ISE I Brief Lecture Notes

ISE I Brief Lecture Notes 1 Partial Differentiation 1.1 Definitions Let f(x, y) be a function of two variables. The partial derivative f/ x is the function obtained by differentiating f with respect to x, regarding y as a constant. Similarly, f/ y is obtained by differentiating f with respect to y, regarding x as a constant. We often use the alternative notation f x = f/ x, f y = f/ y. Example If f(x, y) = x 2 + xy + y 3 1 then f x = 2x + y, f y = x + 3y 2. As a limit, f(x + h, y) f(x, y) f/ x = lim h. h Also, the equation z = f(x, y) defines a surface in 3-dimensional space with x, y, z-axes, and f/ x is the gradient of the tangent at a point in the x- direction. Higher partial derivatives Can differentiate f x, f y partially with respect to x and y to get four second order derivatives: f xx = 2 f/ x 2, f yy = 2 f/ y 2, f xy = 2 f/ x y, f yx = 2 f/ y x Example 1. f(x, y) = x 2 + xy + y 3 1 as above: then f xx = 2, f yy = 6y, f xy = f yx = 1. 2. If f(x, y) = tan 1 (y/x) then we get f xx = 2xy/(x 2 + y 2 ) 2, f yy = 2xy/(x 2 + y 2 ) 2, 1

and also f xy = f yx = (y 2 x 2 )/(x 2 + y 2 ) 2. Notice that in both examples, f xy = f yx. This is a general fact: Theorem If f(x, y) is a function of two variables, and the second order partial derivatives f xy and f yx both exist and are continuous, then f xy = f yx. Here, saying that a function g(x, y) is continuous at a point (a, b) means that lim (x,y) (a,b) g(x, y) = g(a, b) this means that as (x, y) gets closer and closer to (a, b), g(x, y) gets closer and closer to g(a, b). All the functions we meet in this chapter will satisfy the assumptions of the theorem, so from we on we always assume that f xy = f yx. More than 2 variables For a function f(x, y, z,...) of three or more variables, we define the partial derivative f x = f/ x to be the function obtained by differentiating f with respect to x, regarding y, z,... as constants. Similarly for f/ y and f/ z and so on. Example if f(x, y, z) = (x 2 + y 2 + z 2 ) 1/2 then f x = x(x 2 + y 2 + z 2 ) 1/2, f y = y(x 2 + y 2 + z 2 ) 1/2, f z = z(x 2 + y 2 + z 2 ) 1/2. 1.2 Small changes Let f = f(x, y), and make small changes from x to x + δx and from y to y + δy. What is the corresponding change in f, i.e. what is δf = f(x + δx, y + δy) f(x, y)? We can estimate this as follows: write this as δf = [f(x + δx, y + δy) f(x, y + δy)] + [f(x, y + δy) f(x, y)]. From the definitions of f/ x, f/ y as limits, the first of these brackets is roughly equal to f x.δx and the second to f y.δy. So δf f x.δx + f y.δy. Example What is the approximate change in f(x, y) = xe xy if x changes from 1 to 1 + h and y from to k? Ans: δf (e xy + xye xy )δx + x 2 e xy δy. Putting x = 1, y = we get δf h + k. 2

1.3 Chain Rules Recall the chain rule for differentiating functions of one variable: if f = f(u) and u = u(t) then df/dt = df/du.du/dt. Now let f = f(x, y) where x, y are both functions of one variable t say x = x(t), y = y(t). If a small change δt in t gives corresponding changes δx, δy in x and y, then δx dx/dt.δt, δy dy/dt.δt, so by Section 1.2, δf f x.δx + f y.δy f x dx/dt.δt + f y dy/dt.δt. Dividing through by δt and taking the limit as δt, we get Chain Rule I If f = f(x, y) where x = x(t), y = y(t), then df/dt = f x dx/dt + f y dy/dt. Similarly for functions f(x, y, z) with x = x(t), y = y(t), z = z(t) we get df/dt = f x dx/dt + f y dy/dt + f z dz/dt. Example If f(x, y) = x 3 y + sin(x + y) where x = t 2, y = sin t, then df/dt = (3x 2 y + cos(x + y)).2t + (x 3 + cos(x + y)). cos t. (Could of course work this out by substituting for x, y to get f as a function of t, and then differentiating, but would be somewhat unpleasant. Implicit functions An equation of the form f(x, y) = defines y as an implicit function of x. To find dy/dx, we differentiate the equation with respect to x using the Chain Rule I (noting that x and y are both functions of the single variable x). This gives f x + f y dy/dx =, hence dy/dx = f x /f y. For example, if the equation is x 2 y cos x + x 3 y 2 =, then differentiating as above gives 2x + y sin x + 3x 2 y 2 + ( cos x + 2x 3 y)dy/dx =, hence dy/dx = (2x + y sin x + 3x 2 y 2 )/(cos x 2x 3 y). 3

