Analysis/Calculus Review Day 2

Transcription

1 Analysis/Calculus Review Day 2 Arvind Saibaba arvindks@stanford.edu Institute of Computational and Mathematical Engineering Stanford University September 14, 2010

2 Limit Definition Let A R, f : A R and suppose that x 0 is an accumulation point of A. We say that b R is the limit of f at x 0, written lim f(x) = b x x 0 if given any ǫ > 0 there exists δ > 0 (depending on f,x 0 and ǫ) such that for all x A, x x 0, x x 0 < δ implies that f(x) b < ǫ. 1. Define f : R\{0} R f(x) = { 0 x < 0 2 x > 0 {0} is an accumulation point of R\{0} but lim x x0 f(x) does not exist. 2. Define f : R\{0} R { 1 x 0 f(x) = 0 x = 0

3 Directional Limits Suppose that f is defined at least on ]x 0,a] R, for some a > x 0. then lim f(x) = b x x + 0 means the limit of f with domain A =]x 0,a]. In other words, for every ǫ > 0 there is a δ > 0 such that x x 0 < δ,x > x 0 implies f(x) b < ǫ. Thus, we are taking the limit of f as x approaches x 0 from the right. Similarly, we can define lim x x f(x) = b 0 the limit as x approaches x 0 from the left. These are known as one sided limits.

4 Examples

5 Continuity Definition Let A R, f : A R and let x 0 A. We say that f is continuous at x 0 when lim x x 0 f(x) = f(x 0 ) This equivalent to saying that for every convergent sequence x k x 0, we have f(x k ) f(x 0 ).

6 Examples Let f : R R be the identity function x x. Show that f is continuous. Solution Fix x 0 R. By definition we must find δ > 0 for given ǫ > 0 such that x x 0 < δ implies f(x) f(x 0 ) < ǫ. This is trivially true if one chooses δ = ǫ. Hence f is continuous.

7 L Hopital s Rule Suppose f and g are differentiable on ]a,b[ and g (x) 0 for x ]a,b[. Suppose f (x) g (x) A as x a If f(x) 0 and g(x) 0 as x 0, or if g(x) as x a, then Example sinx lim x 0 x f(x) g(x) A as x a d/dx(sinx) cosx = lim = lim x 0 d/dx(x) x 0 1 = 1

8 Sequences of functions and Convergence Definition (Pointwise Convergence) Given a sequence of functions of a set E, namely f n : E R. Suppose that for all x E lim n f n(x) = f(x) then we say that {f n } converges pointwise to f. The disadvantage with pointwise convergence is even if f n are all continuous, f need not be continuous. For example, consider { 0 x 1 f n (x) = k kx +1 0 x < 1 k In this case, for each x [0,1], f k (x) converges. If x 0,f k (x) 0 (since f k (x) = 0 for large k), while if x = 0,f k (x) = 1. The limit is thus { 0 x 0 f(x) = 1 x = 0

9 Uniform Convergence Definition We say {f n } converges uniformly to f (denoted f n f(uniformly), if for all ǫ > 0, there exists an N such that for all n N f n (x) f(x) ǫ, x For example, on R consider the following sequence, { 0 x < k f k (x) = 1 x k Clearly, f k (x) 0 pointwise. (for k large f k (x) = 0). However, f k does not converge to 0 uniformly, because there are points x such that f(x) 0, is not small no matter how large k is.

10 contd. Example Let f n (x) = sinx x,f n : R R Show that f n 0 uniformly as n. Solution We must show that f n (x) 0 = f n (x) gets small independent of x as n. But f n (x) = sinx /n 1/n which gets small independent of x as n. Things to note Uniform convergence implies Pointwise convergence but not vice versa. If {f n } is a sequence of continuous functions and f n f(uniformly), then f is a continuous function.

11 Differentiation For a function f : [a,b] R, f is called differentiable at x 0 ]a,b[ if the limit f f(x 0 +h) f(x 0 ) (x 0 ) = lim h 0 h exists. One also writes df/dx for f (x).

12 Tangents

13 Using the trigonometric identity From First Principles sin(a+b) = sinacosb +cosa+sinb let us compute the derivative of sinx d sin(x +h) sinx sinx = lim dx h 0 h sinx cosh+cosx sinh sinx = lim h 0 h = lim sinx. cosh 1 +cosx. sinh h 0 h h cosh 1 sinh = sinx. lim +cosx. lim h 0 h h 0 h = 0+cosx.1 = cosx

14 Continuity of Derivatives If a function is not countinuous at x but f(x + ) and f(x ) exist, then f has a discontinuity of the first kind. This means that either f(x + ) f(x ), or f(x + ) = f(x ) f(x) If f is not continuous as per definition but these two conditions are met we say f has a discontinuity of the second kind. Suppose that f is differentiable on [a,b] and suppose that f (a) < λ < f (b). Then there exists x ]a,b[ such that f (x) = λ If f is differentiable on [a,b] then f has no simple discontinuities. It may have discontinuities of the second kind.

