going now; multiply both sides by sin\frac{m\pi x}{L} and integrate between -L and L. Using a very similar reasoning to step 2, we’ll end up with:

\[b_n=\frac{1}{L}\int_{-L}^{L}f(x)sin\frac{n\pi x}{L}dx\]

We’ve just found a way to calculate all the coefficients of our Fourier series for :

\[f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n cos\frac{n\pi x}{L}+b_n sin\frac{n\pi x}{L}\right)\]

Where:

\[\begin{aligned} a_0&=\frac{1}{L}\int_{-L}^{L}f(x)dx\\ a_n&=\frac{1}{L}\int_{-L}^{L}f(x)cos\frac{n\pi x}{L}dx\\ b_n&=\frac{1}{L}\int_{-L}^{L}f(x)sin\frac{n\pi x}{L}dx \end{aligned}\]

Conditions on f and convergence of Fourier series

The previous section discusses Fourier series for a function that is well-behaved - but what does that mean? The full answer would lead us deep into analysis, which I’d like to avoid here. So I’ll keep it brief.

We typically assume that is square integrable, which is denoted as L^2. Moreover, we assume that the function is piecewise smooth: each segment of the function has continuous derivatives. A very simple example of a piecewise smooth function is f(x)=|x|. Another is the triangular wave function used in the example below.

These conditions hold for pretty much any reasonable function we want to approximate using Fourier series, so they aren’t a serious burden.

For a function that satisfies these conditions, it’s guaranteed to have a Fourier series that pointwise converges to it. This means that at every continuous point of , the Fourier series converges to it exactly; at every jump point, the Fourier series converges to the mid-point of the jump.

Cosine and Sine series

Sometimes, additional properties of the function can help us simplify the Fourier series for it. If f_e(x) is an even function, then we know that:

\[b_n=\frac{1}{L}\int_{-L}^{L}f(x)sin\frac{n\pi x}{L}dx=0\]

Because the function inside the integral is odd, and integrating an odd function over a symmetric interval results in 0.

Therefore, the Fourier series for such f_e(x) is a cosine series:

\[f_e(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n cos\frac{n\pi x}{L}\]

With coefficients and a_n given as before.

Similarly if f_o(x) is an odd function, then its and a_n are 0, and its Fourier series is a sine series:

\[f_o(x)=\sum_{n=1}^{\infty}b_n sin\frac{n\pi x}{L}\]

Fourier series for a non-periodic function defined on an interval

So far we’ve been talking about 2L-periodic functions that can be faithfully represented by Fourier series. But what if we have a non-periodic function defined on a finite interval?

E.g. suppose we have f(x)=x on the interval [0,L]. Can we approximate it with a Fourier series?

Yes! First, we have to make a choice of how to extend the function to the negative interval [-L,0]. Then, we simply repeat the function every 2L - this is called a periodic extension. Note that the Fourier series calculation only cares about the range [-L,L]. The resulting series will approximate the generated periodic function in its entirety, and in particular will also converge to it in the [0,L] interval (except maybe the endpoints, depending on the mode of extension).

There are several natural ways to extend a function defined on [0,L] into the interval [-L,0] [3]:

Direct periodic repetition: we simply repeat every L: f(x+L)=f(x)\ \forall x.
Even extension: f(|x|)
Odd extension: when x\ge 0 and -f(-x) when x<0.

Here’s an example of extending our sample function f(x)=x onto the full interval [-L,L] and then repeating it periodically every 2L:

Note that the Fourier series for these extended functions will be different. However, they will all converge to in the interval [0,L]. Typically, even and odd extensions have the benefit of producing either cosine or sine series, correspondingly (as discussed in the previous section).

We’ve seen that Fourier series work well for periodic functions and also non-periodic functions defined on a finite domain (because we can extend these periodically). But what about aperiodic functions defined on the entire real line? This is where we’ll have to leave Fourier series behind and move on to their generalization - the Fourier transform.