General chain rule Now suppose f = f(x, y) where x = x(s, t), y = y(s, t). Then f is also a function of s, t and we d like a formula for f/ s, f/ t. Well, if we regard t as a constant then x = x(s), y = y(s), so we can apply Chain Rule I to get Chain Rule II If f = f(x, y) where x = x(s, t), y = y(s, t), then f/ s = f x x/ s + f y y/ s, and similarly f/ t = f x x/ t + f y y/ t. For more than 2 variables the rule is entirely similar: if f = f(x, y, z,...) where x = x(s, t,...), y = y(s, t,...), z = z(s, t,...),... then f/ s = f x x/ s + f y y/ s + f z z/ s + and similarly for f/ t and so on. Example Let f = f(x, y) and let r, θ be polar coordinates, so that x = r cos θ, y = r sin θ. Express the Laplace equation in polar coordinate form. 2 f/ x 2 + 2 f/ y 2 = Step 1 First work out f x, f y in terms of r, θ using the Chain Rule. Well, so by the Chain Rule, r = (x 2 + y 2 ) 1/2, θ = tan 1 (y/x), f x = f r r/ x + f θ θ/ x = f r.x(x 2 + y 2 ) 1/2 f θ.y(x 2 + y 2 ) 1, and so we get Similarly f x = f r cos θ f θ (sin θ)/r. f y = f r sin θ + f θ (cos θ)/r. Step 2 Now we work out f xx in terms of r, θ. By Step 1, f xx = cos θ. / r(f x ) ((sin θ)/r). / θ(f x ) = cos θ. / r(f r cos θ f θ (sin θ)/r) ((sin θ)/r). / θ(f r cos θ f θ (sin θ)/r) = cos θ[cos θf rr + ((sin θ)/r 2 )f θ ((sin θ)/r)f θr ] ((sin θ)/r)[cos θf rθ sin θf r ((cos θ)/r)f θ ((sin θ)/r)f θθ ]. 4

Hence we get f xx = cos 2 θf rr ((2 sin θ cos θ)/r)f θr +((sin 2 θ)/r)f r +((2 sin θ cos θ)/r 2 )f θ +((sin 2 θ)/r 2 )f θθ. Similarly we get f yy = sin 2 θf rr +((2 sin θ cos θ)/r)f θr +((cos 2 θ)/r)f r ((2 sin θ cos θ)/r 2 )f θ +((cos 2 θ)/r 2 )f θθ. Adding these two expressions, we obtain f xx + f yy = f rr + (1/r)f r + (1/r 2 )f θθ. Therefore Laplace s equation in polar form is f rr + (1/r)f r + (1/r 2 )f θθ =. 1.4 Taylor expansions Recall the Taylor expansion of a function f(x) of one variable at a point x = a: this is f(a + h) = f(a) + f (a)h + f (a) 2! where R n (h) is the error term. h 2 + + f (n) (a) h n + R n (h), n! Now we ll get a similar expansion for a function f(x, y) of two variables at a point (a, b). We study f along the line joining (a, b) to (a + h, b + k) for small h, k. A point on this line is (a + th, b + tk), with t between and 1. Define a function F (t) of one variable by F (t) = f(a + th, b + tk) ( t 1). So F () = f(a, b) and F (1) = f(a + th, b + tk). We are interested in the Maclaurin series for F, which is F (t) = F () + F ()t + F () t 2 + (1) 2! Let s work out F () and F () in terms of f and its partial derivatives. By Chain Rule I, F dx (t) = f x dt + f y dy dt where x = a + th, y = b + tk, and hence F (t) = hf x + kf y 5

