Lets go back to the case of a string fixed at 0 0 and L L , its n t h n t h harmonic is y n ( x , t ) = A n sin ( n π x L ) cos ( ω n t − δ n ) y n ( x , t ) = A n sin ( n π x L ) cos ( ω n t − δ n ) In fact all the modes could be permitted, and so any possible motion of the string can be completely specified by: y ( x , t ) = ∑ n = 1 ∞ A n sin ( n π x L ) cos ( ω n t − δ n ) . y ( x , t ) = ∑ n = 1 ∞ A n sin ( n π x L ) cos ( ω n t − δ n ) . This has been rigorously shown by mathematicians but the complete proof is beyond our scope in this course. Lets accept the mathematicians word on this. We could take a snapshot of this function at a time t = t 0 t = t 0 . Then we could write y ( x ) = ∑ n = 1 ∞ B n sin ( n π x L ) y ( x ) = ∑ n = 1 ∞ B n sin ( n π x L ) where B n = A n cos ( ω n t 0 − δ n ) . B n = A n cos ( ω n t 0 − δ n ) . Likewise we could look at one point at space and look at the oscillations as a function of time. In that case we would get. y ( t ) = ∑ n = 1 ∞ C n cos ( ω n t − δ n ) y ( t ) = ∑ n = 1 ∞ C n cos ( ω n t − δ n ) Lets work with the time snapshot, y ( x ) = ∑ n = 1 ∞ B n sin ( n π x L ) y ( x ) = ∑ n = 1 ∞ B n sin ( n π x L ) We need to figure out what the B n B n factors are and this is what Fourier figured out. We can multiply both sides by the sin sin of a particular harmonic y ( x ) s i n ( n i π x L ) = ∑ n = 1 ∞ B n sin ( n π x L ) s i n ( n i π x L ) y ( x ) s i n ( n i π x L ) = ∑ n = 1 ∞ B n sin ( n π x L ) s i n ( n i π x L ) and now we can integrate both sides Recall cos ( θ − φ ) = cos θ cos φ + sin θ sin φ cos ( θ − φ ) = cos θ cos φ + sin θ sin φ cos ( θ + φ ) = cos θ cos φ − sin θ sin φ cos ( θ + φ ) = cos θ cos φ − sin θ sin φ So sin θ sin φ = 1 2 [ c o s ( θ − φ ) − cos ( θ + φ ) ] sin θ sin φ = 1 2 [ c o s ( θ − φ ) − cos ( θ + φ ) ] Thus This is equal to zero at the limits 0 , L 0 , L except for the particular case when n = n i n = n i . In that case ∫ sin ( n π x L ) sin ( n i π x L ) ⅆ x = ∫ sin 2 ( n π x L ) ⅆ x ∫ sin ( n π x L ) sin ( n i π x L ) ⅆ x = ∫ sin 2 ( n π x L ) ⅆ x So you get After all that we should see that for each term in the sum is zero, except the case where n i = n n i = n . Thus we can simplify the equation: ∫ 0 L y ( x ) sin ( n π x L ) ⅆ x = L 2 B n . ∫ 0 L y ( x ) sin ( n π x L ) ⅆ x = L 2 B n . or B n = 2 L ∫ 0 L y ( x ) sin ( n π x L ) ⅆ x B n = 2 L ∫ 0 L y ( x ) sin ( n π x L ) ⅆ x The above is a very specific form of the Fourier Series for a function spanning an interval from 0 0 to L L and passing through zero at x = 0 x = 0 .

In fact Fourier's theorem can be taken to a next step. This is Fourier's integral theorem. That is any function (even if it is not periodic) can be represented by f ( x ) = 1 π ∫ 0 ∞ [ A ( k ) cos ( k x ) + B ( k ) s i n ( k x ) ] d k f ( x ) = 1 π ∫ 0 ∞ [ A ( k ) cos ( k x ) + B ( k ) s i n ( k x ) ] d k where A ( k ) = ∫ − ∞ ∞ f ( x ) cos ( k x ) ⅆ x A ( k ) = ∫ − ∞ ∞ f ( x ) cos ( k x ) ⅆ x B ( k ) = ∫ − ∞ ∞ f ( x ) sin ( k x ) ⅆ x B ( k ) = ∫ − ∞ ∞ f ( x ) sin ( k x ) ⅆ x A A and B B are called the Fourier transforms of f ( x ) f ( x ) Lets look at an example.

Figure 1

Figure 2

Up until now in the course we have been dealing with very simple waves. It turns out that any complicated wave that can possibly exist can be constructed from simple harmonic waves. So while it may seem that an harmonic wave is an over simplification, it can be used in even the most complex cases.