Example

Let’s take the following triangular function t(x) [4]:

t(x) is periodic with period 4. We can define it by starting with a formula on the interval [0,2]:

\[t(x)= \begin{cases} x & 0 \leq x \leq 1 \\ 2-x & 1 < x \leq 2 \\ \end{cases}\]

Then making an odd extension into [-2,0] and repeating it periodically. Now we can go ahead to calculate its Fourier coefficients.

Since this function is odd, we know that we’ll get a sine series, as a_n are going to be 0 for all . Let’s calculate b_n; in our case L=2 (half the period).

\[b_n=\frac{1}{2}\int_{-2}^{2}t(x)sin\frac{n\pi x}{2}dx\]

Since t(x) is odd and so is the sine, we’re integrating an even function over a symmetric interval. Therefore, we only have to integrate on the positive half of the range and multiply the result by two:

\[b_n=\int_{0}^{2}t(x)sin\frac{n\pi x}{2}dx\]

Let’s set k=\frac{n\pi}{2}:

\[b_n=\int_{0}^{2}t(x)sin(kx)dx\]

And split up the integral for the different segments of t(x):

\[b_n=\int_{0}^{1}x\cdot sin(kx)dx+\int_{1}^{2}(2-x)sin(kx)dx\]

The first integral, by the method described in Appendix C:

\[I_1=\int_{0}^{1}x\cdot sin(kx)dx=\bigg[\frac{-x cos(kx)}{k}+\frac{sin(kx)}{k^2} \bigg]_{0}^{1}=\frac{sin(k)}{k^2}-\frac{cos(k)}{k}\]

The second integral can also be split into two:

\[I_2=\int_{1}^{2}2sin(kx)dx - \int_{1}^{2}x\cdot sin(kx)dx\]

The first of these is trivial to calculate; the second can once again use Appendix C. After some tedious but straightforward calculations [5] we’ll get:

\[I_2=\frac{cos(k)}{k}+\frac{sin(k)-sin(2k)}{k^2}\]

Adding I_1+I_2, we get:

\[\begin{aligned} b_n=I_1+I_2&=\frac{sin(k)}{k^2}-\frac{cos(k)}{k}+\frac{cos(k)}{k}+\frac{sin(k)-sin(2k)}{k^2}\\ &=\frac{2sin(k)-sin(2k)}{k^2} \end{aligned}\]

Now let’s substitute k=\frac{n\pi}{2} back. This makes sin(2k) zero because the sine of an integer multiple of \pi is always zero:

\[b_n=\frac{2sin \frac{n\pi}{2}}{\left (\frac{n\pi}{2}\right )^2}=\frac{8sin \frac{n\pi}{2}}{n^2\pi^2}\]

We have b_n, so the Fourier series for our t(x) is:

\[t(x)=\sum_{n=1}^{\infty}\frac{8}{n^2\pi^2}sin\frac{n\pi}{2}sin\frac{n\pi x}{2}\]

Note that for even values of , sin \frac{n\pi}{2} is zero, so only the odd terms remain:

\[t(x)=\frac{8}{\pi^2}\bigg[ sin\frac{\pi x}{2}-\frac{1}{3^2} sin\frac{3\pi x}{2}+\frac{1}{5^2}sin\frac{5\pi x}{2}-\cdots\bigg]\]

Here’s an interactive chart showing how the series t(x) converges to our triangular function. You can set the number of terms in the Fourier series and see the effect (red line). Note that all even coefficients are zero so it will look the same for as for n-1 when is odd.

n (terms in the Fourier series)

Compact formula using a single phase-shifted sinusoid

We’ve written the Fourier series for as follows so far:

\[f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n cos\frac{n\pi x}{L}+b_n sin\frac{n\pi x}{L}\right)\]

We can rewrite this in a somewhat more compact form, using a single sinusoid with a configurable phase at each :

\[f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}q_n\cdot cos\left(\frac{n\pi x}{L}+\theta_n\right)\]

Based on Appendix D, q_n and \theta_n can be computed as follows:

\[\begin{aligned} q_n&=\sqrt{a_n^2+b_n^2}\\ \theta_n&=\operatorname{atan2}(-b_n,a_n) \end{aligned}\]

When Fourier series are used in the context of signal processing, this formulation is easier to reason about because it represents the magnitude and phase shift of each harmonic of in the frequency domain [6]

Complex Fourier series

It should not come as a surprise that the Fourier series, being a combination of trigonometric functions, can also be represented with complex exponential functions.