(evaluated at (a + th, b + tk)). Applying the Chain Rule again, F (t) = h d dt (f x(a + th, b + tk)) + k d dt (f y(a + th, b + tk)) = h(hf xx + kf xy ) + k(hf yx + kf yy ) = h 2 f xx + 2hkf xy + k 2 f yy. Hence F () = f(a, b), F () = hf x (a, b) + kf y (a, b), F () = h 2 f xx (a, b) + 2hkf xy (a, b) + k 2 f yy (a, b). Substituting into (1), we get Taylor expansion of f(x, y) at (a, b): This is f(a + h, b + k) = f(a, b) + hf x (a, b) + kf y (a, b) + 1 2 (h2 f xx (a, b) + 2hkf xy (a, b) + k 2 f yy (a, b)) + terms of degree 3 or more in h, k. For small h, k this gives an approximation to f(a + h, b + k). 1.5 Maxima, minima and stationary points Now we use the previous section to study maxima and minima of functions of 2 variables. For a function f(x, y), we say f has a maximum at a point (a, b) if f(a + h, b + k) < f(a, b) for all small values of h, k (not both ). Similarly, f has a minimum at (a, b) if f(a + h, b + k) > f(a, b) for all small h, k. Finding maxima and minima Spose f has a maximum at (a, b). Then if we fix y = b and vary x, we certainly have f(a + h, b) < f(a, b) for all small h, which says that the function f(x, b) of 1 variable (x) has a maximum at x = a. Hence f/ x = at (a, b), and likewise f/ y = at (a, b). This is also true at a minimum. Hence maxima and minima are stationary points of f, in the following sense: Definition We say (a, b) is a stationary point of f if f x (a, b) = f y (a, b) =. A stationary point may or may not be a max/min. For example, if f(x, y) = xy then (, ) is a stationary point, but it is not a max or a min, as we can make f(h, k) positive or negative for suitable choices of small h, k. We call a stationary point (a, b) a saddle point of f if it is not a max or a min. 6

Example Let f(x, y) = x 2 /2 x + xy 2. Then f x = x 1 + y 2, f y = 2xy. At a stationary point, 2xy = so x = or y =. If x = then 1 + y 2 = so y = ±1. If y = then x 1 = so x = 1. So f has 3 stationary points: (, 1), (, 1), (1, ). Spose (a, b) is a stationary point of f, so f x (a, b) = f y (a, b) =. Write A = f xx (a, b), B = f xy (a, b), C = f yy (a, b). Then the Taylor expansion of f at (a, b) is f(a + h, b + k) = f(a, b) + 1 2 (Ah2 + 2Bhk + Ck 2 ) + higher terms. Write Δ = Ah 2 + 2Bhk + Ck 2. Then we see that (a, b) is a max if Δ < for all small h, k; (a, b) is a min if Δ > for all small h, k; and otherwise (a, b) is a saddle. Suppose now that A. Then Δ = 1 A (A2 h 2 + 2ABhk + ACk 2 ) = 1 A [(Ah + Bk)2 + (AC B 2 )k 2 ]. If AC B 2 > and A > then Δ > for all small h, k, and so (a, b) is a min. If AC B 2 > and A < then Δ < for all small h, k, and so (a, b) is a max. And if AC B 2 < then Δ can be made positive or negative for suitably chosen small h, k, so (a, b) is a saddle. Now suppose A = and B. Then Δ = k(2bh + Ck), which can be made positive or negative for small h, k, so again we have a saddle. Summary Let (a, b) be a stationary point of f(x, y), and let A = f xx (a, b), B = f xy (a, b), C = f y (a, b). The nature of the stationary point is as follows: A AC B 2 nature > > minimum < > maximum any < saddle Example Let f(x, y) = x 2 /2 x + xy 2 as above. We found that f has 3 stationary points, (, 1), (, 1), (1, ). Also f xx = 1, f xy = 2y, f yy = 2x. 7

