Sparse approximation and n-term rates
Select the largest orthonormal coefficients, derive an explicit tail rate, and account for the extra information needed by adaptive compression.
On this page
The question for today
What you’ll understand
- Prove that largest-coefficient selection is optimal in an orthonormal basis.
- Derive an n-term L² rate from a sorted coefficient bound.
- Separate approximation error, recovery from measurements, and encoded storage cost.
A fixed approximation space spends the same coordinates on every target. Sparse approximation lets the target decide which coordinates deserve storage. In an orthonormal basis this nonlinear choice is easy to analyze: retain the largest coefficients and discard the smallest energy. The benefit depends on coefficient decay and basis choice, and the resulting error theorem does not automatically tell us how to identify those coefficients from limited observations.
Choose a subspace after seeing the target
Let f belong to a Hilbert space with a complete orthonormal basis and square-summable coefficient sequence c. An N-term approximant uses at most N basis vectors, but the retained index set may depend on f. For a fixed set, orthogonal projection chooses the original coefficients on those indices and gives residual energy equal to the sum of the remaining squared magnitudes. No different retained coefficient values can do better on that same set.
To minimize over sets, exchange any retained smaller magnitude with an omitted larger one. The retained energy increases, so the residual decreases. Repeating this reasoning shows that N largest magnitudes are optimal, with ties handled arbitrarily. Write the sorted magnitudes in nonincreasing order to express the best error as a tail. Although each fixed-index projection is linear, selecting indices from the target is nonlinear: adding two signals can change which coordinates are largest.
A sorted power bound yields a rate
Suppose the sorted magnitudes satisfy for every positive integer k, where s is strictly greater than one half. Square this bound and sum beyond N. Since the decreasing power tail is bounded by its integral from N to infinity, the squared error is at most . Taking a square root gives the displayed rate. The strict exponent condition is what makes that integral finite.
This is a sufficient bound on a particular coefficient class, with an explicit constant. It does not claim that every square-summable sequence has the same algebraic rate. At the borderline exponent one half, this argument gives no finite tail estimate, even though a specific sequence with extra logarithmic decay might still lie in the Hilbert space. The rate also belongs to the selected basis: a feature sparse in wavelets need not have the same sorted decay in global Fourier coordinates.
Compression needs more than a tail theorem
An adaptive approximation must specify retained locations as well as values. Counting N real coefficients alone therefore omits index storage, finite precision, and metadata such as scale or boundary conventions. Quantization changes each retained value and adds error. In an orthonormal basis, squared quantization errors on retained coordinates add independently to the omitted energy, giving a clean separation between selection and encoding. The optimal balance depends on the actual bit budget, not just the number of mathematical terms.
Finding large coefficients may require access to the full signal or a specialized recovery procedure. Compressed sensing asks a different question: can a sparse or compressible target be recovered from fewer measurements under conditions on the sensing map? A best N-term error is a benchmark for that task, not a proof that recovery succeeds. With nonorthogonal dictionaries, even the largest-coefficient selection rule can fail because atoms interact; sparse optimization then needs assumptions and algorithms beyond this elementary orthogonal argument.
Work it through
Adaptive selection beats a fixed prefix
In a four-dimensional orthonormal basis, take coefficient vector (0.05,3,−0.1,1). Compare the best two-term approximation with keeping the first two coordinates in the original ordering.
The two largest magnitudes are three and one, at the second and fourth indices. Retaining them leaves coefficients 0.05 and −0.1 omitted.
The best squared error is 0.05²+0.1²=0.0125, hence error approximately 0.1118. Orthonormality makes this a direct Euclidean tail calculation; no cross terms appear.
Keeping the first two coordinates instead discards −0.1 and one, giving squared error 1.01 and error approximately 1.005. Both methods retain two coefficients, but the adaptive method must also encode that its selected indices are two and four.
Target-dependent selection reduces squared error from 1.01 to 0.0125 in this example, with an additional need to specify the selected indices.
The living notebook
Explore the model
Interactive lab · spectral
A case where the largest coefficients are already ordered
Pause and Reset, then set N = 3 and p = 2. Predict the retained sine coefficients are 1,1/4,1/9, already decreasing in magnitude. Reset and set both N = 6 and p = 2 to compare three added smaller modes. At time zero, orthogonality on [0,2π] makes added squared energy π times the sum of k^(−4) for k=4,5,6. This finite ordered family makes prefix truncation agree with largest-coefficient selection; it does not show arbitrary adaptive index choices.
- Sum at x = π/2
- 0.867
- Smallest wavelength
- 2π / 5
- Highest-mode coefficient
- 0.2
A finite sum of sin(nx − t)/nᵖ. Playback changes t; increasing the cutoff includes finer spatial scales. This is not a proof about a continuum limit or renormalization. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forSparse selection can differ greatly from a fixed prefix, but they coincide when the basis coefficients already decrease in the chosen order.
Pause & reason
Check your understanding
Your turn
Try it yourself
Assume sorted magnitudes satisfy . Use the derived bound to find the smallest positive integer N for which that bound guarantees L² error at most 0.025. Do not claim the bound is the exact actual error.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Insert C=1 and s=2, then solve 1/(√3 N^(3/2))≤0.025. Compare the neighboring integer candidates directly.
Compare with the solution
The guarantee is . At N=8 it equals , which does not meet the requested guarantee. At N=9 it equals , which does. Since this bound decreases with N, nine is the smallest integer certified by this particular estimate. The actual coefficient tail may be smaller and could meet the tolerance earlier; the calculation establishes only what the stated upper bound proves.
Keep the idea
Take it with you
- Largest-coefficient selection is an exact N-term optimum for orthonormal bases.
- A sorted decay bound with exponent above one half gives an explicit L² tail rate.
- Index storage, quantization, and recovery from measurements add requirements beyond approximation error.
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