Specifically, we’ll show that our can be approximated as follows:

\[f(x)=\sum_{n=-\infty}^{\infty}C_n\cdot e^{in\pi x/L}\]

Let’s calculate C_n. We proceed in a manner similar to before, by multiplying both sides of the equation by e^{-im\pi x/L} and taking an integral in the range [-L,L]:

\[\begin{aligned} \int_{-L}^{L}f(x)e^{-im\pi x/L}dx&=\sum_{n=-\infty}^{\infty}\int_{-L}^{L}C_n\cdot e^{in\pi x/L}e^{-im\pi x/L}dx\\ &=\sum_{n=-\infty}^{\infty}\int_{-L}^{L}C_n\cdot e^{i(n-m)\pi x/L}dx \end{aligned}\]

By Appendix A, the sum elements are all zero when n\neq m. When n=m, we get:

\[\int_{-L}^{L}f(x)e^{-im\pi x/L}dx=\int_{-L}^{L}C_m\cdot 1 \cdot dx=2LC_m\]

Therefore, renaming m to (since it’s just an arbitrary integer constant):

\[C_n=\frac{1}{2L}\int_{-L}^{L}f(x)e^{-in\pi x/L}dx\]

We’ve found an alternative formulation to Fourier series, using complex exponentials instead of trigonometric functions. While this was a direct derivation, another way to achieve the same result is to use the Euler Formula to derive:

\[\begin{aligned} cos\theta&=\frac{e^{i\theta}+e^{-i\theta}}{2}\\ sin\theta&=\frac{e^{i\theta}-e^{-i\theta}}{2i} \end{aligned}\]

And substitute these into the original Fourier series formula. I’ll leave this as an exercise for the diligent reader; eventually, the result will be the same. Moreover, it’s possible to show a direct correspondence between a_n, b_n and C_n, for n>0:

\[\begin{aligned} C_0&=\frac{a_0}{2}\\ C_n&=\frac{a_n-ib_n}{2}\\ C_{-n}&=\frac{a_n+ib_n}{2}\\ \end{aligned}\]

Note that C_{-n}=C_n^* when both a_n and b_n are real (which is the case for a real-valued ). This helps explain why the complex formulation has negative frequencies in the sum; when the function is actually real, each negative frequency is paired up with a positive frequency and the result is real [7]:

\[\begin{aligned} C_n e^{in\pi x/L}+C_{-n} e^{-in\pi x/L}&=C_n e^{in\pi x/L}+C_n^* e^{-in\pi x/L}\\ &=C_n e^{in\pi x/L}+\left(C_{n} e^{in\pi x/L}\right)^*\\ &=2\operatorname{Re}\bigg(C_{n} e^{in\pi x/L}\bigg) \end{aligned}\]

So, for a real function we only need to account for positive frequencies:

\[f(x)=C_0+\sum_{n=1}^{\infty}2\operatorname{Re}\bigg(C_{n} e^{in\pi x/L}\bigg)\]

We can take it further. C_n is a complex number, so let’s represent it in polar form as C_n=\frac{q_n}{2} e^{i\theta_n} (the factor of half will make sense soon). Then:

\[\begin{aligned} \operatorname{Re}\bigg(C_{n} e^{in\pi x/L}\bigg)&=\operatorname{Re}\bigg(\frac{q_n}{2} e^{i\theta_n}e^{in\pi x/L}\bigg)\\ &=\frac{q_n}{2}\operatorname{Re}\bigg(e^{i(n\pi x/L + \theta_n)}\bigg)\\ &=\frac{q_n}{2} cos\bigg(\frac{n\pi x}{L}+\theta_n\bigg) \end{aligned}\]

And substituting back into the sum:

\[f(x)=C_0+\sum_{n=1}^{\infty}q_n cos\bigg(\frac{n\pi x}{L}+\theta_n\bigg)\]

This is precisely the compact formulation from the previous section!