So at (, 1), A = 1, B = 2, C =, so AC B 2 < and this is a saddle. At (, 1), A = 1, B = 2, C =, so AC B 2 < and this is also a saddle. At (1, ), A = 1, B =, C = 2, so AC B 2 > and this is a minimum. Note If AC B 2 = at a stationary point, the point can be a max, min or saddle nothing definite can be said in general. For example, f(x, y) = x 4 + y 4 has a stationary point at (, ); here AC B 2 = and (, ) is clearly a minimum of f. But g(x, y) = x 3 + y 3 also has a stationary point at (, ) with AC B 2 =, and this point is a saddle point of g. Contour sketching A common way to visualise a 3-dimensional surface z = f(x, y) is to sketch some of its contours in the x, y-plane. By a contour I just mean a curve of the form f(x, y) = c for a constant c. In the lectures we saw examples of contours for the simple cases z = x 2 + y 2 and z = xy. For z = x 2 + y 2 there is a min at (, ) and the contours are concentric circles centred at (, ); for z = xy there is a saddle at (, ) and the contours are curves xy = c. Procedure for making contour sketch Suppose we want to make a contour sketch of a surface z = f(x, y). Here are the steps: (1) Find the stationary points of f and determine their nature. (2) Sketch the contours through the saddle points. (3) Fill in some more contours, making sure that near the max/min points the contours look like those of z = x 2 + y 2, and near the saddles they look like those of z = xy. This is of course a pretty rough description, but it is fairly clear what to do in the simple examples in the lectures and problem sheets. 1.6 Exact differential equations Consider an equation u(x, y) = c (c a constant). As we saw earlier, this equation defines y as an implicit function of x, and if we differentiate it with dy respect to x we get u x +u y dx =. So if we put u x = p(x, y) and u y = q(x, y) then we see that the differential equation has solution u(x, y) = c. p(x, y) + q(x, y) dy dx = 8

Now let s think of this the other way round. Suppose we have a differential eqn P (x, y) + Q(x, y) dy dx =. If we can find a function u(x, y) such that u x = P, u y = Q, then we ll be able to solve the eqn the solution will be u(x, y) = c. Notice that if such a function u exists, then u xy = P y and u yx = Q x, so as u xy = u yx, we need to make sure that P y = Q x. This leads to Definition P y = Q x. The differential equation P + Q dy dx = is said to be exact if It turns out that for an exact eqn we can always find a function u such that u x = P, u y = Q, so we can always solve exact eqns. Example Solve 2xy + cos x cos y + (x 2 sin x sin y) dy dx =. Ans Let P = 2xy + cos x cos y, Q = x 2 sin x sin y. Then P y = Q x = 2x cos x sin y, so eqn is exact. To solve, we need to find u = u(x, y) such that (1) u x = 2xy + cos x cos y (2) u y = x 2 sin x sin y. Integrating (1), we get u = x 2 y+sin x cos y+f(y), where f(y) is any function of y. Differentiating wrt y this gives u y = x 2 sin x sin y + f (y). Hence (2) is satisfied if we take f(y) =. So u = x 2 y + sin x cos y satisfies both (1) and (2), and so the solution of the diff eqn is where c is an arbitrary constant. x 2 y + sin x cos y = c Integrating factor Sometimes it is possible to take a non-exact eqn P + Q dy dx = and to find a clever function λ = λ(x, y) to multiply the eqn through by and make it exact. In other words, to make the eqn λp + λq dy dx = an exact eqn. We call such a function λ an integrating factor for the eqn. 9

The only cases where this method is feasible is when we can find an integrating factor of the form λ = λ(x) (a function of x only) or λ(y). When λ = λ(x), the exactness condition is that (λp ) y = (λq) x, which is λp y = λ Q + λq x. This boils down to the following: 1 dλ λ dx = P y Q x Q. If the RHS happens to be a function of x only, then we have a chance of solving this and finding λ. Similarly, to find an integrating factor of the form λ = λ(y) we need to solve 1 dλ λ dy = P y Q x. P Example Solve xy 1 + (x 2 xy) dy dx =. Ans Let P = xy 1, Q = x 2 xy. Then P y = x, Q x = 2x y so eqn is not exact. But note that (P y Q x )/Q = 1/x, so to find an integrating factor λ = λ(x) we need to solve 1 dλ λ dx = 1/x. A solution is λ = 1/x. Multiplying the original eqn by this we get the eqn y 1 x + (x y) dy dx =. This is exact. To solve it we find u = u(x, y) satisfying u x = y 1 x, u y = x y. Integrating the first one gives u = xy log x + f(y), hence u y = x + f (y). So we need to choose f(y) with f (y) = y, so take f(y) = y 2 /2. Hence the solution is xy log x 1 2 y2 = c. 1

