Binomial Expansion, Power Series, Limits, Approximations, Fourier Series

Home , Binomial theorem, Series expansion

Binomial expansion, power series, limits, approximations, Fourier series

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

the lectures

1 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 Pascal’s triangle

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1

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

1.2 Example Let’s expand (a + b)3. The row in Pascal’s 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 Let’s expand (a + b)6. The row starting with 1 and 6 in Pascal’s 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

2 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 example 5! = 5 × 4 × 3 × 2 × 1 = 120 and 8! = 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! 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 8 × 7 × 6 = = = 56 5! 3! (5 × 4 × 3 × 2 × 1)(3 × 2 × 1) 6

1.5 Binomial theorem Pascal’s triangle can be diﬃcult to use if the exponent is very high. In such cases the following binomial theorem is usually better. This states that if n is a positive integer then n(n − 1) n(n − 1)(n − 2) (a + b)n = an + nan−1b + an−2b2 + an−3b3 + ··· + bn 2! 3! An important particular case is when a = 1 and b = x giving n(n − 1) n(n − 1)(n − 2) (1 + x)n = 1 + nx + x2 + x3 + ··· + xn (1.1) 2! 3! which, like the previous result, holds for positive integers n. In the binomial theorem, the general term has the form an−mbm with coeﬃcient n(n − 1)(n − 2) ··· (n − (m − 1)) m! which equals n(n − 1)(n − 2) ··· (n − (m − 1))(n − m)! m!(n − m)! or n! n ! often denoted m!(n − m)! m In terms of the notation introduced above, the binomial theorem can be written as

n ! n ! n ! n ! n n ! (a+b)n = an + an−1b+ an−2b2 +···+ bn = X an−i bi 0 1 2 n i i=0

3 1.6 Example 4 x Expand 2 + 3 . Solution. Using the binomial theorem:

x4 x (4)(3) x (4)(3)(2) x (4)(3)(2)(1) x 2 + = 24 + (4)(23)( ) + (22)( )2 + (2)( )3 + ( )4 3 3 2! 3 3! 3 4! 3 32 8 8 1 = 16 + x + x2 + x3 + x4. 3 3 27 81 1.7 Example 15 x 3 Expand 1 + 3 up to and including the term in x . Solution. By the binomial theorem:

x15 x (15)(14) x (15)(14)(13) x 1 + = 1 + 15( ) + ( )2 + ( )3 + ··· 3 3 2! 3 3! 3 35 455 = 1 + 5x + x2 + x3 + ··· 3 27 1.8 Example Expand (1 − x)3(2 + x)6 up to and including the term in x2. Solution. (6)(5) ! (1 − x)3(2 + x)6 = (1 − x)3 26 + (6)(25)x + (24)x2 + ··· 2!   (3)(2)  2 3  2 = 1 + 3(−x) + (−x) + (−x)  (64 + 192x + 240x + ···) 2! | {z } redundant = (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 is a positive integer only. For the case when the number n is not a positive integer the binomial theorem becomes, for −1 < x < 1,

n(n − 1) n(n − 1)(n − 2) (1 + x)n = 1 + nx + x2 + x3 + ··· (1.2) 2! 3! This might look the same as the binomial expansion given by expression (1.1), but let us make the following important distinctions between (1.1) and (1.2):

4 • the expansion for positive integer powers (expansion (1.1)) terminates, i.e. it has only a ﬁnite number of terms. However, for powers that are not positive integers 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.

• expression (1.2) cannot be applied to something of the form (a + x)n. Such an expression must ﬁrst be rewritten as follows: xn xn (a + x)n = a 1 + = an 1 + a a | {z } apply binomial to this

1.10 Example √ Expand 1 + 2x and state what values of x the series is valid. Solution. √ 1 + 2x = (1 + 2x)1/2 1 ( 1 )(− 1 ) ( 1 )(− 1 )(− 3 ) ( 1 )(− 1 )(− 3 )(− 5 ) = 1 + (2x) + 2 2 (2x)2 + 2 2 2 (2x)3 + 2 2 2 2 (2x)4 + ··· 2 2! 3! 4! 1 1 5 = 1 + x − x2 + x3 − x4 + ··· 2 2 8

1 1 This series is valid when −1 < 2x < 1. i.e. when − 2 < x < 2 .

