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derivatives is irrelevant and the process of integration“by“parts can only be blocked after
an odd number of integrations.
Third, if the function is periodic and differentiable to all orders, we can integrate“by“
parts an arbitrary number of times. In that case, the theorem implies that for large n, the
Fourier coef¬cients are decreasing faster than any ¬nite power of n. This is the property of
“in¬nite order” or “exponential” convergence de¬ned above.
Fourth, since a Chebyshev series in x becomes a Fourier series in y only after we
make the substitution x = cos(y), a Chebyshev series “ after the change of variable “
is automatically periodic in y. Therefore, a CHEBYSHEV SERIES ALWAYS HAS THE
NON-PERIODIC (in x, the Chebyshev argument) if all its derivatives are bounded on the
expansion interval, x ∈ [’1, 1].
Fifth, the constant F increases with k, the number of times we integrate-by-parts. Con-
sequently, we cannot take the limit of k ’ ∞ for ¬xed n. (Taking this limit with the false
assumption of k-independent F would imply that all the coef¬cients are zero!) Rather, the
theorem tells us how rapidly the coef¬cients decrease for large n as n ’ ∞ for ¬xed k.
Sixth, this theorem can be strengthened to the statement that if f (x) is in the Lipschitz
space L» “on the circle”, then an , bn ∼ O(n’» ). A function is in the Lipschitz space L» (for
0 < » < 1) if |f (x) ’ f (y)| = O(|x ’ y|» ) for all x, y on the interval; Lipschitz spaces for
» > 1 are de¬ned by taking higher order differences of f . The phrase “on the circle” means
that continuity is analyzed as if the interval x ∈ [’π, π] is bent into a circle so that x = π is
continuous with x = ’π; for example, f (’π) = f (π) implies that the function is not in any
Lipschitz space with » > 0.

EXAMPLE: Let us calculate the index of convergence of the Fourier series solution to

uxx + Q(x) u = f (x); u(0) = u(π) = 0

Since the boundary conditions do not impose periodicity, the solution to this problem is
usually not periodic even if the functions Q(x) and f (x) are periodic themselves.
Because of the lack of periodicity, we really ought to use Chebyshev polynomials for
this problem. Nonetheless, many old papers have applied Fourier series to non-periodic

problems; certain analytical techniques, such as the method of separation-of-variables, fail
otherwise. It therefore is useful to see just how bad this choice of basis functions is.
Because the boundary conditions demand u(0) = 0, we must exclude the cosine terms
(and the constant) from the Fourier series and use only the sine functions:

u(x) = bn sin(nx)

Since the sum of the sine series and the individual sine functions are both antisymmetric
with respect to x = 0, we halve the interval of integration to [0, π] and double the result.
This is natural for this problem since [0, π] is the interval between the boundaries. Thus,
bn = (2/π) u(x) sin(nx)dx

One integration-by-parts gives
bn = ’(2/[nπ]) u(x) cos(nx) + (2/[nπ]) u(1) (x) cos(nx) (2.62)
0 0

The boundary conditions u(0) = u(π) guarantee that the boundary term in (2.62) is zero
independent of Q(x) and f (x). Integrating“by“parts a second time gives
(x) sin(nx) ’ (2/[n π])
2 (1) 2
u(2) (x) sin(nx)dx (2.63)
bn = (2/[n π]) u
0 0

Since sin(0) = sin(nπ) = 0, it does not matter whether u(1) (0) = u(1) (π); the boundary term
in (2.63) is 0 anyway.
Integrating-by-parts a third time gives
(x) cos(nx) ’ (2/[n π])
3 (2) 3
u(3) (x) cos(nx)dx (2.64)
bn = (2/[n π]) u
0 0

Since cos(nπ) = cos(0) for odd n, we are stuck with k = 3 as the algebraic index of conver-
gence unless u(2) (0) = u(2) (π). Since one can only impose one pair of boundary conditions
on the solution of a second order differential equation, we would seem to be out of luck.
Evaluating the differential equation at x = 0, however, gives

= ’Q(0) u(0) + f (0)
u(2) (0)
= f (0)

since the boundary conditions on the differential equation require u(0) = 0. Similarly,
u(2) (π) = f (π). We see that if

f (0) = f (π) = 0,

then the combination of the differential equation with its associated boundary conditions
forces the boundary term in (2.64) to vanish. We can integrate twice more before being
defeated by the fact that u(4) (0) = u(4) (π). Thus, whether the index of convergence is k
= 3 or 5 depends entirely upon whether f (x) satis¬es condition (2.66) (as is often true in
practical applications). A k = 5 geophysical example is Eliasen (1954), who used a Fourier
sine series to solve the “barotropic instability problem” for a cosine jet.
The important point is that we deduced the index of convergence simply from the form
of the differential equation and the boundary conditions. We did not have to specify Q(x)

at all, and f (x) only at two points to compute k. The integration“by“parts theorem is a
very powerful tool because we can apply it before we solve the differential equation.
This theorem is also important because it generalizes to all other spectral basis sets. That
is to say, integration“by“parts can be used to provide similar coef¬cient bounds for any
of the standard basis functions including Chebyshev and Legendre polynomials, Hermite
functions and spherical harmonics.