Fourier orthogonal basis in Hilbert space

The most beautiful aspect of Fourier theory is that it doesn’t just happen to work by chance, and is deeply connected to linear algebra. Please read my post on Hilbert space before proceeding.

The space of real-valued square integrable functions L^2 forms a Hilbert space, in which we can define the inner product (assuming real functions):

\[\langle f,g \rangle=\int_{-L}^{L}f(x)g(x) dx\]

We’ve demonstrated that the family of functions:

\[1,\qquad cos\frac{n\pi x}{L},\qquad sin\frac{n\pi x}{L}\]

Are all mutually orthogonal, because their pairwise inner products are zero! We’ve also shown that any function in L^2 can be represented as a weighted sum of these functions:

\[f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n cos\frac{n\pi x}{L}+b_n sin\frac{n\pi x}{L}\right)\]

So these functions form a basis for L^2. When we think of these functions as vectors (in an infinite Hilbert space), much of what we did in this post starts feeling like "normal" linear algebra. For example, when we have a set of basis vectors and we want to know how to represent some vector in this basis, we usually find the coefficients by projecting it onto the basis. E.g. with a basis vector e_1, the coefficient of :

\[c=\frac{\langle v, e_1\rangle}{\langle e_1, e_1\rangle}\]

Similarly, when we calculate the Fourier coefficient b_n for some function , we project onto the basis vector sin\frac{n\pi x}{L} by calculating:

\[b_n=\frac{\langle f(x), sin\frac{n\pi x}{L}\rangle}{\langle sin\frac{n\pi x}{L}, sin\frac{n\pi x}{L}\rangle}\]

From Appendix B, we know that the denominator is L, and we’ve just denoted:

\[\langle f(x), sin\frac{n\pi x}{L}\rangle=\int_{-L}^{L}f(x)sin\frac{n\pi x}{L}dx\]

So we get:

\[b_n=\frac{1}{L}\int_{-L}^{L}f(x)sin\frac{n\pi x}{L}dx\]

Which should look familiar!

This is the core linear-algebra idea behind Fourier series: the functions 1, cos\frac{n\pi x}{L}, and sin\frac{n\pi x}{L} play the role of orthogonal basis vectors, while the Fourier coefficients are coordinates of in this basis. The integral formulas for a_n and b_n are not mysterious tricks; they are projections, just like dot products with basis vectors in ordinary Euclidean space.

Fourier series therefore let us decompose a function into independent orthogonal directions, much like decomposing a vector into its x, , and z components.

Appendix A: Integrals of sinusoids

For any integer n\neq 0 and an arbitrary constant L, we have:

\[\begin{aligned} \int_{-L}^{L}cos\frac{n\pi x}{L}dx&=\bigg[\frac{L}{n\pi}sin\frac{n\pi x}{L}\bigg]_{-L}^{L}\\ &=\frac{L}{n\pi}(sin(n\pi)-sin(-n\pi))=0 \end{aligned}\]

Similarly:

\[\begin{aligned} \int_{-L}^{L}sin\frac{n\pi x}{L}dx&=\bigg[\frac{-L}{n\pi}cos\frac{n\pi x}{L}\bigg]_{-L}^{L}\\ &=\frac{-L}{n\pi}\left(cos(n\pi)-cos(-n\pi)\right)=0 \end{aligned}\]

Using these, we can calculate the integral of a complex exponential function for an integer n\neq 0:

\[\begin{aligned} \int_{-L}^{L}e^{in\pi x/L}dx=\int_{-L}^{L}\bigg[ cos\frac{n\pi x}{L} +i\cdot sin\frac{n\pi x}{L}\bigg] dx=0 \end{aligned}\]

Appendix B: Integrals of products of sinusoids

We’ll start with the product of two sines, for any positive integers m and :

\[ss=\int_{-L}^{L}sin\frac{m\pi x}{L}\cdot sin\frac{n\pi x}{L}dx\]

Using the trigonometric identity for a product of sines, we can write:

\[\begin{aligned} ss&=\frac{1}{2}\int_{-L}^{L}\bigg(cos\frac{(m-n)\pi x}{L}-cos\frac{(m+n)\pi x}{L}\bigg)dx\\ &=\frac{1}{2}\int_{-L}^{L}cos\frac{(m-n)\pi x}{L}dx-\frac{1}{2}\int_{-L}^{L}cos\frac{(m+n)\pi x}{L}dx \end{aligned}\]

Now let’s focus on two different scenarios, m\neq n and m=n.

