Advanced Mathematical Methods for Scientists and Engineers Episode 4 Part 5 pdf

28.6 Fourier Sine Series If f x is an odd function, f −x = −f x, then there will not be any cosine terms in the Fourier series.. 28.7 Complex Fourier Series and Parseval’s Theorem By wri

We integrate Equation 28.1 from −π to π to determine a0.

Z π

−π

f (x) dxMultiplying by cos(mx) and integrating will enable us to solve for am

Z π

−π

f (x) cos(mx) dx = πam

am = 1π

Z π

−π

f (x) cos(mx) dx

Note that this formula is valid for m = 0, 1, 2,

Similarly, we can multiply by sin(mx) and integrate to solve for bm The result is

bm = 1π

Z π

−π

f (x) sin(mx) dx

an and bn are called Fourier coefficients

Although we will not show it, Fourier series converge for a fairly general class of functions Let f (x−) denote theleft limit of f (x) and f (x+) denote the right limit

Example 28.2.1 For the function defined

f (x) =

(

0 for x < 0,

x + 1 for x ≥ 0,the left and right limits at x = 0 are

Periodic Extension of a Function Let g(x) be a function that is arbitrarily defined on −π ≤ x < π TheFourier series of g(x) will represent the periodic extension of g(x) The periodic extension, f (x), is defined by the twoconditions:

Figure 28.1: The Periodic Extension of g(x) = x2.Limited Fluctuation A function that has limited total fluctuation can be written f (x) = ψ+(x) − ψ−(x), where

ψ+ and ψ− are bounded, nondecreasing functions An example of a function that does not have limited total fluctuation

is sin(1/x), whose fluctuation is unlimited at the point x = 0.

Functions with Jump Discontinuities Let f (x) be a discontinuous function that has a convergent Fourierseries Note that the series does not necessarily converge to f (x) Instead it converges to ˆf (x) = 1

2(f (x−) + f (x+)).Example 28.2.2 Consider the function defined by

π − 2x for 0 < x < π

The function ˆf (x) is plotted in Figure28.2

28.3 Least Squares Fit

Approximating a function with a Fourier series Suppose we want to approximate a 2π-periodic function

f (x) with a finite Fourier series

is it possible to choose coefficients αn and βn such that

-3 -2 -1 1 2 3

-3 -2 -1

1 2

Figure 28.2: Graph of ˆf (x)

would give a better approximation?

Least squared error fit The most common criterion for finding the best fit to a function is the least squares fit.The best approximation to a function is defined as the one that minimizes the integral of the square of the deviation.Thus if f (x) is to be approximated on the interval a ≤ x ≤ b by a series

the best approximation is found by choosing values of cn that minimize the error E.

E ≡

Z b a

2

dx

Generalized Fourier coefficients We consider the case that the φn are orthogonal For simplicity, we alsoassume that the φn are real-valued Then most of the terms will vanish when we interchange the order of integrationand summation

E =

Z b a

E =

Z b a

φnφmdx

E =

Z b a

φ2ndx

E =

Z b a

φ2ndx − 2cn

Z b a

We call these the generalized Fourier coefficients.

For such a choice of the cn, the error is

E =

Z b a

φ2ndx

Since the error is non-negative, we have

Z b a

φ2ndx

This is known as Bessel’s Inequality If the series in Equation 28.2 converges in the mean to f (x), lim N → ∞E = 0,then we have equality as N → ∞

Z b a

φ2ndx

This is Parseval’s equality

Fourier coefficients Previously we showed that if the series,

an= 1π

Now we show that by choosing the coefficients to minimize the squared error, we obtain the same result We applyEquation 28.3 to the Fourier eigenfunctions.

