f(x) = Sn=0,¥ {an cos(nx) + bn sin(nx)}      [1]

I.e.,

f(x) = a0 cos(0x) + b0 sin(0x) + a1 cos(1x) + b1 cos(1x) + a2 cos(2x) + b2 cos(2x) ... + an cos(nx) + bn sin(nx)  ,  n ® ¥;     [1']

The coefficients a_n & b_n which accomplish this (all other variables are known) are :

a0 = 1/2p -p,p f(x) cos(0x) dx = 1/2p -p,p f(x) dx     [2]

an = 1/p -p,p f(x) cos(nx) dx  ,  n = 1, 2,...,¥     [3]

bn = 1/p -p,p f(x) sin(nx) dx  ,  n = 1, 2,...,¥     [4]

Note that 1/2p = 1/(2p). No parenthesis in the denominator for such a combined quantity appears less cluttered. b₀ in [1] is arbitrary because sin(0x) = 0. Thus [2], [3], & [4] express the relevant coefficients a_n & b_n in [4]. This is defined only for the interval x = [-p,p].

Alternate form

Because 1/2p precedes the integral expressing a₀ in [2] and 1/p precedes the integrals expressing the other coefficients in [3] & [4], the factor 1/2 is often moved to [1] in the a₀ term. Thus [1] is often expressed as (noting that cos(0x) = 1) :

f(x) = a0/2 + Sn=1,¥ {an cos(nx) + bn sin(nx)}     [5]

for which

an = 1/p -p,p f(x) cos(nx) dx  ,  n = 0, 1, 2,...,¥     [6]

bn = 1/p -p,p f(x) sin(nx) dx  ,  n = 1, 2,...,¥     [7]

The sets of equations [5], [6], & [7] or [1], [2], [3], & [4] with b₀ arbitrarily chosen are the Fourier series representation of a function f(x).

Variable interval

If an interval other than that of 2p is desired, a change of variables must be made. A substitution variable which converts the x interval [-p,p] to the desired interval is either used when solving the red or blue sets of equations, or placed directly in those equations as a general formula. These are equivalent, and the form above makes the characteristics of the series more clear, so I do the former in the example below.

Truncated series

Because Fourier series are infinite sums, you may hope that the terms become smaller as n ® ¥, such that a series truncated after several terms :

SN(x) = a0/2 + Sn=1,N {an cos(nx) + bn sin(nx)}     [8]

may be reasonably accurate. This is indeed so, though this rate of convergence is generally not fast for non-curved and non-periodic functions.

Example

A simple though rather involved and good example is the Fourier series for the linear function f(x) = 3x for the interval x = 0 to 5 (see graph below). The substitution variable mentioned above is used in the red equations. Calling this variable u,

u = p (2x/5 - 1)  ,  so  x = 5 (p + u)/2p

Thus,

f(u) = f(x) = 3 x = 3 (5 (p + u)/2p) = 15 (p + u)/2p

Note that u varies from -p to p as x does from 0 to 5, and f(u) varies from 0 to 15 on the interval (-p,p) as f(x) does on the interval (0,5), the required transformation.

The coefficients can then be calculated :

a0 = 1/2p -p,p f(u) cos(0u) du  =  1/2p -p,p (15 (p + u)/2u) (1) du  =  15/4p -p,p du  +  15/4p2 -p,p u du

an = 1/p -p,p f(u) cos(nu) du  =  1/p -p,p (15 (p + u)/2p)) cos(nu) du  =  15/2p -p,p cos(nu) du  +  15/2p2 -p,p u cos(nu) du

bn = 1/p -p,p f(u) sin(nu) du  =  1/p -p,p (15 (p + u)/2p)) sin(nu) du  =  15/2p -p,p sin(nu) du  +  15/2p2 -p,p u sin(nu) du

The constant term a₀ is :

a0  =  15/4p u½-p,p  +  15/4p2 (u2/2)½-p,p  =  15/4p (p - (-p))  +  15/8p2 (p2 - p2)  =  15/2

Concerning other coefficients, solutions to the trigonometric integrals are :

-p,p cos(nu) du  =  sin(nu)/n½-p,p  =  0

-p,p sin(nu) du  =  - cos(nu)/n½-p,p  =  0

-p,p u cos(nu) du  =  {u sin(nu)/n + cos(nu)/n2}½-p,p  =  0

-p,p u sin(nu) du  =  {sin(nu)/n2 - u cos(nu)}/n½-p,p  =  - (-1)n (2p/n)

for which the (-1)ⁿ implies that the terms alternate signs for odd & even values of n (cos(np) is 1 if n is even & -1 if n is odd). Thus, the only non-zero coefficients if n > 0 are b_n :

bn = - (15/2p2) (-1)n (2p/n)  =  -(-1)n (15/np)