If m\neq n, then each of the integrals constituting ss are 0 (see on Appendix A), so ss=0.

If m=n, then the second integral is still 0, but the first one isn’t:

\[\begin{aligned} ss&=\frac{1}{2}\int_{-L}^{L}cos\frac{0\pi x}{L}dx\\ &=\frac{1}{2}\int_{-L}^{L}1dx=L \end{aligned}\]

Therefore:

\[ss=\int_{-L}^{L}sin\frac{m\pi x}{L}\cdot sin\frac{n\pi x}{L}dx= \begin{cases} L & m = n \\ 0 & m \neq n \end{cases}\]

We can use exactly the same approach to show that:

\[cc=\int_{-L}^{L}cos\frac{m\pi x}{L}\cdot cos\frac{n\pi x}{L}dx= \begin{cases} L & m = n \\ 0 & m \neq n \end{cases}\]

One more variant to cover:

\[sc=\int_{-L}^{L}sin\frac{m\pi x}{L}\cdot cos\frac{n\pi x}{L}dx\]

Since sine is an odd function and cosine is an even function, their product is an odd function. And the integral of an odd function over a symmetric interval is 0 (see this post for more details).

Therefore:

\[sc=\int_{-L}^{L}sin\frac{m\pi x}{L}\cdot cos\frac{n\pi x}{L}dx=0\]

Appendix C: A useful integral

Let’s calculate the indefinite integral:

\[I=\int x\cdot sin(kx) dx\]

For some constant k. We’ll use integration by parts:

\[\int u\cdot dv =u\cdot v - \int v\cdot du\]

Here u=x, so du=dx. Also dv=sin(kx), so v=-\frac{cos(kx)}{k}.

Putting it together:

\[I=\frac{-x\cdot cos(kx)}{k}+\int \frac{cos(kx)}{k} dx=\frac{-x\cdot cos(kx)}{k}+\frac{sin(kx)}{k^2}\]

Appendix D: Sinusoid with phase as a sum of sin and cos

Let’s take a general sinusoid with magnitude q, frequency and phase :

\[s(x)=q\cdot cos(wx+\theta)\]

We’re going to show that s(x) can be represented as a sum of a sine and a cosine with no phase. This is related to my earlier post on the sum of same-frequency sinusoids.

Let’s start by expanding s(x) using a trigonometric identity:

\[s(x)=q\cdot cos(\theta)cos(wx)-q\cdot sin(\theta)sin(wx)\]

Now we’ll denote: a=q\cdot cos(\theta) and b=-q\cdot sin(\theta), so:

\[s(x)=a\cdot cos(wx)+b\cdot sin(wx)\]

We have a and b in terms of q and , but what about the other way around?

Let’s take the equations:

\[\begin{aligned} a&=q\cdot cos(\theta)\\ b&=-q\cdot sin(\theta) \end{aligned}\]

Square both of them and add together:

\[\begin{aligned} a^2+b^2&=q^2\cdot(cos^2(\theta)+sin^2(\theta))=q^2\\ &\Rightarrow q=\sqrt{a^2+b^2} \end{aligned}\]

Now we’ll take the equations for b and a and divide one by the other:

\[\begin{aligned} \frac{b}{a}&=\frac{-sin(\theta)}{cos(\theta)}\\ &\Rightarrow\theta=\operatorname{atan2}(-b,a) \end{aligned}\]

Where the atan2 function is careful to take into account the sign of both numerator and denominator. Also it’s worth mentioning that is determined up to additions of 2\pi.

To conclude, for any q, and :

\[q\cdot cos(wx+\theta)=a\cdot cos(wx)+b\cdot sin(wx)\]

With the aforementioned conversion formulas for a, b.

