BINOMIAL Series

7/29/2019 BINOMIAL Series

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Binomial expansion, power series,

limits,

approximations, Fourier series

Notice: this material must not be used as a substitute for attending

the lectures

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1 Binomial expansion

We know that

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

The question is (at this stage): what about (a + b)n where n is any positive integer?

1.1 Pascals triangle

11 1

1 2 11 3 3 11 4 6 4 1

1 5 10 10 5 11 6 15 20 15 6 1

To expand (a + b)n we look for the row starting with 1 and n.

1.2 Example

Lets expand (a + b)3. The row in Pascals triangle starting with 1 and 3 is

1 3 3 1

Therefore the expansion of (a + b)3 is

(a + b)3 = a3 + 3a2b + 3ab2 + b3

1.3 Example

Lets expand (a + b)6.The row starting with 1 and 6 in Pascals triangle is the row

1 6 15 20 15 6 1

This means that the expansion of (a + b)6 is

(a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6

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1.4 Factorial notation

The factorial n! of a positive integer n is defined by

n! = n(n 1)(n 2) (3)(2)(1)

so for example5! = 5 4 3 2 1 = 120

and8! = 8 7 6 5 4 3 2 1 = 40320

We work with the convention that

1! = 1 and 0! = 1

Expressions involving factorials can often be simplified as shown in the example below:

8!5!3!

= 8 7 6 5 4 3 2 1(5 4 3 2 1)(3 2 1) = 8 7 66 = 56

1.5 Binomial theorem

Pascals triangle can be difficult to use if the exponent is very high. In such cases thefollowing binomial theorem is usually better. This states that if n is a positiveinteger then

(a + b)n = an + nan1b +n(n 1)

2!an2b2 +

n(n 1)(n 2)3!

an3b3 + + bn

An important particular case is when a = 1 and b = x giving

(1 + x)n = 1 + nx +n(n 1)

2!x2 +

n(n 1)(n 2)3!

x3 + + xn (1.1)

which, like the previous result, holds for positive integers n.In the binomial theorem, the general term has the form anmbm with coefficient

n(n 1)(n 2) (n (m 1))m!

which equalsn(n 1)(n 2) (n (m 1))(n m)!

m! (n m)!or

n!

m! (n m)! often denoted

nm

In terms of the notation introduced above, the binomial theorem can be written as

(a +b)n =

n0

an+

n1

an1b+

n2

an2b2+ +

nn

bn =

ni=0

ni

ani bi

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1.6 Example

Expand

2 + x3

4.

Solution. Using the binomial theorem:

2 + x34 = 24 + (4)(23)( x

3) + (4)(3)

2!(22)( x

3)2 + (4)(3)(2)

3!(2)(x

3)3 + (4)(3)(2)(1)

4!(x

3)4

= 16 +32

3x +

8

3x2 +

8

27x3 +

1

81x4.

1.7 Example

Expand

1 + x3

15up to and including the term in x3.

Solution. By the binomial theorem:

1 + x315

= 1 + 15(x

3) +

(15)(14)

2!(

x

3)2 +

(15)(14)(13)

3!(

x

3)3 +

= 1 + 5x +

35

3x2 +

455

27x3 +

1.8 Example

Expand (1 x)3(2 + x)6 up to and including the term in x2.Solution.

(1 x)3(2 + x)6 = (1 x)3

26 + (6)(25)x +(6)(5)

2!(24)x2 +

=

1 + 3(x) + (3)(2)2!

(x)2 + (x)3 redundant

(64 + 192x + 240x2 + )= (1 3x + 3x2 x3)(64 + 192x + 240x2 + )= 64 + (192 (64)(3))x + (3(64) 3(192) + 240)x2= 64 144x2 +

1.9 Powers that are NOT positive integers

The binomial expansion as discussed up to now is for the case when the exponent isa positive integer only.