2 Laplace Transforms If f(t) is a function of one variable, we define its Laplace transform to be the function F (s) defined by F (s) = e st f(t)dt. Often we also write L(f(t)) or just L(f) for the Laplace transform of f. Examples 1. Laplace transform of the constant function f(t) = 1 is L(1) = e st dt = [ 1 s e st ] = 1 s (s > ). 2. L(e at ) = e at e st dt = 1 s a (s > a). 3. Let L(sin at) = F (s) = e st sin atdt. Then if we integrate by parts twice we get F (s) = a s 2 a2 s 2 F (s), and hence a L(sin at) = s 2 + a 2. Similarly L(cos at) = s s 2 + a 2. 4. To work out L(t n ) = e st t n dt, integrate by parts to get L(t n ) = n s L(tn 1 ). Repeating, we get L(t n ) = n s L(tn 1 ) = n(n 1) s 2 L(t n 1 n(n 1)... 2.1 ) = = s n L(1) = n! s n+1. Note The definition Laplace transform implies easily that L(f + g) = L(f) + L(g) and L(cf) = cl(f) (c a constant). For example, L(2t 3 3 sin t) = 2L(t 3 ) 3L(sin t) = 12 s 4 3 s 2 +1. Use of Laplace transforms 11

First we make the crucial observaton that if y is a function of t then we can express L( dy dt ) in terms of L(y) as follows: which by parts is equal to so we get the formula L( dy dt ) = [e st y] + st dy e dt dt, se st ydt, L( dy ) = y() + sl(y). (2) dt Another piece of notation: if L(f(t)) = F (s) we say that f is the inverse Laplace transform of F, and write f(t) = L 1 (F (s)). Now here s an example illustrating how we can use transforms to solve a differential eqn. More complicated examples will appear later when we ve done a bit more theory. Example Solve the differential eqn dy dt + 2y = cos t, with y() = 1. Ans Taking Laplace transforms of both sides gives L( dy dt )+2L(y) = L(cos t), so using (2) we get This gives 1 + sl(y) + 2L(y) = L(y) = s s 2 + 1. s (s + 2)(s 2 + 1) + 1 s + 2. So y is the inverse Laplace transform of the RHS, or y = L 1 s ( (s + 2)(s 2 + 1) + 1 s + 2 ). To work this out, express the RHS in partial fractions that A = 3 5, B = 2 5, C = 1 5. So y = L 1 3 ( 5(s + 2) ) + L 1 ( 2s + 1 5(s 2 + 1) ). 12 A s+2 + Bs+C s 2 +1. We find

Hence the solution is y = 3 5 e 2t + 2 5 cos t + 1 sin t. 5 More Laplace Transforms We can get lots more from Shift Rule I If L(f(t)) = F (s), then L(e at f(t)) = F (a + s). Proof L(e at f(t)) = e st e at f(t)dt = e (s+a)t f(t)dt = F (s + a). Hence for example we get L(e at t n ) = L(e at sin bt) = n! (s + a) n+1, b (s + a) 2 + b 2. As another example, suppose we want to work out Well, f(t) = L 1 2s + 3 ( s 2 + 2s + 5 ). 2s + 3 s 2 + 2s + 5 = 2s + 3 2(s + 1) (s + 1) 2 = + 4 (s + 1) 2 + 4 + 1 (s + 1) 2 + 4. Hence f(t) = 2e t cos 2t + 1 2 e t sin 2t. There is another shift rule, based on the Heaviside step function H a (t), defined as follows, where a > is a constant: {, if t < a H a (t) = 1, if t a First observe that More generally, we have L(H a (t)) = a 13 e st dt = e as s.

