Suppose the function values and the derivatives, and are known at the four corners,,, and of the unit square. The interpolated surface can then be written as The interpolation problem consists of determining the 16 coefficients. Matching with the function values yields four equations:
Likewise, eight equations for the derivatives in the and the directions:
And four equations for the mixed partial derivative:
The expressions above have used the following identities: This procedure yields a surface on the unit square that is continuous and has continuous derivatives. Bicubic interpolation on an arbitrarily sized regular grid can then be accomplished by patching together such bicubic surfaces, ensuring that the derivatives match on the boundaries. Grouping the unknown parameters in a vector and letting the above system of equations can be reformulated into a matrix for the linear equation. Inverting the matrix gives the more useful linear equation, where which allows to be calculated quickly and easily. There can be another concise matrix form for 16 coefficients: or where
Extension to rectilinear grids
Often, applications call for bicubic interpolation using data on a rectilinear grid, rather than the unit square. In this case, the identities for and become where is the spacing of the cell containing the point and similar for. In this case, the most practical approach to computing the coefficients is to let then to solve with as before. Next, the normalized interpolating variables are computed as where and are the and coordinates of the grid points surrounding the point. Then, the interpolating surface becomes
Finding derivatives from function values
If the derivatives are unknown, they are typically approximated from the function values at points neighbouring the corners of the unit square, e.g. using finite differences. To find either of the single derivatives, or, using that method, find the slope between the two surrounding points in the appropriate axis. For example, to calculate for one of the points, find for the points to the left and right of the target point and calculate their slope, and similarly for. To find the cross derivative, take the derivative in both axes, one at a time. For example, one can first use the procedure to find the derivatives of the points above and below the target point, then use the procedure on those values to obtain the value of for the target point. At the edges of the dataset, when one is missing some of the surrounding points, the missing points can be approximated by a number of methods. A simple and common method is to assume that the slope from the existing point to the target point continues without further change, and using this to calculate a hypothetical value for the missing point.
Bicubic spline interpolation requires the solution of the linear system described above for each grid cell. An interpolator with similar properties can be obtained by applying a convolution with the following kernel in both dimensions: where is usually set to −0.5 or −0.75. Note that and for all nonzero integers. This approach was proposed by Keys, who showed that produces third-order convergence with respect to the sampling interval of the original function. If we use the matrix notation for the common case, we can express the equation in a more friendly manner: for between 0 and 1 for one dimension. Note that for 1-dimensional [|cubic convolution] interpolation 4 sample points are required. For each inquiry two samples are located on its left and two samples on the right. These points are indexed from −1 to 2 in this text. The distance from the point indexed with 0 to the inquiry point is denoted by here. For two dimensions first applied once in and again in :
The bicubic algorithm is frequently used for scaling images and video for display. It preserves fine detail better than the common bilinear algorithm. However, due to the negative lobes on the kernel, it causes overshoot. This can cause clipping, and is an artifact, but it increases acutance, and can be desirable.