For the case when the number n is not a positive integer the binomial theorembecomes, for 1 < x < 1,

(1 + x)n = 1 + nx +n(n 1)

2!x2 +

n(n 1)(n 2)3!

x3 + (1.2)

This might look the same as the binomial expansion given by expression (1.1), butlet us make the following important distinctions between (1.1) and (1.2):

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the expansion for positive integer powers (expansion (1.1)) terminates, i.e. ithas only a finite number of terms. However, for powers that are not positiveintegers the series (1.2) is an infinite series that goes on forever.

it can be mathematically proven that the series (1.2) is valid only for

1 < x 1:

1 +

1

x1/2

= 1 +1

2 1

x+

(12

)(12

)

2! 1

x2

+(12

)(12

)(32

)

3! 1

x3

+

= 1 + 12x

18x2

+ 116x3

+

valid for |x| > 1.The above expansion is no good if |x| < 1. For this case the following trick

produces a valid expansion:

1 +

1

x

1/2=

x + 1

x

1/2=

1

x1/2(1 + x)1/2 expand this

=

1

x1/2 1 +1

2x +

(12

)(

12

)

2! x2

+

(12

)(

12

)(

32

)

3! x3

+ =

1

x1/2

1 +

1

2x 1

8x2 +

1

16x3 +

=1

x1/2+

1

2x1/2 1

8x3/2 +

1

16x5/2 +

Note that this is actually defined only for 0 < x < 1.

1.14 Example

Expand(1+x)2

(1x/2)3 up to and including the term in x2

.Solution.

(1 + x)2

(1 x/2)3 = (1 + x)2

1 x2

3

= (1 + 2x + x2)

1 + (3)

x

2

+

(3)(4)2!

x

2

2+

= (1 + 2x + x2)

1 +

3x

2+

3x2

2+

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= 1 +

3

2+ 2

x +

1 + 2

3

2

+

3

2

x2 +

= 1 +7x

2+

11x2

2

2 Taylor and Maclaurin series

2.1 Taylor series

The idea is to expand a function f(x) about a point a in the form of a sum of powersof (x a), i.e. to form a series of the form

f(x) = a0 + a1(x a) + a2(x a)2 + a3(x a)3 + =n=0

an(x a)n (2.3)

we want to know the coefficients an, n = 0, 1, 2, . . . in the above expansion.

If we differentiate expression (2.3) again and again, we get the following expres-sions for the first, second, third, etc derivatives of f(x):

f(x) = a1 + 2a2(x a) + 3a3(x a)2 + 4a4(x a)3 + f(x) = 2a2 + (3)(2)a3(x a) + (4)(3)a4(x a)2 + f(x) = (3)(2)a3 + (4)(3)(2)a4(x a) +

......

Putting x = a in these expressions gives

f(a) = a1

a1 = f

(a)

f(a) = 2a2 a2 = 12

f(a)

f(a) = (3)(2)a3 a3 = 1(2)(3)

f(a)

Spotting the pattern, we see that the general formula for the coefficient an will be

an =1

n!f(n)(a)

where f(n)(a) means the nth derivative of f(x), evaluated at the value x = a.This gives us what we call the Taylor expansion of a function f(x) valid for

values of x near to a:

f(x) = f(a) + (x a)f(a) + (x a)2

2!f(a) +

(x a)33!

f(a) + (2.4)

The series carries on to infinity, and has general term (xa)n

n!f(n)(a).

Taylors expansion, and the related Maclaurin expansion discussed below, areused in approximations. In practice usually only the first few terms in the series arekept and the rest are discarded. The idea is that the resulting truncated expansionshould provide a good approximation to the function f(x) for values ofx close to theparticular value a. The more terms we keep, the better the approximation.

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2.2 Maclaurin series

There is also the Maclaurin expansion, which is just the Taylor expansion in theparticular case when a = 0, i.e.

f(x) = f(0) + xf

(0) +x2

2! f

(0) +x3

3! f

(0) + (2.5)or, in summation notation

f(x) =n=0

f(n)(0)xn

n!

Not all functions have Taylor or Maclaurin expansions but most do.

2.3 Example

Let us find the Maclaurin series of ex.