1.11 Example −5 x Expand 1 − 2 . For what values of x is the expansion valid? Solution. x−5 x (−5)(−6) x2 (−5)(−6)(−7) x3 1 − = 1 + (−5) − + − + − + ··· 2 2 2! 2 3! 2 5 15 35 = 1 + x + x2 + x3 + ··· 2 4 8 x This is valid when −1 < − 2 < 1, i.e. when −2 < x < 2.

1.12 Example

− 1 Expand (3 + x) 2 . Solution. Remember that when the power is not a positive integer your expression has to be of the form (1 + something)power. Deal with this as follows:

− 1 − 1 − 1 x 2 − 1 x 2 (3 + x) 2 = 3 1 + = 3 2 1 + 3 3 | {z } expand this

5 1 3 ! − 1 1 x (− 2 )(− 2 ) x 2 = 3 2 1 + (− )( ) + ( ) + ··· 2 3 2! 3 1 x x2 ! = √ 1 − + + ··· 3 6 24

This is valid when −1 < x/3 < 1, i.e. when −3 < x < 3.

1.13 Example

1/2 1 Find expansions for 1 + x for the cases (i) |x| > 1 and (ii) 0 < x < 1. Solution. the following calculation produces an expansion which will be valid when 1/|x| < 1, i.e. |x| > 1:

1 1/2 1 1 ( 1 )(− 1 ) 1 2 ( 1 )(− 1 )(− 3 ) 1 3 1 + = 1 + + 2 2 + 2 2 2 + ··· x 2 x 2! x 3! x 1 1 1 = 1 + − + + ··· 2x 8x2 16x3 valid for |x| > 1. The above expansion is no good if |x| < 1. For this case the following trick produces a valid expansion:

1/2 1/2 1 x + 1 1 1/2 1 + = = 1/2 (1 + x) x x x | {z } expand this 1 1 ( 1 )(− 1 ) ( 1 )(− 1 )(− 3 ) ! = 1 + x + 2 2 x2 + 2 2 2 x3 + ··· x1/2 2 2! 3! 1 1 1 1 = 1 + x − x2 + x3 + ··· x1/2 2 8 16 1 1 1 1 = + x1/2 − x3/2 + x5/2 + ··· x1/2 2 8 16 Note that this is actually deﬁned only for 0 < x < 1.

1.14 Example

(1+x)2 2 Expand (1−x/2)3 up to and including the term in x . Solution. (1 + x)2 x−3 = (1 + x)2 1 − (1 − x/2)3 2 x (−3)(−4) x2 ! = (1 + 2x + x2) 1 + (−3) − + − + ··· 2 2! 2 3x 3x2 ! = (1 + 2x + x2) 1 + + + ··· 2 2

6 3 3 3 = 1 + + 2 x + 1 + 2 + x2 + ··· 2 2 2 7x 11x2 = 1 + + ··· 2 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 powers of (x − a), i.e. to form a series of the form ∞ 2 3 X n f(x) = a0 + a1(x − a) + a2(x − a) + a3(x − a) + ··· = an(x − a) (2.3) n=0

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 expressions for the first, second, third, etc derivatives of f(x):

0 2 3 f (x) = a1 + 2a2(x − a) + 3a3(x − a) + 4a4(x − a) + ··· 00 2 f (x) = 2a2 + (3)(2)a3(x − a) + (4)(3)a4(x − a) + ··· 000 f (x) = (3)(2)a3 + (4)(3)(2)a4(x − a) + ··· . . . . Putting x = a in these expressions gives

0 0 f (a) = a1 ⇒ a1 = f (a) 1 f 00(a) = 2a ⇒ a = f 00(a) 2 2 2 1 f 000(a) = (3)(2)a ⇒ a = f 000(a) 3 3 (2)(3)

Spotting the pattern, we see that the general formula for the coefficient an will be 1 a = f (n)(a) n n! 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: (x − a)2 (x − a)3 f(x) = f(a) + (x − a)f 0(a) + f 00(a) + f 000(a) + ··· (2.4) 2! 3! (x−a)n (n) The series carries on to infinity, and has general term n! f (a). Taylor’s expansion, and the related Maclaurin expansion discussed below, are used in approximations. In practice usually only the first few terms in the series are kept and the rest are discarded. The idea is that the resulting truncated expansion should provide a good approximation to the function f(x) for values of x close to the particular value a. The more terms we keep, the better the approximation.