Shift Rule II If L(f(t)) = F (s), then L(H a (t)f(t a)) = e as F (s). Proof The LHS = a e st f(t a)dt. Put u = t a. Then LHS = e s(u+a) f(u)du = e as e su f(u)du = e as F (s). Examples 1. L(H a (t) sin(t a)) = e as L(sin t) = e as s 2 +1. 2. What is L 1 ( e 2s )? Well, we know L(t 2 ) = 2, so by the shift rule, s 2 s 3 L 1 ( e 2s s 2 ) = 1 2 H 2(t)(t 2) 2. Higher order differential equations Recall (2): L( dy dt ) = y() + sl(y). Now we find L( d2 y ). Well, dt 2 L( d2 y dt 2 ) = e st d2 y dy = [e st dt2 dt ] Substituting for L( dy dt ) gives + s st dy e dt dt = y () + sl( dy dt ). L( d2 y dt 2 ) = y () sy() + s 2 L(y). (3) Example Use Laplace transforms to solve with y() =, y () = 1. d 2 y dt 2 2dy dt + y = et Answer Take Laplace trans of both sides: y () sy() + s 2 L(y) 2( y() + sl(y)) + L(y) = L(e t ) = 1 s 1. So 1 + (s 2 2s + 1)L(y) = 1 s 1 which gives (s 1)2 L(y) = 1 s 1 + 1, so L(y) = 1 (s 1) 2 + 1 (s 1) 3. 14

So y = L 1 1 ( (s 1) 2 ) + 1 L 1 ( (s 1) 3 ) = tet + t2 e t 2. Simultaneous differential equations This is where Laplace transforms come into their own. Example Find functions y, z of t satisfying the simultaneous differential eqns d 2 y dt 2 + 2y z = d 2 z dt 2 + 2z y = with z() = 2, y() = z () = y () =. Ans Taking Laplace trans and using (3), get (s 2 + 2)L(y) L(z) = (s 2 + 2)L(z) L(y) = 2s. Eliminate L(y) by taking the first eqn plus (s 2 + 2) times the second, to get By partial fracs the RHS is Similarly L(z) = s s 2 +1 + 2s(s 2 + 2) (s 2 + 3)(s 2 + 1). s s 2 +3. So z = L 1 s ( s 2 + 1 ) + s L 1 ( s 2 + 3 ) = cos t + cos( 3t). y = cos t cos( 3t). Integration and Laplace transforms Rule If g(t) = t f(u)du, then L(g) = 1 s L(f). Proof By the Fundamental theorem of Calculus (from last term this says integration is the reverse of differentiation), g (t) = f(t). Hence L(f(t)) = L(g (t)) = g() + sl(g). As g() = f(u)du =, L(f) = sl(g). 15

Example Solve the differential/integral eqn with y() = 1. Ans Take Laplace trans: This gives L(y) = dy dt + 2y + t y(u)du =, 1 + sl(y) + 2L(y) + 1 L(y) =. s s 1, which by partial fracs is (s+1) 2 s+1 1. Hence (s+1) 2 y = L 1 1 ( s + 1 ) 1 L 1 ( (s + 1) 2 ) = e t te t. 16

3 Fourier Series A function f(x) is periodic with period P if f(x + P ) = f(x) for all x. Such a function is determined completely once we know its values on any interval of length P. Fourier series are defined for periodic functions of period 2π. Such a function is determined by its values for π x < π. Definition Let f(x) be a function defined for π x < π. The Fourier series of f(x) is defined to be the series a 2 + a n cos nx + b n sin nx, where a = π π f(x)dx, a n = π π f(x) cos nxdx, b n = π π f(x) sin nxdx. The numbers a, a n, b n are called the Fourier coefficients. In the lectures I explained why they are defined as above. A basic question is: under what conditions is a function equal to its Fourier series? A very good answer is provided by Dirichlet s Theorem Let f(x) be defined for π x < π. Suppose f(x) is continuous except at a finite number of points, and also f(x) has a left-hand and a right-hand derivative at every point. Then (1) at points x where f is continuous, f(x) is equal to its Fourier series (2) at points x = a where f is not continuous, the Fourier series is equal to 1 2 (f(a)+ + f(a) ), where f(a) + is the limit of f(x) as x tends to a from the right, and f(a) is the limit of f(x) as x tends to a from the left. In the lectures I explained these conditions in more detail, with examples. Example of calculation of Fourier series Let f(x) = x 2, π x < π We ll find the Fourier series of f(x). f(x) sin nx is odd, and so b n = 1 π π π First, f(x) is an even function, so f(x) sin nxdx =. 17