Solution. Let f(x) = ex.Then f(0) = 1.Also f(x) = ex so f(0) = 1.f(x) = ex so f(0) = 1. Clearly in this particular example f(n)(0) = 1 for all

n = 1, 2, 3, . . .. Putting these values for f(0), f(0), f(0), etc, into (2.5) gives us theMaclaurin series for the particular function f(x) = ex, namely

ex = 1 + x +x2

2!+

x3

3!+ (2.6)

or, in summation notation, ex

=

n=0

xn

n!

2.4 Example

Deduce the Maclaurin series of e5x from that for ex.Solution. Just replace every x by 5x in expression (2.6) above to get

e5x = 1 + 5x +(5x)2

2!+

(5x)3

3!+

= 1 + 5x +25x2

2

+125x3

6

+

2.5 Example

Find the Maclaurin series of cos x.Solution. Let f(x) = cos x.Then f(0) = 1.Also f(x) = sin x so f(0) = 0.f(x) = cos x so f(0) = 1.f(x) = sin x so f(0) = 0.

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f(x) = cos x so f(0) = 1.f(x) = sin x so f(0) = 0.We see the pattern emerging. The values f(0), f(0), f(0), f(0), etc, cy-

cle through the values 1, 0,1, 0, 1, 0,1, 0, . . .. Putting these values into the gen-eral Maclaurin expansion (2.5) gives the Maclaurin expansion for the function cos x,

namely

cos x = 1 x2

2!+

x4

4!

or, in summation notation,

cos x =n=0

(1)nx2n(2n)!

Similarly, it can be shown that the Maclaurin expansion of sin x is

sin x = x x3

3!+

x5

5!

2.6 Example

Find the Taylor series of the function f(x) = 1/x about x = 2.Solution. We are asked for a Taylor series here, not the Maclaurin one. The

relevant formula is therefore (2.4) in the case when a = 2. So we need to work outthe values f(2), f(2), f(2), etc. We do this next:

f(2) = 12

.f(x) = 1

x2so f(2) = 1

4.

f(x) = 2x3

so f(2) = 14

.f(x) =

6

x4 so f

(2) =

3

8,

and so on. The Taylor series about the value x = 2 is

f(x) = f(2) + (x 2)f(2) + (x 2)2

2!f(2) +

(x 2)33!

f(2) +

which becomes, since f(x) = 1/x,

1

x=

1

2 1

4(x 2) + 1

8(x 2)2 1

16(x 2)3 +

What this means, is that the first few terms of the above series expansion will con-

stitute a good approximation to 1/x for values of x close to 2.Note that the function f(x) = 1/x does not have a Taylor series expansion about

the point x = 0. This is because this function goes to infinity as x 0, so we couldhardly expect the function to have an approximation for small values of x as a seriesof powers of x. Had we attempted to find f(0), f(0), f(0), etc, they would all turnout to be infinity.

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2.7 Example

Find the first three non-zero terms of the Maclaurin series of e2x sin x.Solution. One way to do this would be to write down the Maclaurin series for

e2x (which can be inferred from the one for ex by replacing every x by 2x) andthe series for sin x and then multiplying the series together and expanding out. Theapproach below is a direct approach not requiring such advance knowledge of the twoseparate Maclaurin expansions.

Let f(x) = e2x sin x. Then f(0) = 0.f(x) = e2x cos x 2e2x sin x so f(0) = 1. Differentiating again

f(x) = e2x( sin x) 2e2x cos x 2(e2x cos x 2e2x sin x)= 3e2x sin x 4e2x cos x

andf(x) = 3(e2x cos x

2e2x sin x)

4(

e2x sin x

2e2x cos x)

From these expressions we get f(0) = 4 and f(0) = 11. Putting these values intothe general Maclaurin series (2.5) gives the following expression for our particularfunction f(x) = e2x sin x:

e2x sin x = x 2x2 + 11x3

6+

which will constitute a good approximation to e2x sin x provided x is reasonablysmall.

2.8 Example

Find the binomial expansion of (1x2)1/2 and deduce from it a power series expan-sion for sin1 x.

Solution. First we find the expansion of (1 + x)1/2.