7 2.2 Maclaurin series There is also the Maclaurin expansion, which is just the Taylor expansion in the particular case when a = 0, i.e.

x2 x3 f(x) = f(0) + xf 0(0) + f 00(0) + f 000(0) + ··· (2.5) 2! 3! or, in summation notation ∞ xn f(x) = X f (n)(0) n=0 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 0(x) = ex so f 0(0) = 1. f 00(x) = ex so f 00(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(0), f 00(0), etc, into (2.5) gives us the Maclaurin series for the particular function f(x) = ex, namely

x2 x3 ex = 1 + x + + + ··· (2.6) 2! 3! ∞ xn or, in summation notation, ex = X n=0 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

(5x)2 (5x)3 e5x = 1 + 5x + + + ··· 2! 3! 25x2 125x3 = 1 + 5x + + + ··· 2 6 2.5 Example Find the Maclaurin series of cos x. Solution. Let f(x) = cos x. Then f(0) = 1. Also f 0(x) = − sin x so f 0(0) = 0. f 00(x) = − cos x so f 00(0) = −1. f 000(x) = sin x so f 000(0) = 0.

8 f 0000(x) = cos x so f 0000(0) = 1. f 00000(x) = − sin x so f 00000(0) = 0. We see the pattern emerging. The values f(0), f 0(0), f 00(0), f 000(0), etc, cy- cle through the values 1, 0, −1, 0, 1, 0, −1, 0,.... Putting these values into the general Maclaurin expansion (2.5) gives the Maclaurin expansion for the function cos x, namely x2 x4 cos x = 1 − + ··· 2! 4! or, in summation notation, ∞ (−1)nx2n cos x = X n=0 (2n)! Similarly, it can be shown that the Maclaurin expansion of sin x is

x3 x5 sin x = x − + − · · · 3! 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 out the values f(2), f 0(2), f 00(2), etc. We do this next: 1 f(2) = 2 . 0 1 0 1 f (x) = − x2 so f (2) = − 4 . 00 2 00 1 f (x) = x3 so f (2) = 4 . 000 6 000 3 f (x) = − x4 so f (2) = − 8 , and so on. The Taylor series about the value x = 2 is

(x − 2)2 (x − 2)3 f(x) = f(2) + (x − 2)f 0(2) + f 00(2) + f 000(2) + ··· 2! 3! which becomes, since f(x) = 1/x, 1 1 1 1 1 = − (x − 2) + (x − 2)2 − (x − 2)3 + ··· x 2 4 8 16 What this means, is that the first few terms of the above series expansion will constitute 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 could hardly expect the function to have an approximation for small values of x as a series of powers of x. Had we attempted to find f(0), f 0(0), f 00(0), etc, they would all turn out to be infinity.

9 2.7 Example Find the ﬁrst three non-zero terms of the Maclaurin series of e−2x sin x. Solution. One way to do this would be to write down the Maclaurin series for e−2x (which can be inferred from the one for ex by replacing every x by −2x) and the series for sin x and then multiplying the series together and expanding out. The approach below is a direct approach not requiring such advance knowledge of the two separate Maclaurin expansions. Let f(x) = e−2x sin x. Then f(0) = 0. f 0(x) = e−2x cos x − 2e−2x sin x so f 0(0) = 1. Diﬀerentiating again

f 00(x) = e−2x(− sin x) − 2e−2x cos x − 2(e−2x cos x − 2e−2x sin x) = 3e−2x sin x − 4e−2x cos x and f 000(x) = 3(e−2x cos x − 2e−2x sin x) − 4(−e−2x sin x − 2e−2x cos x) From these expressions we get f 00(0) = −4 and f 000(0) = 11. Putting these values into the general Maclaurin series (2.5) gives the following expression for our particular function f(x) = e−2x sin x:

11x3 e−2x sin x = x − 2x2 + + ··· 6 which will constitute a good approximation to e−2x sin x provided x is reasonably small.