15 Properties of Derivatives Let f,g be differentiable on [a,b]. Then, (cf) (x) = cf (x) (f +g) (x) = f (x)+g (x) Product Rule. Quotient Rule Chain Rule (f g) (x) = f (x)g(x)+f(x)g (x) (f/g) (x) = f (x) g(x) g (x)f(x) g 2 (x) h(x) = g(f(x)) h (x) = g (f(x))f (x)

16 Examples 1. Compute derivative of g(x) = sin(x 2 ). Call f(x) = x 2 and recall Chain Rule for g(f(x)) g (x) = g (f(x)) f (x) = cos(x 2 ) (2x) 2. Compute derivative of f(x) = { x sin 1 x x 0 0 x = 0 Apply Product Rule and Chain Rule f (x) = sin 1 ( x +x 1 ) x 2 cos 1 x = sin 1 x 1 x cos 1 x When x = 0 appeal to the definition f(h) f(0) h = sin 1 x The limit does not exist and therefore the derivative does not exist at x = 0.

17 Rolle s Theorem Theorem If f : [a,b] R is continuous, f is differentiable on ]a, b[ and f(a) = f(b) = 0 then there is a number c ]a,b[ such that f (c) = 0.

18 Mean Value Theorem Theorem If f : [a,b] R is continuous, f is differentiable on ]a,b[, there is a point c ]a,b[ such that f(b) f(a) = f (c)(b a) Corollary If f is differentiable on ]a,b[ and f (x) = 0 for all x ]a,b[ then f is a constant on ]a,b[.

19 Integration Partition [a,b] as a = x 0 < x 1 < < x n 1 < x n = b U(f,P) = L(f,P) = n 1 [sup{f(x) : x [x i,x i+1 ]}](x i+1 x i ) i=0 n 1 [inf{f(x) : x [x i,x i+1 ]}](x i+1 x i ) i=0 We say that f is Riemann integrable if inf{u(f,p)} = sup{l(f,p)} for any partition.

20 Notation If f is Riemann integrable, f = f(x)dx = sup{l(f,p)} = inf{u(f,p)} A A

21 Properties of Integration For integrable functions f 1,f 2,f f 1 +f 2 is integrable and cf is integrable. (Linearity) If f 1 f 2, Let a b c. Then, c a b a f(x)dx = Suppose sup f(x) M, then f 1 (x)dx b a b a f 2 (x)dx c f(x)dx + f(x)dx b f 1 f 2 is integerable. If f(x) is integrable b a f(x)dx M(b a) b f(x)dx b a a f(x) dx

22 Fundamental Theorem of Calculus If f is integrable over [a,b], define F(x) = x a f(t)dt Then, F is continuous. If f is continuous at x 0 (a,b) then F is differentiable at x 0 and F (x 0 ) = f(x 0 ). F is called the anti-derivative of f. Theorem Let f : [a,b] R be continuous. Then f has an antiderivative F and b a f(x)dx = F(b) F(a) If G is any other antiderivative of f, then we also have that f(x)dx = G(b) G(a). b a

23 Standard Integrals

24 Method of Substitution f(g(x))g (x)dx = F(g(x))+C where, F is the anti-derivative of f. This follows from d dx F(g(x)) = F (g(x))g (x) = f(g(x))g (x) Example Integrate 1 1+x 2dx Plugging in x = tanθ, dx = sec 2 θdθ. 1 1+x 2dx = 1 sec 2 θ sec2 θdθ = dθ = θ +C = tan 1 x +C

25 Integration by parts The product rule for differentiation is (f g) (x) = f (x)g(x)+f(x)g (x) Thus, we can write f(x)g (x)dx = f(x)g(x) f (x)g(x) Example Integrate log xdx We use the above formula with f(x) = logx and g (x) = 1. Therefore, 1 logxdx = (logx)x x xdx = x logx x

26 Differentiation under the Integral sign Theorem (Leibniz Rule) If f is continuous on [a,b] and u(x),v(x) are differentiable functions of x whose values lie in [a,b], then d dx v(x) u(x) f(t)dt = f(v(x))v (x) f(u(x))u (x)

27 Fubini s theorem Theorem Let A be the triangle described by a x b, c y d and let f : A R be continuous. Then, ( b ) d ( d ) b f = f(x,y)dy dx = f(x,y)dx dy A a c c a Corollary Let φ,ψ : [a,b] R be continuous maps such that φ(x) ψ(x) for all x [a,b] and let A = {(x,y) : a x b,φ(x) y ψ(x)}. Let f : A R be continuous. Then, ( b ) ψ(x) f = f(x,y)dy dx A a φ(x)

28 Example For the region R in the figure R sinx x da 1 0 ( x 0 ) sinx x dy dx = = ( y sinx ) x y=x y=0 dx sin x dx = cos(1)+1 whereas, we can t compute y sinx x dydx

29 Change of Variables Theorem Let A be an open bounded set with volume and let g : A R n be a C 1 mapping which is one-to-one. Let B = g(a) have volume. For f : B R bounded and integrable, f g J(g) is integrable on A and B f(y 1,...,y n )dy 1...dy n = A f(g(x 1,...,x n )) (g 1,...,g n ) (x 1,...,x n ) dx 1...dx n In addition, we require Jg(x) 0 for all x A and Jg(x), 1/ Jg(x) are bounded on A. For example, (f 1,f 2 ) f (x,y) = 1 f 1 x y f 2 x f 2 y

30 Polar Coordinates Define the mapping g(r,θ) : (x,y) (r cosθ,r sinθ) Jg(r,θ) = cosθ r sinθ sinθ r cosθ = r cos2 θ +r sin 2 θ = r

31 r = x 2 +y 2 θ = tan 1 y x