(1 + x)1/2 = 1 + (12

)x +(1

2)(3

2)

2!x2 +

(12

)(32

)(52

)

3!x3 +

= 1 12

x +3

8x2 5

16x3 +

In the above, we now replace every x by x2 to deduce that

(1 x2)1/2 = 1 12

(x2) + 38

(x2)2 516

(x2)3 +

= 1 +1

2x2 +

3

8x4 +

5

16x6 +

Now

sin1 x =x0

dt1 t2

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=x0

(1 t2)1/2 dt

=x0

1 +

1

2t2 +

3

8t4 +

5

16t6 +

dt

= x +1

6x3 +

3

40x5 +

5

112x7 +

3 Applications to working out limits

The notationlimxa

f(x)

means the value (if any) that f(x) approaches, when x approaches a. The word limmeans limit.

3.1 Important issues to do with limitsTwo trivial examples of working out limits would be

limx2

(x2 3) = 1, limx0

cos x = 1

In the above examples we can just put the value in. But in many situations we cannotdo this because we end up with the mathematically meaningless expression 0

0which

could be anything.For example, lets work out

limx

2

x2 4x 2

In this example we cannot put x = 2 into the expression otherwise we get 00

whichcould be anything. But we can simplify the expression by factorising and cancellingfactors to get

limx2

x2 4x 2 = limx2

(x 2)(x + 2)x 2 = limx2(x + 2) = 4

Similarly, lets work out

limx1

x2 + x 2x2

x

Again we cannot just put x = 1 into this expression or we would get 00

. But we canfactorise and simplify as follows:

limx1

x2 + x 2x2 x = limx1

(x 1)(x + 2)x(x 1) = limx1

x + 2

x= 3.

It is not always possible to work out limits simply by looking for factors and sim-plifying as in the above examples. We now want to add binomial expansion andTaylor/Maclaurin series to our list of methods for working out limits.

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3.2 Example

Lets work out

limx0

(1 + x/2)5/7 1x

Again, we cannot put x = 0 into this expression as it stands. But we can use binomialexpansion, as follows;

(1 + x/2)5/7 1x

=

1 + ( 5

7)(x

2) +

( 57)( 2

7)

2!(x2

)2 + 1

x

=514

x 5196

x2 + x

=5

14 5

196x +

We can let x 0 in the above expression to deduce that

limx0

(1 + x/2)5/7 1x

= 514

3.3 Example

Lets work out

limx0

sin x

xand lim

x0

sin2x

xSolution. We mentioned earlier that

sin x = x x3

3!+

x5

5! x

7

7!+

Hencesin x

x= 1 x

2

3!+

x4

5! x

6

7!+

We can let x 0 in this to deduce thatlimx0

sin x

x= 1

From the Maclaurin expansion for sin x given above, we can deduce the expansion forsin2x to be

sin2x = 2x

(2x)3

3!

+(2x)5

5! = 2x 4x

3

3+

32x5

120

Hencesin2x

x= 2 4x

2

3+

Letting x 0 we deduce thatlimx0

sin2x

x= 2

It is in fact a general result that limx0sinkxx

= k for any constant k.

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3.4 Example

Find

limx0

sin2 x x2 cos xx4

Solution. Recall thatsin x = x x

3

3!+

x5

5! x

7

7!+

and

cos x = 1 x2

2!+

x4

4! x

6

6!+

Squaring the formula for sin x gives

sin2 x =

x x

3

6+

x5

120

x x

3

6+

x5

120

= x2

x4

6

+ (something) x6

x4

6

+ (something) x6

= x2 x4

3+ (something) x6

Hence, using also the expansion for cos x given above, we have

sin2 x x2 cos xx4

=

x2 x4

3+ (something) x6 +

x2

1 x2

2+ x

4

24+

x4

=16

x4 + (something) x6 + even higher powers of x

x4

=1

6+ (something) x2 +

Let x 0 in the above to get

limx0

sin2 x x2 cos xx4

=1

6

3.5 Example

Findlimx

x(e1/x 1)Solution. To deal with x going to infinity, we shall let y = 1/x and let y 0. Thisgives

limx

x(e1/x 1) = limy0

1

y(ey 1)

= limy0

1

y

1 + (y) + (y)

2

2!+

1

= limy0

1 + y

2!+

= 1

where we have used the Maclaurin expansion for the exponential, given by (2.6).