2.8 Example Find the binomial expansion of (1 − x2)−1/2 and deduce from it a power series expansion for sin−1 x. Solution. First we ﬁnd the expansion of (1 + x)−1/2.

1 (− 1 )(− 3 ) (− 1 )(− 3 )(− 5 ) (1 + x)−1/2 = 1 + (− )x + 2 2 x2 + 2 2 2 x3 + ··· 2 2! 3! 1 3 5 = 1 − x + x2 − x3 + ··· 2 8 16 In the above, we now replace every x by −x2 to deduce that 1 3 5 (1 − x2)−1/2 = 1 − (−x2) + (−x2)2 − (−x2)3 + ··· 2 8 16 1 3 5 = 1 + x2 + x4 + x6 + ··· 2 8 16 Now Z x dt sin−1 x = √ 0 1 − t2

10 Z x = (1 − t2)−1/2 dt 0 Z x 1 3 5 = 1 + t2 + t4 + t6 + ··· dt 0 2 8 16 1 3 5 = x + x3 + x5 + x7 + ··· 6 40 112 3 Applications to working out limits

The notation lim f(x) x→a means the value (if any) that f(x) approaches, when x approaches a. The word “lim” means limit.

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

lim(x2 − 3) = 1, lim cos x = 1 x→2 x→0 In the above examples we can just put the value in. But in many situations we cannot 0 do this because we end up with the mathematically meaningless expression 0 which could be anything. For example, let’s work out x2 − 4 lim x→2 x − 2 0 In this example we cannot put x = 2 into the expression otherwise we get 0 which could be anything. But we can simplify the expression by factorising and cancelling factors to get

x2 − 4 (x − 2)(x + 2) lim = lim = lim(x + 2) = 4 x→2 x − 2 x→2 x − 2 x→2 Similarly, let’s work out x2 + x − 2 lim x→1 x2 − x 0 Again we cannot just put x = 1 into this expression or we would get 0 . But we can factorise and simplify as follows:

x2 + x − 2 (x − 1)(x + 2) x + 2 lim = lim = lim = 3. x→1 x2 − x x→1 x(x − 1) x→1 x

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 and Taylor/Maclaurin series to our list of methods for working out limits.

11 3.2 Example Let’s work out (1 + x/2)5/7 − 1 lim x→0 x Again, we cannot put x = 0 into this expression as it stands. But we can use binomial expansion, as follows;

( 5 )(− 2 ) 1 + ( 5 )( x ) + 7 7 ( x )2 + ··· − 1 (1 + x/2)5/7 − 1 7 2 2! 2 = x x 5 x − 5 x2 + ··· = 14 196 x 5 5 = − x + ··· 14 196 We can let x → 0 in the above expression to deduce that (1 + x/2)5/7 − 1 5 lim = x→0 x 14 3.3 Example Let’s work out sin x sin 2x lim and lim x→0 x x→0 x Solution. We mentioned earlier that x3 x5 x7 sin x = x − + − + ··· 3! 5! 7! Hence sin x x2 x4 x6 = 1 − + − + ··· x 3! 5! 7! We can let x → 0 in this to deduce that sin x lim = 1 x→0 x From the Maclaurin expansion for sin x given above, we can deduce the expansion for sin 2x to be (2x)3 (2x)5 sin 2x = 2x − + − · · · 3! 5! 4x3 32x5 = 2x − + − · · · 3 120 Hence sin 2x 4x2 = 2 − + ··· x 3 Letting x → 0 we deduce that sin 2x lim = 2 x→0 x sin kx It is in fact a general result that limx→0 x = k for any constant k.