[1]	A function is called periodic if there exists some constant P>0 such that f(x+P)=f(x)\ \forall x. In our case, we denote the period as P=2L to make presentation and calculations neater.

[2]	Since f(x)\in L^2, we know that it’s integrable on a finite interval

[3]	There’s also an infinite number of less natural ways to extend the function; they will still work, but may make the calculation needlessly complicated

[4]	I chose this one because the more common ones like a square wave and a sawtooth wave have many derivations available online, including on Wikipedia.

[5]	Try it for yourself - it’s a good exercise.

[6]	When q_n=0 for some , the corresponding \theta_n can be chosen arbitrarily.

[7]	For a complex-valued , this conjugate symmetry no longer holds in general. a_n and b_n may be complex, so C_n and C_{-n} need not be conjugates.

Scaling, stretching and shifting sinusoids

2026-05-02T07:17:00-07:00

This is a brief and simple [1] explanation of how to adjust the standard sinusoid sin(x) to change its amplitude, frequency and phase shift. More precisely, given the general function:

\[s(x)=A\cdot sin(w\cdot x+\theta)\]

We’ll see how adjusting the parameters , and affect the shape of s(x). Each section below covers one of these aspects mathematically, and you can use the demo at the bottom to experiment with the topic visually.

Scaling

Scaling is conceptually the simplest change; we adjust to increase or decrease the amplitude (maximal height) of s(x). Setting A=2 will make the value twice as large (in both the positive and negative direction) as the original function.

Stretching

Stretching changes the frequency of sin(x), which is inverse proportional to its period. The baseline function sin(x) has a period of 2\pi, meaning it repeats every 2\pi. In other words, sin(x)=sin(x+2\pi) for any .

If we set w=2, we get sin(2x). This function repeats itself twice as fast as sin(x), because is multiplied by 2 before being fed into the sinusoid. If changes by \pi, the sinusoid’s input changes by 2\pi. Therefore, the period of sin(2x) is \pi, the period of sin(4x) is \frac{\pi}{2} and so on. [2]

More generally, the period of sin(wx) is \frac{2\pi}{w}. Play with the demo below to see this in action, by changing and observing how the waveform changes.

If we know the period p we want, we can easily calculate the that gives us this period:

\[p=\frac{2\pi}{w} \implies w=\frac{2\pi}{p}\]

Shifting

The final parameter we discuss is ; it’s called the phase of the sinusoid. In the baseline sin(x), . The sinusoid is 0 at x=0, achieves its positive peak at x=\frac{\pi}{2}, crosses 0 again at x=\pi, negative peak at x=\frac{3\pi}{2} and returns to its original position at x=2\pi where the repetition begins.

By adding a non-zero , we don’t affect the sinusoid’s amplitude or frequency, but we do shift it right or left along the axis. For example, suppose we use the function sin(x+\theta) with \theta=\frac{\pi}{2}. Then when x=0, we have sin(\frac{\pi}{2}), so the sinusoid is already at its positive peak; at x=\frac{\pi}{2}, the sinusoid crosses 0 into the negatives, etc. Everything happens earlier (by exactly the value of \theta=\frac{\pi}{2}) than in the baseline sinusoid. In other words, we’ve shifted the function left by \frac{\pi}{2}. Similarly, when is negative, everything happens later, and the function is shifted right.

Putting it all together

We’ve now gone over all the parameters for the function:

\[s(x)=A\cdot sin(w\cdot x+\theta)\]

controls the scaling factor (amplitude).
is the frequency and controls the repetition period
controls the phase - how much the sinusoid is shifted left or right

Use the demo below to adjust these parameters and observe their effect on the sinusoid:

A ω θ

[1]	The math level of this post is high-school, at best. My main goal here is to test how to integrate interactive demos into my blog posts.

[2]

This can be a bit counter-intuitive at first; we scale by 2, but the period scales by half. Why? The reason is that affects the sinusoid’s domain, while the period is a property of its range. Therefore, an inverse relation is reasonable, once we put more thought into it. In fact, is often called the angular frequency of the sinusoid, and frequency is inverse proportional to the period.