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4 LHopitals rule

Another way of working out a limit when in a 00

situation is the following result:

if f(a) = 0 and g(a) = 0 then limxa

f(x)

g(x)= lim

xa

f(x)

g(x)

The above result is called LHopitals rule.It is absolutely crucial to check the condition f(a) = 0 and g(a) = 0 before using

the rule, because it does not work otherwise.

4.1 Example

limx0

3x sin xx

would be 00

if we put x = 0 in, so use LHopital

= limx0

3 cos x1

no longer 00

= 3 cos01= 2

4.2 Example

limx0

1 cos xx + x2

would be 00

if we put x = 0 in, so use LHopital

= limx0 sin x1 + 2x no longer00

= 01

= 0

4.3 Example

limx2

x 2x2 4 = limx2

1

2x=

1

4

Sometimes we have to apply LHopitals rule more than once to get an answer, as thenext example illustrates:

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4.4 Example

limx0

x sin xx3

00

so use LHopital

= limx01

cos x

3x2 still0

0 so use LHopital again

= limx0

sin x

6xstill 0

0so use LHopital again

= limx0

cos x

6no longer 0

0

= 16

4.5 Example

limx0

lncos x

lncos3x0

0 so use LHopital

= limx0

sinx

cosx

3sin3x

cos3x

now simplify this= lim

x0

tan x

3tan3xstill 0

0so use LHopital again

= limx0

sec2 x

9sec2 3xno longer 0

0

= 19

5 Fourier Series

A Fourier Series is an expansion of a periodic function as an infinite sum of sinesand cosines.

Simple examples of periodic functions (other than sin and cos) are the square waveand sawtooth functions. An example of a square wave function (of period 4 in thisparticular case) is the periodic function

f(t) =

1, 2 < t < 0,

1, 0 < t < 2,

with f(t + 4) = f(t).An example of a sawtooth function of period 2 would be the periodic function

of period 2 such that f(t) = t for t (, ). Since this function has period 2 wemight suppose that it has an expansion in terms of the functions cos t, cos 2t, cos 3t , . . .and the functions sin t, sin 2t, sin 3t , . . . since these functions also have period 2. Suchan expansion does indeed exist and in fact any periodic function of period 2 has anexpansion in terms of these trigonometric functions.

If the period is T rather than 2 this is no particular problem. All we have to dois modify the period of the cos and sin functions we work with, i.e. we instead seek

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an expansion in terms of the functions cos 2ntT

and sin 2ntT

for n = 1, 2, 3 . . ., ratherthan cos nt and sin nt. Letting f(t) be a T0 periodic function, this expansion, calledthe Fourier series of f(t), turns out to be

f(t) = 12

a0

+

n=1

an cos2nt

T0+ b

nsin

2nt

T0 (5.7)where

an =2

T0

T00

f(t)cos2nt

T0dt, n = 0, 1, 2, 3, . . . (5.8)

bn =2

T0

T00

f(t)sin2nt

T0dt, n = 1, 2, 3, . . . (5.9)

A number of important points need to be made:

When working out the integrals in (5.8,5.9) you can in fact use any interval of

length T0. As a consequence, the alternative formulae:

an =2

T0

T0/2T0/2

f(t)cos2nt

T0dt, n = 0, 1, 2, 3, . . . (5.10)

bn =2

T0

T0/2T0/2

f(t)sin2nt

T0dt, n = 1, 2, 3, . . . (5.11)

will work just as well.

to work out a0 in (5.7) you use the an formula (either (5.8) or (5.10)) withn = 0. You will sometimes find that the n = 0 case needs to be dealt with

separately from the other an coefficients due to division by zero problems.

the quantity T0 is the period of the wave so the frequency would be 1/T0,usually measured in cycles per second. It is, however, more usual to define thefrequency to be the quantity 0 defined by

0 =2

T0rather than 1

T0

often we want to work out the Fourier series of a periodic function that containspoints of discontinuity (the abovementioned square wave and sawtooth functions

being examples). It is known that, at a point of discontinuity (at x = a, say)the Fourier series of the function converges to

12

(f(a+) + f(a))rather than to f(a). This applies regardless of how f(t) is defined (if it is definedat all) at the point a itself. In the above formula the notation f(a+) meansthe value just to the right of the discontinuity and f(a) means the value tothe left. More formally, f(a+) is the limit of f(a + h) as h tends to zero fromabove, and f(a) is the limit of f(a h) as h tends to zero from above.