12 3.4 Example Find sin2 x − x2 cos x lim x→0 x4 Solution. Recall that x3 x5 x7 sin x = x − + − + ··· 3! 5! 7! and x2 x4 x6 cos x = 1 − + − + ··· 2! 4! 6! Squaring the formula for sin x gives x3 x5 ! x3 x5 ! sin2 x = x − + − · · · x − + − · · · 6 120 6 120 x4 x4 = x2 − + (something) x6 − + (something) x6 6 6 x4 = x2 − + (something) x6 3 Hence, using also the expansion for cos x given above, we have

4 2 4 sin2 x − x2 cos x x2 − x + (something) x6 + ··· − x2 1 − x + x + ··· = 3 2 24 x4 x4 1 x4 + (something) x6 + even higher powers of x = 6 x4 1 = + (something) x2 + ··· 6 Let x → 0 in the above to get sin2 x − x2 cos x 1 lim = x→0 x4 6 3.5 Example Find lim x(e−1/x − 1) x→∞ Solution. To deal with x going to inﬁnity, we shall let y = 1/x and let y → 0. This gives 1 lim x(e−1/x − 1) = lim (e−y − 1) x→∞ y→0 y 1 ( (−y)2 ) ! = lim 1 + (−y) + + ··· − 1 y→0 y 2! y = lim −1 + + ··· y→0 2! = −1 where we have used the Maclaurin expansion for the exponential, given by (2.6).

13 4 L’Hopital’s rule

0 Another way of working out a limit when in a 0 situation is the following result: f(x) f 0(x) if f(a) = 0 and g(a) = 0 then lim = lim x→a g(x) x→a g0(x)

The above result is called L’Hopital’s 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

3x − sin x lim would be 0 if we put x = 0 in, so use L’Hopital x→0 x 0 3 − cos x = lim no longer 0 x→0 1 0 3 − cos 0 = 1 = 2

4.2 Example

1 − cos x lim would be 0 if we put x = 0 in, so use L’Hopital x→0 x + x2 0 sin x = lim no longer 0 x→0 1 + 2x 0 0 = 1 = 0

4.3 Example

x − 2 1 1 lim = lim = x→2 x2 − 4 x→2 2x 4 Sometimes we have to apply L’Hopital’s rule more than once to get an answer, as the next example illustrates:

14 4.4 Example

x − sin x lim 0 so use L’Hopital x→0 x3 0 1 − cos x = lim still 0 so use L’Hopital again x→0 3x2 0 sin x = lim still 0 so use L’Hopital again x→0 6x 0 cos x = lim no longer 0 x→0 6 0 1 = 6 4.5 Example

ln cos x lim 0 so use L’Hopital x→0 ln cos 3x 0 − sin x = lim cos x now simplify this x→0 3 sin 3x − cos 3x tan x = lim still 0 so use L’Hopital again x→0 3 tan 3x 0 sec2 x = lim no longer 0 x→0 9 sec2 3x 0 1 = 9 5 Fourier Series

A Fourier Series is an expansion of a periodic function as an inﬁnite sum of sines and cosines. Simple examples of periodic functions (other than sin and cos) are the square wave and sawtooth functions. An example of a square wave function (of period 4 in this particular case) is the periodic function

( −1, −2 < t < 0, f(t) = 1, 0 < t < 2,

with f(t + 4) = f(t). An example of a sawtooth function of period 2π would be the periodic function of period 2π such that f(t) = t for t ∈ (−π, π). Since this function has period 2π we might 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 2π. Such an expansion does indeed exist and in fact any periodic function of period 2π has an expansion in terms of these trigonometric functions. If the period is T rather than 2π this is no particular problem. All we have to do is modify the period of the cos and sin functions we work with, i.e. we instead seek

15 2nπt 2nπt an expansion in terms of the functions cos T and sin T for n = 1, 2, 3 ..., rather than cos nt and sin nt. Letting f(t) be a T0 periodic function, this expansion, called the Fourier series of f(t), turns out to be

∞ 1 X 2nπt 2nπt f(t) = 2 a0 + an cos + bn sin (5.7) n=1 T0 T0 where

2 Z T0 2nπt an = f(t) cos dt, n = 0, 1, 2, 3,... (5.8) T0 0 T0 2 Z T0 2nπt bn = f(t) sin dt, n = 1, 2, 3,... (5.9) T0 0 T0 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:

2 Z T0/2 2nπt an = f(t) cos dt, n = 0, 1, 2, 3,... (5.10) T0 −T0/2 T0 2 Z T0/2 2nπt bn = f(t) sin dt, n = 1, 2, 3,... (5.11) T0 −T0/2 T0 will work just as well.