Z π

−π

f (x) sin(nx) dx

28.4 Fourier Series for Functions Defined on Arbitrary Ranges

If f (x) is defined on c − d ≤ x < c + d and f (x + 2d) = f (x), then f (x) has a Fourier series of the form

d



Since

Z c+d c−d

sin2 nπ(x + c)

d



dx = d,the Fourier coefficients are given by the formulas

an= 1d

Z c+d c−d

f (x) cos nπ(x + c)

d

dx

bn= 1d

Z c+d c−d

f (x) sin nπ(x + c)

d

dx

Example 28.4.1 Consider the function defined by

This function is graphed in Figure 28.3

The Fourier series converges to ˆf (x) = (f (x−) + f (x+))/2,

f (x) is also graphed in Figure 28.3

The Fourier coefficients are

an= 13/2

= 23

= 23

Z 3/2 1/2

(x − 1/2) cos 2nπx

3

dx

+23

Z 5/2 3/2

(4 − 2x) cos 2nπx

3

dx

= − 1(nπ)2 sin 2nπ

3

h2(−1)nnπ + 9 sin

nπ3

i

-1 -0.5 0.5 1 1.5 2

-1 -0.5

0.5

-1 -0.5 0.5

Figure 28.3: A Function Defined on the range −1 ≤ x < 2 and the Function to which the Fourier Series Converges

bn= 13/2

= 23

= 23

Z 3/2 1/2

(x − 1/2) sin 2nπx

3

dx

+23

Z 5/2 3/2

(4 − 2x) sin 2nπx

3

dx

= − 2(nπ)2sin2nπ

3

 h2(−1)nnπ + 4nπ cosnπ

28.5 Fourier Cosine Series

If f (x) is an even function, (f (−x) = f (x)), then there will not be any sine terms in the Fourier series for f (x) TheFourier sine coefficient is

bn = 1π

Z π

−π

f (x) sin(nx) dx

Since f (x) is an even function and sin(nx) is odd, f (x) sin(nx) is odd bn is the integral of an odd function from −π

to π and is thus zero We can rewrite the cosine coefficients,

an = 1π

Z π

−π

f (x) cos(nx) dx

= 2π

Z π 0

Z π/2 0

x cos(nx) dx + 2

π

Z π π/2

In Figure28.4the even periodic extension of f (x) is plotted in a dashed line and the sum of the first five nonzero terms

in the Fourier cosine series are plotted in a solid line

-3 -2 -1 1 2 3

0.25 0.5 0.75 1 1.25

Figure 28.4: Fourier Cosine Series

28.6 Fourier Sine Series

If f (x) is an odd function, (f (−x) = −f (x)), then there will not be any cosine terms in the Fourier series Since

f (x) cos(nx) is an odd function, the cosine coefficients will be zero Since f (x) sin(nx) is an even function,we canrewrite the sine coefficients

bn = 2π

Z π 0

f (x) sin(nx) dx

Example 28.6.1 Consider the function defined on [0, π) by

Z π/2 0

x sin(nx) dx + 2

π

Z π π/2



In Figure28.5 the odd periodic extension of f (x) is plotted in a dashed line and the sum of the first five nonzero terms

in the Fourier sine series are plotted in a solid line

28.7 Complex Fourier Series and Parseval’s Theorem

By writing sin(nx) and cos(nx) in terms of eınx and e−ınx we can obtain the complex form for a Fourier series

-3 -2 -1 1 2 3

-1.5 -1 -0.5

0.5 1

Figure 28.5: Fourier Sine Series.The functions { , e−ıx, 1, eıx, eı2x, }, satisfy the relation

0 for n 6= m.Starting with the complex form of the Fourier series of a function f (x),

we multiply by e−ımx and integrate from −π to π to obtain

where ¯z denotes the complex conjugate of z

Assume that f (x) has a uniformly convergent Fourier series

4 +

12

Result 28.7.1 Parseval’s Theorem If f (x) has the Fourier series

28.8 Behavior of Fourier Coefficients

Before we jump hip-deep into the grunge involved in determining the behavior of the Fourier coefficients, let’s take astep back and get some perspective on what we should be looking for