Next, integrating by parts, a n = 1 π π π x2 cos nxdx = 1 nπ [x2 sin nx] π π 1 π nπ π 2x sin nxdx = + 2 n 2 π [x cos nx]π π + 2 π n 2 π π cos nxdx 4 cos nπ = = 4 ( 1)n. n 2 n 2 Also Hence the Fourier series is a = 1 π π π π 2 3 + 4 x 2 dx = 2π2 3. ( 1) n cos nx n 2. Since f(x) is continuous for all x, by Dirichlet s Theorem f(x) is equal to its Fourier series for all x. For example, putting x = π we get and hence π 2 = π2 3 + 4 Or putting x = we get so ( 1) n cos nπ n 2 1 n 2 = π2 6. = π2 3 + 4 ( 1) n n 2 = π2 3 + 4 1 n 2, ( 1) n n 2, = π2 12. Even and odd functions Let f(x) be a function defined for π x < π. Recall that f is an even function if f( x) = f(x) for all x. Such a function is determined by its values for x < π. If f is even then f(x) sin nx is odd so the Fourier coefficient b n =. Likewise, f is an odd function if f( x) = f(x) for all x. If f is odd then the Fourier coefficient a n =. 18

Summary (1) An even function f(x) ( x < π) has a Fourier cosine series a 2 + a n cos nx, where a = 2 π π f(x)dx, a n = 2 π π f(x) cos nxdx. (2) An odd function f(x) ( x < π) has a Fourier sine series b n cos nx, where b n = 2 π π f(x) sin nxdx. Example Find the Fourier cosine series for the function f(x) = sin x ( x < π). (This means the Fourier series of the even function taking these values.) Answer The Fourier coefficients are a n = 2 π π = 2 π π a = 2 π π which works out as if n is odd, and cosine series is 2 π 4 π sin xdx = 4 π, sin x cos nxdx (sin(n + 1)x sin(n 1)x)dx 1 2 4 if n is even. Hence the Fourier π(n 2 1) cos 2nx 4n 2 1. Parseval s Formula Let f(x) ( π x < π) be a function and suppose it is continuous and differentiable, so by Dirichlet is equal to its Fourier series: f(x) = a 2 + a n cos nx + b n sin nx. Multiplying both sides by f(x) and integrating we get π π f(x)2 dx = π a π 2 f(x)dx + π a n π f(x) cos nxdx + π b n π f(x) sin nxdx = a 2.πa + a n.πa n + b n.πb n (since π π f(x) cos nxdx = πa n, π π f(x) sin nxdx = πb n from the definition of Fourier coeffs a n, b n ). Hence we get 19

Parseval s Formula If f(x) ( π x < π) is equal to its Fourier series, then 1 π f(x) 2 dx = a2 π π 2 + a 2 n + b 2 n. Example In our first example of Fourier series for f(x) = x 2 ( π x < π) we found x 2 = π2 3 + 4 ( 1) n cos nx n 2. So a = 2π2 3 and a n = 4 ( 1)n, and by Parseval s formula we get n 2 1 π x 4 dx = 2π4 π π 9 + 16 n 4. The left hand side is 2π4 5, and hence we find 1 n 4 = π4 9. Changing the period Sometimes one wants to get a Fourier series for a periodic function of period different to 2π. Here s how. Let f(x) be periodic of period 2r, say. So f is completely determined by its values for r x < r. If we define another function F (x) by F (x) = f( rx π ), then F (x + 2π) = f( rx π + 2r) = f(rx π ) = F (x), so F has period 2π. So F has a Fourier series a 2 + a n cos nx + b n sin nx, where and similarly and a = 1 r a n = 1 π π π = 1 π π π f(rx π = 1 r r r r r f(y)dy. F (x) cos nxdx ) cos nxdx nπy f(y) cos( r )dy (substituting y = rx π ) b n = 1 r r r f(y) sin( nπy r )dy 2

Summary If f(x) is defined for r x < r then the Fourier series of f is a 2 + a n cos nπx r + b n sin nπx r, where a = 1 r r r f(x)dx, a n = 1 r nπx r r f(x) cos( r )dx, b n = 1 nπx r f(x) sin( r )dx. r r 21