• to work out a0 in (5.7) you use the an formula (either (5.8) or (5.10)) with n = 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 the frequency to be the quantity ω0 deﬁned by

2π 1 ω0 = rather than T0 T0

• often we want to work out the Fourier series of a periodic function that contains points 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

1 2 (f(a+) + f(a−)) rather than to f(a). This applies regardless of how f(t) is deﬁned (if it is deﬁned at all) at the point a itself. In the above formula the notation f(a+) means the value just to the right of the discontinuity and f(a−) means the value to the left. More formally, f(a+) is the limit of f(a + h) as h tends to zero from above, and f(a−) is the limit of f(a − h) as h tends to zero from above.

16 5.1 Example Let ( −1 −π < t < 0 f(t) = 1 0 < t < π with f(t + 2π) = f(t). Find the Fourier series of f(t). Solution. In this case the period T0 is given by T0 = 2π. Let us ﬁnd an ﬁrst. The following formulae will be useful cos nπ = (−1)n and sin nπ = 0, n = 0, ±1, ±2, ±3,... We have

2 Z T0/2 2nπt an = f(t) cos dt T0 −T0/2 T0 1 Z π = f(t) cos nt dt π −π 1 Z 0 1 Z π = (−1) cos nt dt + cos nt dt π −π π 0 1 − sin nt0 1 sin ntπ = + π n −π π n 0 an = 0

Because of the n in the denominator of the above calculations we need to ﬁnd a0 separately, 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 let’s ﬁnd bn. we have 1 Z π bn = f(t) sin nt dt π −π 1 Z 0 Z π = (−1) sin nt dt + sin nt dt π −π 0 1 "cos nt0 − cos ntπ# = + π n −π n 0 1 = (1 − cos(−nπ) + (− cos nπ + 1)) πn 1 = (2 − 2(−1)n) πn and so 2 b = (1 − (−1)n) n πn With all the an, n = 0, 1, 2,..., equal to zero, and also recalling that the period T0 = 2π, the Fourier series becomes ∞ X f(t) = bn sin nt n=1 = b1 sin t + b2 sin 2t + b3 sin 3t + ···

17 i.e. 4 4 4 f(t) = sin t + sin 3t + sin 5t + ··· π 3π 5π 6 Even and odd functions

If a function is an even function or an odd function then certain simpliﬁcations are possible in the calculations required for computing the Fourier series. But note that plenty 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 is symmetrical 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 that

Z a Z a f(t) dt = 2 f(t) dt −a 0

6.2 Odd functions f(t) is said to be an odd function if f(−t) = −f(t). This means the graph has 180o rotational symmetry about the origin. Examples of odd functions are f(t) = t, f(t) = t3, f(t) = t5,... (all odd powers of t); also f(t) = sin t and f(t) = sinh t. An odd function has the important property that

Z a f(t) dt = 0 −a

6.3 Useful rules of even and odd functions

even × even = even even × odd = odd odd × even = odd odd × odd = even

6.4 Fourier series of an even function Suppose that f(t) is an even function and we want it’s Fourier series. Since sin t is an odd function we might anticipate that the Fourier series of an even function will contain no sine terms. We shall show that this is indeed the case. With f(t) being

18 even the bn Fourier coeﬃcient is given by

2 Z T0/2 2nπt bn = f(t) sin dt T0 −T0/2 |{z} T0 even | {z } odd 2 Z T0/2 = (something odd) dt T0 −T0/2 = 0

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

2 Z T0/2 2nπt an = f(t) cos dt T0 −T0/2 |{z} T0 even | {z } even 4 Z T0/2 2nπt = f(t) cos dt T0 0 T0 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

2 Z T0/2 2nπt an = f(t) cos dt T0 −T0/2 |{z} T0 odd | {z } even 2 Z T0/2 = (something odd) dt T0 −T0/2 = 0

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

6.6 Example Find 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 (including n = 0) and we only need to calculate bn.