Thus the coefficients are :

a0 = 15/2 an = 0  ,  n = 0, 1, 2,...,¥

bn = - (-1)n (15/np)  ,  n = 0, 1, 2,...,¥

So that the Fourier series for f(u) is :

f(u)  =  15/2 + Sn=1,¥ {0 cos(nu) - (-1)n (15/np) sin(nu)}  =  15/2 - Sn=1,¥ (-1)n (15/np) sin(nu)

Calculating for f(x) on the interval 0 to 5 requires no more than substituting for u in the Fourier series :

f(x)  =  15/2 - Sn=1,¥ (-1)n (15/np) sin(np (2x/5 - 1))  =  15/2 + (15/1p) sin(1p (2x/5 - 1)) - (15/2p) sin(2p (2x/5 - 1)) ...

Basic properties

A graph of the the truncated series S_N(x) of the above example for N = 0-6 & 101 reveals several basic properties of Fourier series :


The constant term a₀ (or a₀/2 if blue equations are used), which is S₀(x), is 15/2 - the average of f(x) on the interval. Thus, a first approximation to the function is its average. Further terms create an oscillatory solution nearer f(x). This series converges quite slowly (O(1/n)) because a general combination of circular functions are solving for a straight (linear) function. Even 102 terms are insufficient for completely surpressing the endpoint problem. If f(x) were a curved function, convergence would be much better; and (of course) if f(x) were a trigonometric function, it would be calculated exactly as that function. A paradox for this example (and many others) is that although the solution approaches f(x) as n ® ¥, it can never be at the endpoints - it is always 15/2 there ! I suppose the reasoning is that the endpoints vanish as n ® ¥. These properties are certainly not exhaustive, but are the most important ones.

Symmetry

Note in the above example that all a_nterms are 0 for n > 0. This is because f(x) is an oddly symmetric function (see also relevant graph below) :


The b_n sine terms are oddly symmetric; but the a_n cosine terms are evenly symmetric, thus contribute nothing to the solution. The statement is often made that if you know the symmetry of a function which interests you, you can keep only the relevant coefficients, multiply the equations calculating them with 2, and halve the calculation interval to p :

Even symmetry - Fourier cosine series :

f(x) = Sn=0,¥ an cos(nx)     [9]

a0 = 1/p 0,p f(x) dx     [10]

an = 2/p 0,p f(x) cos(nx) dx  ,  n = 1, 2,...,¥     [11]

Odd symmetry - Fourier sine series :

f(x) = a0 + Sn=0,¥ bn sin(nx)     [12]

a0 = 1/p 0,p f(x) dx     [10]

bn = 2/p 0,p f(x) sin(nx) dx  ,  n = 1, 2,...,¥     [13]

though an improved solution is sometimes obtained forcing a type of symmetry if possible. For example, consider the function f(x) = 3x shown above. Though this is oddly symmetric on an infinite interval, you can proverbially convince the Fourier series that it is an evenly symmetric function, using a Fourier cosine series. As illustrated, a cosine function with amplitude equaling f(x) on the interval [0,5] (its negative value) is rather similar with f(x) - @ least beginning with a minimum & ending with a maximum; but the sine function with a maximum in the middle and minima of 0 at the endpoints causes osciallation and endpoint problems. The 'attempt' of the cosine series' convergence to portray the sharp V turn at x = 0 toward evenly symmetric values, provides faster convergence (O(1/n²)) (the function as defined does not extend past that toward negative values, but the series doesn't know that).

Solving the Fourier cosine series (calculations not shown) yields :

f(x)  =  15/2 - 60/p2 Sn=1,3,5,...,¥ cos(np x/5)/n2

A plot of its truncated series for N = 0, 1, 3, 5, & 101 is shown below for comparison with the previous solution.

Fourier Interpolation

My purpose of discussing Fourier series was actually as a prerequisite for its use for interpolation. If you know a function f(x), calculating its Fourier series can be rather absurd (though it can reveal some properties of the function of which you might else be unaware) - why calculate what you already know ? A more practical use though is calculating such a series for a presumed unknown function.