2.10 Asymptotic Calculation of Fourier Coef¬cients
Theorem 5 (STRIP OF CONVERGENCE FOR FOURIER SERIES) Let z = x + iy and let
ρ denote the absolute value of the imaginary part of the location of that singularity of f (z) which is
nearest the real z-axis. In other words, if zj , j = 1, 2, . . . denotes the location of the singularities of
f (z), then

ρ = min | (zj )| . (2.67)

Then the Fourier series converges uniformly and absolutely within the STRIP in the complex z-
plane, centered on the real axis, which is de¬ned by

| y |< ρ, (2.68)
x = arbitrary [convergence]

and diverges outside the strip,

| y |> ρ, (2.69)
x = arbitrary [divergence]

If the limit exists, then the asymptotic coef¬cients of the Fourier series
∞ ∞
f (z) = a0 + an cos(nz) + bn sin(nz)
n=1 n=1

are related to ρ, the half-width of the strip of convergence, by

lim sup log | an /an+1 |= ρ (2.71)

and similarly for the sine coef¬cients bn .
The asymptotic rate of convergence for real z is given by µ = ρ.
If the series has a ¬nite algebraic index of convergence, then ρ = 0 and the series converges only
on the real axis.
If the function is periodic and entire, i. e., has no poles or branch points except at in¬nity, then
usually ρ = ∞ and the coef¬cients are O[exp(’(n/k) log(n) + O(n))] for some constant k. (The
exceptions with ¬nite ρ are entire functions that grow very rapidly with | z | such as exponentials-
of-exponentials, as discussed in Boyd (1994b).)
Note: In this context, “singularity” includes not only poles and branch points, but also discon-
tinuities caused by a lack of periodicity, as for the saw-tooth function.

PROOF: We omit a rigorous proof, but offer the following argument. Let z = x + iy so
that | exp(i n z) | = exp(’ n y). Thus, in the complex form of the Fourier series
∞ ∞
f (z) = cn exp(inz) = cn exp(inx) exp(’ny)
n=’∞ n=’∞

all terms with negative n grow exponentially as exp(| n |y) in the upper half-plane while
those with n > 0 grow in the lower half-plane. It follows that the series can converge at
| y | = ρ if and only if | c±n | decay at least as fast as exp(’ n ρ).
On the other hand, if f (z) has a pole at y = ρ, the series must converge to ever-larger
values as (z) ’ ρ. This implies that the coef¬cients cannot decrease faster than exp(’ nρ)
(modulo the usual algebraic factor of n). (Note that if the coef¬cients could be bounded by
exp(’n(ρ ’ )) for some small positive , for example, then one could bound the Fourier
series, term-by-term, by a convergent geometric series to show it was ¬nite even at the
Thus, the series coef¬cients must decrease as exp(’ n ρ) when f (z) has a singularity at
| (z) | = µ. Q. E. D.
The theorem shows that “geometric” convergence is normal for a Fourier series. It is only
when the function has a discontinuity or other singularity for real z that the series will have
sub-geometric or algebraic convergence.

2.11 Convergence Theory for Chebyshev Polynomials
Since a Chebyshev polynomial expansion is merely a Fourier cosine series in disguise, the
convergence theory is very similar to that for Fourier series. Every theorem, every identity,
of Chebyshev polynomials has its Fourier counterpart. Nonetheless, the mapping does
produce some noteworthy consequences.
The mapping is

z = cos(θ)

and then

Tn (z) ≡ cos(nθ) (2.74)

The following two series are then equivalent under the transformation:

f (z) = an Tn (z)

f (cos θ) = an cos(nθ)

In other words, the coef¬cients of f (z) as a Chebyshev series are identical with the Fourier
cosine coef¬cients of f (cos(θ)).
Even if f (z) itself is not periodic in z, the function f (cos θ) must inevitably be periodic
in θ with period 2π. As we vary θ over all real θ, the periodicity of cos(θ) implies that z (=
cos[θ]) merely oscillates between -1 to 1 as shown in Fig. 2.15. Since f (cos θ) is periodic, its
Fourier series must have exponential convergence unless f (z) is singular for z ∈ [’1, 1]. It
does not matter if f (z) is periodic in z nor does it matter if f (z) has singularities for real z
outside the interval [-1, 1]. The Fourier cosine series in (2.76) sees f (cos θ) only as a periodic
function. For real θ, the Fourier series sees only those variations in f (z) that occur for
z ∈ [’1, 1]. The exponential convergence of the Fourier series (2.76) then implies equally
fast convergence of the Chebyshev series since the sums are term“by“term identical.





’6 ’4 ’2 0 2 4 6

Figure 2.15: The Chebyshev polynomials, Tn (z), are related to the terms of a Fourier co-
sine series through the identity Tn (cos[θ]) = cos( n θ). The graph shows the relationship
between z and θ, z = cos(θ).

This is implicit in (2.76). Since cos(θ) is symmetric about θ = 0, f (cos θ) is forced to be
symmetric in θ, too, even if f (z) has no symmetry whatsoever with respect to z. Conse-
quently, we need only the cosines.1

Theorem 6 (CHEBYSHEV TRUNCATION THEOREM) The error in approximating f (z) by
the sum of its ¬rst N terms is bounded by the sum of the absolute values of all the neglected coef¬-
cients. If

fN (z) ≡ (2.77)
an Tn (z)


ET (N ) ≡| f (z) ’ fN (z) |¤ | an | (2.78)
n=N +1


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