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

2 Z T0/2 2nπt bn = f(t) sin dt T0 −T0/2 T0 2 Z 2 2nπt = f(t) sin dt 4 −2 4 1 Z 2 nπt = t sin dt odd × odd = even 2 −2 |{z} 2 odd | {z } odd Z 2 nπt = t sin dt 0 2 2 "−t cos nπt # Z 2 cos nπt = 2 + 2 dt nπ/2 0 0 nπ/2 4 2 "sin nπt #2 = − cos nπ + 2 nπ nπ nπ/2 0 so that 4 b = − (−1)n n nπ

Recalling that T0 = 4, the Fourier series is ∞ 4 nπt f(t) = X − (−1)n sin n=1 nπ 2 4 ∞ (−1)n nπt = − X sin π n=1 n 2 or, in expanded form, 4 πt 1 3πt 1 f(t) = − − sin + 1 sin πt − sin + sin 2πt + ··· π 2 2 3 2 4

6.7 Example Find the Fourier series of the function such that

f(t) = t2 + t for − π < t < π

with f(t + 2π) = f(t) for all t. Solution. This function is neither even nor odd. It is 2π-periodic so T0 = 2π. Using this value for T0 the formula for an becomes 1 Z π an = f(t) cos nt dt π −π 1 Z π = (t2 + t) cos nt dt π −π 1 " sin ntπ Z π sin nt # = (t2 + t) − (2t + 1) dt if n 6= 0 π n −π −π n

20 1 Z π = − (2t + 1) sin nt dt πn −π 1 " cos ntπ Z π 2 cos nt # = − −(2t + 1) + dt πn n −π −π n 1 " (2π + 1)(−1)n (2(−π) + 1)(−1)n 2 sin ntπ # = − − + + πn n n n n −π 1 " 4π(−1)n # 4 = − − = (−1)n πn n n2

We have shown that 4 a = (−1)n if n 6= 0 n n2

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

Z π " 3 2 #π 1 2 1 t t a0 = (t + t) dt = + π −π π 3 2 −π 2π2 a = 0 3

Next we ﬁnd bn. We have

Z π 1 2 bn = (t + t) sin nt dt π −π 1 " cos ntπ Z π cos nt # = −(t2 + t) + (2t + 1) dt π n −π −π n 1 " (−1)n (−1)n 1 Z π # = −(π2 + π) + (π2 − π) + (2t + 1) cos nt dt π n n n −π 1 " (−1)n 1 " sin ntπ Z π 2 sin nt ## = −2π + (2t + 1) − dt π n n n −π −π n 1 " (−1)n 2 Z π # = −2π − 2 sin nt dt π n n2 −π

so that 2 b = − (−1)n n n

Putting the formulae for a0, an and bn into the general Fourier series expansion, which when T0 = 2π is ∞ 1 X f(t) = 2 a0 + (an cos nt + bn sin nt) n=1 gives 2 ∞ π X 4 n 2 n f(t) = + 2 (−1) cos nt − (−1) sin nt 3 n=1 n n

21 7 Complex form of a Fourier series

Recall that √ j = −1 Recall also, Euler’s formula ejt = cos t + j sin t (7.12) and the other useful version of it: e−jt = cos t − j sin t (7.13) It is easy to see where (7.12) comes from. We simply expand ejt using Maclaurin series: (jt)2 (jt)3 ejt = 1 + jt + + + ··· 2! 3! t2 t4 ! t3 t5 ! = 1 − + − · · · + j t − + − · · · 2! 4! 3! 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: ejt + e−jt cos t = (7.14) 2 Similarly, subtracting (7.13) from (7.12) gives ejt − e−jt sin t = (7.15) 2j Since cos t and sin t are periodic of period 2π, so is ejt. Furthermore, so are the functions enjt for any integer n (positive or negative). This leads us to suppose that any 2π 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 ∞ X 2jnπt/T0 f(t) = cn e (7.16) n=−∞ where 1 Z T0/2 −2jnπt/T0 cn = f(t)e dt for n = 0, ±1, ±2, ±3,... (7.17) T0 −T0/2

Expression (7.16) with coeﬃcients cn given by (7.17) is called the complex form of the Fourier series. Note that the sum is over all integer values of n including negative ones. The convergence properties of the complex form are the same as for the form we discussed 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 midpoint of the jump (regardless of the actual value of f(t) at such a point).