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5.1 Example

Let

f(t) =

1 < t < 0

1 0 < t <

with f(t + 2) = f(t). Find the Fourier series of f(t).Solution. In this case the period T0 is given by T0 = 2. Let us find an first. The

following formulae will be useful

cos n = (1)n and sin n = 0, n = 0,1,2,3, . . .We have

an =2

T0

T0/2T0/2

f(t)cos2nt

T0dt

=1

f(t)cos ntdt

= 1

0

(1) cos ntdt + 1

0

cos ntdt

=1

sin ntn

0

+1

sin nt

n

0

an = 0

Because of the n in the denominator of the above calculations we need to find a0separately, but it turns out also to be zero. We would warn you in advance, however,that in plenty of other situations a separate calculation for a0 is absolutely essential

for a correct Fourier series.

Now lets find bn. we have

bn =1

f(t)sin ntdt

=1

0

(1)sin ntdt +0

sin ntdt

=1

cos nt

n

0

+ cos nt

n

0

=1

n(1 cos(n) + ( cos n + 1))

=1

n(2

2(

1)n)

and so

bn =2

n(1 (1)n)

With all the an, n = 0, 1, 2, . . ., equal to zero, and also recalling that the periodT0 = 2, the Fourier series becomes

f(t) =n=1

bn sin nt

= b1 sin t + b2 sin2t + b3 sin3t +

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i.e.

f(t) =4

sin t +

4

3sin3t +

4

5sin5t +

6 Even and odd functions

If a function is an even function or an odd function then certain simplifications arepossible in the calculations required for computing the Fourier series. But note thatplenty of functions are neither even nor odd, eg f(t) = t2 + t.

6.1 Even functions

f(t) is said to be an even function if f(t) = f(t). This means the graph issymmetrical about the y-axis.

Examples of even functions are f(t) = constant, f(t) = t2, f(t) = t4, f(t) = t6, . . .

(all even powers of t); also f(t) = cos t and f(t) = cosh t.An even function has the important property thataa

f(t) dt = 2a0

f(t) dt

6.2 Odd functions

f(t) is said to be an odd function iff(t) = f(t). This means the graph has 180orotational symmetry about the origin.

Examples of odd functions are f(t) = t, f(t) = t3, f(t) = t5, . . . (all odd powersof t); also f(t) = sin t and f(t) = sinh t.

An odd function has the important property thataa

f(t) dt = 0

6.3 Useful rules of even and odd functions

even even = eveneven odd = oddodd

even = odd

odd odd = even

6.4 Fourier series of an even function

Suppose that f(t) is an even function and we want its Fourier series. Since sin t isan odd function we might anticipate that the Fourier series of an even function willcontain no sine terms. We shall show that this is indeed the case. With f(t) being

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even the bn Fourier coefficient is given by

bn =2

T0

T0/2T0/2

f(t)

even

sin2nt

T0

odddt

= 2T0

T0/2T0/2

(something odd) dt

= 0

So iff(t) is even we can declare from the outset that the bn terms are all zero, and weonly need to work out an, n = 0, 1, 2, . . ., a separate calculation often being neededfor a0. With f(t) being even we get an alternative formula for an as follows:

an =2

T0

T0/2T0/2

f(t)

evencos

2nt

T0

evendt

=4

T0

T0/20

f(t)cos2nt

T0dt

This can save us time and effort.

6.5 Fourier series of an odd function

If f(t) is odd then the an coefficients (including a0) are zero because

an =2

T0 T0/2

T0/2

f(t)odd

cos2nt

T0 even

dt

=2

T0

T0/2T0/2

(something odd) dt

= 0

Thus the Fourier series of an odd function contains only sine terms. Moreover, cal-culation of the bn coefficients of these sine terms can be simplified by exploiting theoddness property.

