Mathematical Tools for Physics

8—Multivariable Calculus 228

These two equations determine the parametersαandβ.

α

∫L

0

dxsin^2

πx

L

=

∫L

0

dxf(x) sin

πx

L

β

∫L

0

dxsin^2

2 πx

L

=

∫L

0

dxf(x) sin

2 πx

L

The other integrals vanish because of the orthogonality ofsinπx/Landsin 2πx/Lon this interval. What you
get is exactly the coefficients of the Fourier series expansion off. The Fourier series is the best fit (in the least
square sense) of a sum of orthogonal functions tof. See section11.6for more on this
Is it a minimum? Yes. Look at the coefficients ofα^2 andβ^2 in Eq. ( 21 ). They are positive;+α^2 +β^2 has
a minimum, not a maximum or saddle point.
The distance function Eq. ( 20 ) is simply (the square of) the norm in the vector space sense of the difference
of the two vectorsfandg. Equations(6.8) and (6.4) here become

‖f−g‖^2 =

〈

f−g,f−g

〉

=

∫b

a

dx

∣

∣f(x)−g(x)

∣

∣^2

~e 1

~e 2

shortest distance

to the plane

The geometric meaning of Eq. ( 21 ) is that~e 1 and~e 2 provide a basis for the two dimensional space

α~e 1 +β~e 2 =αsin

πx

L

+βsin

2 πx

L

The plane is the set of all linear combinations of the two vectors, and for a general vector not in this plane, the
shortest distance to the plane defines the vectorinthe plane that is the best fit to the given vector. It’s the
one that’s closest. Because the vectors~e 1 and~e 2 are orthogonal it makes it easy to find the closest vector. You
require that the difference,~v−α~e 1 −β~e 2 have only an~e 3 component. That is Fourier series.