22 7.1 Example Find the complex Fourier series of the function f(t) such that

 0 −π < t < −π/2  f(t) = 1 −π/2 < t < π/2  0 π/2 < t < π

with f(t + 2π) = f(t) for all t. Solution. The period is 2π so T0 = 2π. With this value of T0 expressions (7.16) and (7.17) reduce to ∞ X jnt f(t) = cn e n=−∞ where Z π 1 −jnt cn = f(t)e dt 2π −π 1 Z π/2 1 "e−jnt #π/2 = e−jnt dt = 2π −π/2 2π −jn −π/2 1 h i = e−jnπ/2 − ejnπ/2 −2πjn 1 "ejnπ/2 − e−jnπ/2 # = πn 2j

ejx−e−jx Recalling the formula sin x = 2j , the above formula for cn can be put into the form 1 nπ c = sin for n = ±1, ±2, ±3,... n πn 2 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 for cn, with n = 0, gives 1 Z π 1 Z π/2 1 c0 = f(t) dt = dt = 2π −π 2π −π/2 2 So the complex Fourier series of the function is

∞ X jnt f(t) = cn e n=−∞

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 −1 ∞ 1 X jnt X jnt f(t) = 2 + cne + cne n=−∞ n=1

23 1 in which the 2 at the front is the n = 0 term of the sum. Making the substitution n = −m in the ﬁrst sum of the above expression gives us

∞ ∞ 1 X −jmt X jnt f(t) = 2 + c−me + cne m=1 n=1 and we can now simply replace m by n in the ﬁrst sum, since they are dummy variables playing a similar role to the variable in a deﬁnite integral. This observation gives

∞ ∞ 1 X −jnt X jnt f(t) = 2 + c−ne + cne insert expression for cn n=1 n=1 ∞ 1 X 1 −nπ −jnt 1 nπ jnt = 2 + sin e + sin e n=1 −πn 2 πn 2 ∞ 1 nπ = 1 + X sin e−jnt + ejnt 2 πn 2 n=1 | {z } =2 cos nt and so ∞ 1 X 2 nπ f(t) = 2 + sin cos nt n=1 πn 2 We have converted the complex form into a real form. In the above sum the terms with n = 2, 4, 6,... are all zero. Replacing n by 2n − 1 has the eﬀect of removing these zero terms to give

∞ 1 X 2 (2n − 1)π f(t) = 2 + sin cos(2n − 1)t n=1 π(2n − 1) 2 ∞ n+1 1 X 2(−1) = 2 + cos(2n − 1)t n=1 π(2n − 1) which is more computationally eﬃcient.

8 Amplitude and phase spectrum

Recall that the Fourier series of a periodic function of period T0 (and frequency ω0 = 2π/T0) is ∞ 1 X 2nπt 2nπt f(t) = 2 a0 + an cos + bn sin n=1 T0 T0

with certain formulae, namely (5.8) and (5.9), for the coeﬃcients an and bn. We can write the Fourier series in terms of ω0 as

∞ 1 X f(t) = 2 a0 + (an cos nω0t + bn sin nω0t) n=1 ∞ 1 X = 2 a0 + An cos(nω0t + δn) n=1

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

An cos(nω0t + δn) = An cos nω0t cos δn − An sin nω0t sin δn

Comparing the right hand side of the above expression with

an cos nω0t + bn sin nω0t we see that we need

An cos δn = an and An sin δn = −bn

These expressions imply that q 2 2 An = an + bn and bn tan δn = − an

The numbers δn, n = 1, 2, 3,... are called the phase spectrum and the numbers An, n = 1, 2, 3,... are called the amplitude spectrum. These two sets of numbers together form the spectrum and (with the frequency ω0) constitute one way of describing a periodic function.