Suppose you have a set of periodic data points such as average monthly precipitation (periodic because they vary from the January to December values, then the next value would be January's again - a period of a year) and you want values between these points which do not vary abruptly. Though rarely bad, linear interpolation does not accomplish this - abrupt shifts likely at the data points :

The periodic nature suggests that a Fourier series is a natural solution. A truncated quasi-Fourier series which interpolates (passes thru) the data points is the presumed function. The solution differs slightly depending whether an odd or even number of points is chosen.

For an odd number 2N + 1 of equally-spaced interpolation points x_k on a periodic interval of 2 p :

xk = 2p k/(2N+1)  ,  k = 1, 2,...,2N+1     [14]

with values at those points f(x_k), the interpolating series F(x) is :

F(x) = Sn=0,N {an cos(nx) + bn sin(nx)}     [15]

a0 = 1/(2N+1) Sk=1,2N+1 f(xk) cos(0xk) = 1/(2N+1) Sk=1,2N+1 f(xk)     [16]

an = 2/(2N+1) Sk=1,2N+1 f(xk) cos(nxk)  ,  n = 1, 2,...,N     [17]

bn = 2/(2N+1) Sk=1,2N+1 f(xk) sin(nxk)  ,  n = 0, 1, 2,...,N     [18]

For an even number 2N of equally-spaced interpolation points x_k on a periodic interval of 2 p :

xk = pk/N  ,  k = 1, 2,...,2N     [19]

with values at those points f(x_k), the interpolating series F(x) is :

F(x) = Sn=0,N-1 {an cos(nx) + bn sin(nx)} + 1/2 {aN cos(Nx) + bN sin(Nx)}     [20]

a0 = 1/2N Sk=1,2N f(xk) cos(nxk) = 1/2N Sk=1,2N f(xk)     [21]

an = 1/N Sk=1,2N f(xk) cos(nxk)  ,  n = 1, 2,...N     [22]

bn = 1/N Sk=1,2N f(xk) sin(nxk)  ,  n = 0, 1, 2,...N     [23]

These forms are analogous with the red equations above. Note that the b₀ & b_N terms are 0 because sin(0x_k) = sin(Nx_k) = 0, so they can be removed from the equations. I used the word quasi-Fourier series above because though F(x) has the same general properties - e.g., a₀ is the average of all f(x_k), convergence is best for smoothly varying data, similar adjustments for symmetry can be made - which you may feel should exist even if the data deviates from it - a few differences exist. The sums for the coefficients using the trapezoidal rule compute them exactly, though obviously fewer are used for interpolation than the infinite number available if the function were known.

Variable change

 Month  k     xk     f(xk)
  JAN   1    1p/6   1.861944
  FEB   2    2p/6   1.663889
  MAR   3    3p/6   2.468889
  APR   4    4p/6   3.047500
  MAY   5    5p/6   2.882778
  JUN   6    6p/6   3.510278
  JUL   7    7p/6   3.162500
  AUG   8    8p/6   3.355833
  SEP   9    9p/6   2.915000
  OCT  10   10p/6   2.158056
  NOV  11   11p/6   2.678056
  DEC  12   12p/6   2.650833

For this, f(x_k) are data for months k = 1, 2,...,12 (thus 2N = 12) January to December, such that the interval begins and ends (x_k = 0 & 2p) at December's monthly value. Any beginning/ending point can be chosen - one closely corresponding with a solstice may be best; though the series doesn't know about solstices either - the best is
probably that for which the plotted data (from beginning to end) most closely resembles the sines & cosines. This is a long calculation, though easy for a computer program. Using the above equations for an even number of interpolation points yields the following coefficients :

  n        an            bn
  0    2.696296          0
  1    -.5756244    -.2815022
  2     .1595369    -.1042437
  3     .2869913    -.1082867
  4     .1899539    -.05059842
  5    -.1410878    -.04984074
  6     .06953704        0

Using which F(x) can be calculated. A diagram of such is :

with the abscissa adjusted, but interpolation points x_k remaining at the center of each month. This method can be used even for irregular data intervals, though the interpolation points must be equally-spaced. The precipitation data is actually an example of this, because monthly durations differ. Below are the days of an average year (assuming 1 leap year for every 4 years, for which most data will be valid) :

 Month   Days of year   Midpoint
  JAN         0-31       15.5
  FEB        31-59.25    45.125
  MAR     59.25-90.25    74.75
  APR    90.25-120.25   105.25
  MAY   120.25-151.25   135.75
  JUN   151.25-181.25   166.25
  JUL   181.25-212.25   196.75
  AUG   212.25-243.25   227.75
  SEP   243.25-273.25   258.25
  OCT   273.25-304.25   288.75
  NOV   304.25-334.25   319.25
  DEC   334.25-365.25   349.75