6.6 ExampleFind the Fourier series of the sawtooth function given by

f(t) = t when 2 < t < 2

with f(t + 4) = f(t) for all t (i.e. the function has period 4).Solution. This function is odd. Its graph has 180o rotational symmetry about

the origin. Since it is odd, we can immediately say that an = 0 for all n (includingn = 0) and we only need to calculate bn.

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Also note that since the period is 4, we have in this case T0 = 4. We now find bn:

bn =2

T0

T0/2T0/2

f(t)sin2nt

T0dt

=2

4 2

2f(t)sin

2nt

4dt

=1

2

22

todd

sinnt

2 odd

dt odd odd = even

=20

t sinnt

2dt

=

t cos nt2

n/2

20

+20

cos nt2

n/2dt

=

4

n

cos n +2

n sin nt

2

n/2 2

0

so that

bn = 4n

(1)n

Recalling that T0 = 4, the Fourier series is

f(t) =n=1

4

n(1)n

sin

nt

2

= 4

n=1(1)n

nsin

nt

2

or, in expanded form,

f(t) = 4

sin t

2+ 1

2sin t 1

3sin

3t

2+

1

4sin2t +

6.7 Example

Find the Fourier series of the function such that

f(t) = t2 + t for < t <

with f(t + 2) = f(t) for all t.Solution. This function is neither even nor odd. It is 2-periodic so T0 = 2.

Using this value for T0 the formula for an becomes

an =1

f(t)cos ntdt

=1

(t2 + t)cos ntdt

=1

(t2 + t)

sin nt

n

(2t + 1)sin nt

ndt

if n = 0

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= 1n

(2t + 1)sin ntdt

= 1n

(2t + 1) cos nt

n

+

2cos nt

ndt

= 1

n(2 + 1)(1)

n

n +(2(

) + 1)(

1)n

n +2

nsin nt

n

= 1

n

4(1)

n

n

=

4

n2(1)n

We have shown that

an =4

n2(1)n if n = 0

A separate calculation has to be done for a0, since we obviously cannot put n = 0into the above formula for an. Putting n = 0 into the original an integral gives

a0 = 1

(t2 + t) dt = 1

t33

+ t22

a0 =22

3

Next we find bn. We have

bn =1

(t2 + t)sin ntdt

=1

(t2 + t)

cos nt

n

+

(2t + 1)

cos nt

n

dt=

1

(2 + ) (1)

n

n+ (2 )(1)

n

n+

1

n

(2t + 1) cos ntdt

=1

2 (1)

n

n+

1

n

(2t + 1)

sin nt

n

2sin nt

ndt

=1

2 (1)

n

n 2

n2

2sin ntdt

so that

bn =

2

n

(

1)n

Putting the formulae for a0, an and bn into the general Fourier series expansion, whichwhen T0 = 2 is

f(t) = 12

a0 +n=1

(an cos nt + bn sin nt)

gives

f(t) =2

3+

n=1

4

n2(1)n cos nt 2

n(1)n sin nt

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7 Complex form of a Fourier series

Recall thatj =

1Recall also, Eulers formula

ejt = cos t + j sin t (7.12)

and the other useful version of it:

ejt = cos t j sin t (7.13)It is easy to see where (7.12) comes from. We simply expand ejt using Maclaurinseries:

ejt = 1 + jt +(jt)2

2!+

(jt)3

3!+

= 1 t2

2!

+t4

4! + j t t3

3!

+t5

5! = cos t + j sin t

Formula (7.13) follows from (7.12) when we replace t by t.Now, if we add equations (7.12) and (7.13) the sin t terms disappear and we get

the following result:

cos t =ejt + ejt

2(7.14)

Similarly, subtracting (7.13) from (7.12) gives

sin t =ejt ejt

2j

(7.15)

Since cos t and sin t are periodic of period 2, so is ejt . Furthermore, so are thefunctions enjt for any integer n (positive or negative). This leads us to supposethat any 2 periodic function f(t) can be represented as a sum of the functions enjt

involving all integers n (positive and negative).With suitable adjustments to the period it can be shown that a similar statement

can be made for a function f(t) of period T0. In fact

f(t) =

n=

cn e2jnt/T0 (7.16)

wherecn =

1

T0

T0/2T0/2

f(t)e2jnt/T0 dt for n = 0,1,2,3, . . . (7.17)Expression (7.16) with coefficients cn given by (7.17) is called the complex formof the Fourier series. Note that the sum is over all integer values of n includingnegative ones.

