# Real-valued 2D Fourier series?

For a (well-behaved) one-dimensional function $f: [-\pi, \pi] \rightarrow \mathbb{R}$, we can use the Fourier series expansion to write $$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n\sin(nx) \right)$$

For a function of two variables, Wikipedia lists the formula

$$f(x,y) = \sum_{j,k \in \mathbb{Z}} c_{j,k} e^{ijx}e^{iky}$$

In this formula, $f$ is complex-valued. Is there a similar series representation for real-valued functions of two variables?

• Substitute $e^{i\omega} = \cos\omega + i\sin\omega$ and $c_{j,k} = a_{j,k} + ib_{j,k}$ in the formula you get from Wikipedia, and look only at the real value of the result. The formula gets a bit unwieldy due to the 4 $\sin\cos$ combinations you get, but it works...
– fgp
Oct 12, 2012 at 15:06
• Following fgp's step, you can actually get a compact formula if you allow the index running from negative infinity to positive inifinity (rather than the positive index usually used when expanded in terms of sine and cosine). Mar 23, 2023 at 1:52

The full real-valued 2D Fourier series is: \begin{align} f(x, y) & = \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\alpha_{n,m}cos\left(\frac{2\pi n x}{\lambda_x}\right)cos\left(\frac{2\pi m y}{\lambda_y}\right) \\ & + \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\beta_{n,m}cos\left(\frac{2\pi n x}{\lambda_x}\right)sin\left(\frac{2\pi m y}{\lambda_y}\right) \\ & + \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\gamma_{n,m}sin\left(\frac{2\pi n x}{\lambda_x}\right)cos\left(\frac{2\pi m y}{\lambda_y}\right) \\ & + \sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\delta_{n,m}sin\left(\frac{2\pi n x}{\lambda_x}\right)sin\left(\frac{2\pi m y}{\lambda_y}\right) \\ \end{align}

The coefficients are found with: $$\alpha_{n,m} = \frac{\kappa}{\lambda_x \lambda_y}\int_{y_0}^{y_0+\lambda_y}\int_{x_0}^{x_0+\lambda_x}f(x,y)cos\left(\frac{2\pi n x}{\lambda_x}\right)cos\left(\frac{2\pi m y}{\lambda_y}\right)dx dy \\ \beta_{n,m} = \frac{\kappa}{\lambda_x \lambda_y}\int_{y_0}^{y_0+\lambda_y}\int_{x_0}^{x_0+\lambda_x}f(x,y)cos\left(\frac{2\pi n x}{\lambda_x}\right)sin\left(\frac{2\pi m y}{\lambda_y}\right)dx dy \\ \gamma_{n,m} = \frac{\kappa}{\lambda_x \lambda_y}\int_{y_0}^{y_0+\lambda_y}\int_{x_0}^{x_0+\lambda_x}f(x,y)sin\left(\frac{2\pi n x}{\lambda_x}\right)cos\left(\frac{2\pi m y}{\lambda_y}\right)dx dy \\ \delta_{n,m} = \frac{\kappa}{\lambda_x \lambda_y}\int_{y_0}^{y_0+\lambda_y}\int_{x_0}^{x_0+\lambda_x}f(x,y)sin\left(\frac{2\pi n x}{\lambda_x}\right)sin\left(\frac{2\pi m y}{\lambda_y}\right)dx dy$$ \begin{align} \text{Where } \kappa & = 1 \text{ if } n = 0 \text{ and } m = 0 \\ & = 2 \text{ if } n = 0 \text{ or } m = 0\\ & = 4 \text{ if } n> 0 \text{ and } m > 0 \end{align} Example plot

• are there any references for this? Apr 11, 2019 at 14:26
• @vlizana Davis, John C. Statistics and Data Analysis in Geology. Wiley, 1973. Apr 11, 2019 at 23:44
• @apprehensivebob If I am correct, then this representation has no assumption of Dirichlet condition at the boundaries of the rectangle and is the most general form ? Jun 8, 2020 at 6:54
• The formula is derived directly from the Fourier expansion in terms of sine and cosine basis functions, so you need to handle the edge case of m=0 and/or n=0. If you start from Fourier expansion in terms of exp() and then take the real part, you will get a more compact formula, with its index ranging from -infinity to +infinity, and you need not handle any edge case. Mar 23, 2023 at 2:00

Yes! And these types of expansions occur in a variety of applications, e.g., solving the heat or wave equation on a rectangle with prescribed boundary and initial data.

As a specific example, we can think of the following expansion as a two dimensional Fourier sine series for $f(x,y)$ on $0<x<a$, $0<y<b$: $$f(x,y)=\sum_{n=1}^\infty \sum_{m=1}^\infty c_{nm}\sin\left({n\pi\, x\over a}\right)\sin\left({m\pi\, y\over b}\right), \quad 0<x<a,\ 0<y<b,$$ where the coefficients (obtained from the same type of orthogonality argument as in the 1D case) are given by \begin{align} c_{nm}&={\int_0^b \int_0^a f(x,y)\sin\left({n\pi\, x\over a}\right)\sin\left({m\pi\, y\over b}\right)\,dx\,dy\over \int_0^b \int_0^a \sin^2\left({n\pi\, x\over a}\right)\sin^2\left({m\pi\, y\over b}\right)\,dx\,dy}\\ &={4\over a b}\int_0^b \int_0^a f(x,y)\sin\left({n\pi\, x\over a}\right)\sin\left({m\pi\, y\over b}\right)\,dx\,dy, \quad n,m=1,2,3,\dots \end{align}

For example, the picture below shows (left) the surface $$f(x,y)=30x y^2 (1-x)(1-y)\cos(10x)\cos(10y), \quad 0<x<1,\ 0<y<1,$$ and a plot of the two dimensional Fourier sine series (right) of $f(x,y)$ for $n,m,=1,\dots,5$:

Finally, keep in mind that we are not limited just to double sums of the form sine-sine. We could have any combination we like so long as they form a complete orthogonal family on the domain under discussion.

• This representation seems to be valid only if the values of the function on the boundary of the rectangle are zero. Nov 13, 2013 at 11:59
• Yes, that's why I said, "As a specific example..." The sine functions used there are the eigenfunctions obtained when solving the heat equation on a rectangle where zero boundary conditions are specified. Nov 14, 2013 at 0:47
• What about the case where cosine terms are required? This answer is useless because it does not address the general case. Dec 21, 2013 at 14:19
• @JohnD I know this is very old, but can you add a more general representation which is not restricted to the function having Dirichlet conditions on the boundaries. Should I ask a separate question ? Jun 8, 2020 at 6:42
• @Avrana You can add a phase for x and y, and also add a rotation, to get a more general expression. Oct 28, 2023 at 7:45

@JohnD only details about the coefficients. The correct formula is:

$$c_{n,m} = \frac{\int_{0}^a \int_0^b f(x,y)sin(\frac{n\pi x}{a})sin(\frac{n\pi y}{b})dxdy}{{\int_{0}^a \int_0^b sin^2(\frac{n\pi x}{a})sin^2(\frac{n\pi y}{b})dxdy}{}}$$

the impression that $c_{n,m}$ is $1$. That's not true. Cheers!

• Yes, it was a typo (I had left off the squares in the denominator). Fixed now. Mar 6, 2013 at 4:48