One of the important questions is whether the Fourier series converges uniformly From Result12.2.1we know that

a uniformly convergent series represents a continuous function Thus we know that the Fourier series of a discontinuousfunction cannot be uniformly convergent From Section 12.2 we know that a series is uniformly convergent if it can bebounded by a series of positive terms If the Fourier coefficients, an and bn, are O(1/nα) where α > 1 then the seriescan be bounded by (const)P∞

n=11/nα and will thus be uniformly convergent

Let f (x) be a function that meets the conditions for having a Fourier series and in addition is bounded Let(−π, p1), (p1, p2), (p2, p3), , (pm, π) be a partition into a finite number of intervals of the domain, (−π, π) such that

on each interval f (x) and all it’s derivatives are continuous Let f (p−) denote the left limit of f (p) and f (p+) denotethe right limit

f (p−) = lim

→0 +f (p − ), f (p+) = lim

→0 +f (p + )Example 28.8.1 The function shown in Figure 28.6 would be partitioned into the intervals

(−2, −1), (−1, 0), (0, 1), (1, 2)

-2 -1 1 2

-1 -0.5 0.5

Figure 28.6: A Function that can be Partitioned

Suppose f (x) has the Fourier series

Using integration by parts,

= 1nπ

h

1π

where

An= 1π

and the b0n are the sine coefficients of f0(x)

Since f (x) is bounded, An = O(1) Since f0(x) is bounded,

b0n = 1π

Now we repeat this analysis for the sine coefficients.

and the a0n are the cosine coefficients of f0(x)

Since f (x) and f0(x) are bounded, Bn, a0n= O(1) and thus bn = O(1/n) as n → ∞

With integration by parts on the Fourier coefficients of f0(x) we could find that

where Bn0 = (−1)π n[f0(−π) − f0(π)] −π1 Pm

j=1cos(npj)[f0(p−j) − f0(p+j )] and the a00n are the cosine coefficients of f00(x)

Now we can rewrite an and bn as

then the Fourier coefficients will be O(1/n2) A sufficient condition for this is that the periodic extension of f (x) is

continuous We see that if the periodic extension of f0(x) is continuous then A0n and Bn0 will be zero and the Fourier

coefficients will be O(1/n3)

Result 28.8.1 Let f (x) be a bounded function for which there is a partition of the range

(−π, π) into a finite number of intervals such that f (x) and all it’s derivatives are continuous

on each of the intervals If f (x) is not continuous then the Fourier coefficients are O(1/n).

If f (x), f0(x), , f(k−2)(x) are continuous then the Fourier coefficients are O(1/nk).

If the periodic extension of f (x) is continuous, then the Fourier coefficients will be O(1/n2) The seriesP∞

n=1|ancos(nx)bnsin(nx)|can be bounded by MP∞

n=11/n2 where M = max

n (|an| + |bn|) Thus the Fourier series converges to f (x) uniformly

Result 28.8.2 If the periodic extension of f (x) is continuous then the Fourier series of f (x)

will converge uniformly for all x.

If the periodic extension of f (x) is not continuous, we have the following result

Result 28.8.3 If f (x) is continuous in the interval c < x < d, then the Fourier series is uniformly convergent in the interval c + δ ≤ x ≤ d − δ for any δ > 0.

Example 28.8.2 Different Rates of Convergence

A Discontinuous Function Consider the function defined by

f1(x) =

(

−1 for − 1 < x < 0

1, for 0 < x < 1

This function has jump discontinuities, so we know that the Fourier coefficients are O(1/n)

Since this function is odd, there will only be sine terms in it’s Fourier expansion Furthermore, since the function issymmetric about x = 1/2, there will be only odd sine terms Computing these terms,

bn= 2

Z 1 0

sin(nπx) dx

= 2 −1

nπ cos(nπx)

1 0

The function and the sum of the first three terms in the expansion are plotted, in dashed and solid lines respectively,

in Figure 28.7 Although the three term sum follows the general shape of the function, it is clearly not a goodapproximation

-1 -0.5 0.5 1

-1 -0.5

0.5

-0.4 -0.2

0.2 0.4

Figure 28.7: Three Term Approximation for a Function with Jump Discontinuities and a Continuous Function

A Continuous Function Consider the function defined by

-1 -0.5 0.5 1

-0.2 -0.1

0.1 0.2

1 0.25

0.1 0

0.1 1 1

(−x + 1) sin(nπx) dx

= 2

Z 1/2 0

x sin(nπx) dx + 2

Z 1 1/2

(1 − x) sin(nπx) dx

(nπ)2sin(nπ/2)