The convergence properties of the complex form are the same as for the form wediscussed earlier (expression (5.7)), i.e. the infinite sum in (7.16) converges to f(t)unless there us a jump discontinuity in which case the sum converges to the midpointof the jump (regardless of the actual value of f(t) at such a point).

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7.1 Example

Find the complex Fourier series of the function f(t) such that

f(t) = 0 < t < /21 /2 < t < /20 /2 < t <

with f(t + 2) = f(t) for all t.Solution. The period is 2 so T0 = 2. With this value of T0 expressions (7.16)

and (7.17) reduce to

f(t) =

n=

cn ejnt

where

cn =

1

2

f(t)e

jnt

dt

=1

2

/2/2

ejnt dt =1

2

ejnt

jn

/2/2

=1

2jnejn/2 ejn/2

=1

n

ejn/2 ejn/2

2j

Recalling the formula sin x = ejxejx

2j, the above formula for cn can be put into the

formcn =

1

nsin

n

2for n = 1,2,3, . . .

The above calculation does not work if n = 0, because it has n in the denominator,so we do a separate calculation for n = 0. The first line of the above calculation forcn, with n = 0, gives

c0 =1

2

f(t) dt =1

2

/2/2

dt =1

2

So the complex Fourier series of the function is

f(t) =

n=

cn ejnt

with the above expressions for cn and c0.It is possible to convert the complex form into a real form, as follows. We can

write it in the form

f(t) = 12

+1

n=

cnejnt +

n=1

cnejnt

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in which the 12

at the front is the n = 0 term of the sum. Making the substitutionn = m in the first sum of the above expression gives us

f(t) = 12

+

m=1cme

jmt +

n=1cne

jnt

and we can now simply replace m by n in the first sum, since they are dummy variablesplaying a similar role to the variable in a definite integral. This observation gives

f(t) = 12

+n=1

cnejnt +

n=1

cnejnt insert expression for cn

= 12

+n=1

1

n sinn

2ejnt +

1

nsin

n

2ejnt

= 12

+

n=11

nsin

n

2 ejnt + ejnt

=2cosntand so

f(t) = 12

+n=1

2

nsin

n

2cos nt

We have converted the complex form into a real form. In the above sum the termswith n = 2, 4, 6, . . . are all zero. Replacing n by 2n 1 has the effect of removingthese zero terms to give

f(t) = 12

+

n=12

(2n

1)

sin(2n 1)

2cos(2n 1)t

= 12

+n=1

2(1)n+1(2n 1) cos(2n 1)t

which is more computationally efficient.

8 Amplitude and phase spectrum

Recall that the Fourier series of a periodic function of period T0 (and frequency0 = 2/T0) is

f(t) = 12a0 +n=1

an cos 2ntT0

+ bn sin 2ntT0

with certain formulae, namely (5.8) and (5.9), for the coefficients an and bn.

We can write the Fourier series in terms of 0 as

f(t) = 12

a0 +n=1

(an cos n0t + bn sin n0t)

= 12

a0 +n=1

An cos(n0t + n)

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where An is called the amplitude and n the phase. To find formulae for the numbersAn and n, n = 1, 2, 3, . . ., we use elementary trigonometry. Now

An cos(n0t + n) = An cos n0t cos n An sin n0t sin n

Comparing the right hand side of the above expression with

an cos n0t + bn sin n0t

we see that we need

An cos n = an and An sin n = bnThese expressions imply that

An =

a2n + b2n

and

tan n = bnan

The numbers n, n = 1, 2, 3, . . . are called the phase spectrum and the numbersAn, n = 1, 2, 3, . . . are called the amplitude spectrum. These two sets of num-bers together form the spectrum and (with the frequency 0) constitute one way ofdescribing a periodic function.

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