=( 4 (nπ) 2(−1)(n−1)/2 for odd n

The function and the sum of the first three terms in the expansion are plotted, in dashed and solid lines respectively,

in Figure 28.7 We see that the convergence is much better than for the function with jump discontinuities

A Function with a Continuous First Derivative Consider the function defined by

f3(x) =

(x(1 + x) for − 1 < x < 0x(1 − x) for 0 < x < 1

Since the periodic extension of this function is continuous and has a continuous first derivative, the Fourier coefficientswill be O(1/n3) We see that the Fourier expansion will contain only odd sine terms

x(1 − x) sin(nπx) dx

= 2

Z 1 0

x(1 − x) sin(nπx) dx

= 4(1 − (−1)

n)(nπ)3

=

( 4 (nπ) 3 for odd n

0 for even n

The function and the sum of the first three terms in the expansion are plotted in Figure 28.8 We see that the firstthree terms give a very good approximation to the function The plots of the function, (in a dashed line), and the threeterm approximation, (in a solid line), are almost indistinguishable

In Figure 28.8 the convergence of the of the first three terms to f1(x), f2(x), and f3(x) are compared In the lastgraph we see a closeup of f3(x) and it’s Fourier expansion to show the error

For any fixed x, the series converges to 12(f (x−) + f (x+)) For any δ > 0, the convergence is uniform in the intervals

−1 + δ ≤ x ≤ −δ and δ ≤ x ≤ 1 − δ How will the nonuniform convergence at integral values of x affect the Fourierseries? Finite Fourier series are plotted in Figure 28.9 for 5, 10, 50 and 100 terms (The plot for 100 terms is closeup

of the behavior near x = 0.) Note that at each discontinuous point there is a series of overshoots and undershootsthat are pushed closer to the discontinuity by increasing the number of terms, but do not seem to decrease in height

In fact, as the number of terms goes to infinity, the height of the overshoots and undershoots does not vanish This isknown as Gibb’s phenomenon

28.10 Integrating and Differentiating Fourier Series

Integrating Fourier Series Since integration is a smoothing operation, any convergent Fourier series can beintegrated term by term to yield another convergent Fourier series

Example 28.10.1 Consider the step function

f (x) =

(

π for 0 ≤ x < π

−π for − π ≤ x < 0

1 1

Z π 0

f (x) ∼

∞

X

n=1 oddn

4

nsin nxIntegrating this relation,

4n

4n

4

∞

X

n=1 oddn

−1

n2 = 2π

Z π 0

F (x) dx = −1

π.

conver-of the function is continuous.

Result 28.10.1 The Fourier series of a function f (x) can be differentiated only if the periodic extension of f (x) is continuous.

Example 28.10.2 Consider the function defined by

cos(nx),

which is false The series does not converge This is as we expected since the Fourier series for f (x) is not uniformlyconvergent.

This is called Parseval’s equation.

2 Find the Fourier series for f (x) = x on −π ≤ x < π (and repeating periodically) Use this to show

Consider the Fourier series of f (x) = x on −π ≤ x < π as found above Investigate the convergence at the points ofdiscontinuity

1 Let SN be the sum of the first N terms in the Fourier series Show that

1 Choose an appropriate periodic extension and find a Fourier series solution

2 Solve directly and find the Fourier series of the solution (using the same extension) Compare the result to theprevious step and verify the series agree

Exercise 28.4

Consider the boundary value problem on 0 < x < π

y00+ 2y = sin x y0(0) = y0(π) = 0

1 Find a Fourier series solution

2 Suppose the ODE is slightly modified: y00+ 4y = sin x with the same boundary conditions Attempt to find aFourier series solution and discuss in as much detail as possible what goes wrong

1 Compute the Fourier sine series for the function

πsin(πν) =

Exercise 28.11

1 Show that

ln

cos

x2

= − ln 2 −

Hint: use the identity

sin((x + ξ)/2)sin((x − ξ)/2)

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-1 -0 .5 0 .5 1 .5 2

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Figure 28 .5: Fourier Sine