Newton s Method and Linear Approximations

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1 Newton s Method and Linear Approximations

2 Curves are tricky. Lines aren t. Newton s Method and Linear Approximations

3 Newton s Method for finding roots Goal: Where is f (x) = 0? f (x) = x 7 + 3x 3 + 7x

4 Newton s Method for finding roots Goal: Where is f (x) = 0? f (x) = x 7 + 3x 3 + 7x x0-2

5 Newton s Method for finding roots Goal: Where is f (x) = 0? f (x) = x 7 + 3x 3 + 7x x0-2

6 Newton s Method for finding roots Goal: Where is f (x) = 0? f (x) = x 7 + 3x 3 + 7x x0

7 Newton s Method for finding roots Goal: Where is f (x) = 0? f (x) = x 7 + 3x 3 + 7x x1 x0

8 Newton s Method for finding roots Goal: Where is f (x) = 0? f (x) = x 7 + 3x 3 + 7x x1 x0

9 Newton s Method for finding roots Goal: Where is f (x) = 0? f (x) = x 7 + 3x 3 + 7x x1

10 Newton s Method for finding roots Goal: Where is f (x) = 0? f (x) = x 7 + 3x 3 + 7x 2 1 x x1

11 f (x) = x 7 + 3x 3 + 7x 2 1 f (x) = 7x 6 + 9x x i x i f (x i ) f (x i ) tangent line x-intercept

12 f (x) = x 7 + 3x 3 + 7x 2 1 f (x) = 7x 6 + 9x x i x i f (x i ) f (x i ) tangent line x-intercept y = (x 0.5)

13 f (x) = x 7 + 3x 3 + 7x 2 1 f (x) = 7x 6 + 9x x i x i f (x i ) f (x i ) tangent line x-intercept y = (x 0.5)

14 f (x) = x 7 + 3x 3 + 7x 2 1 f (x) = 7x 6 + 9x x i x i f (x i ) f (x i ) tangent line x-intercept y = (x 0.5) y = (x 0.379)

15 f (x) = x 7 + 3x 3 + 7x 2 1 f (x) = 7x 6 + 9x x i x i f (x i ) f (x i ) tangent line x-intercept y = (x 0.5) y = (x 0.379)

16 f (x) = x 7 + 3x 3 + 7x 2 1 f (x) = 7x 6 + 9x x i x i f (x i ) f (x i ) tangent line x-intercept y = (x 0.5) y = (x 0.379) y = (x 0.353)

17 f (x) = x 7 + 3x 3 + 7x 2 1 f (x) = 7x 6 + 9x x i x i f (x i ) f (x i ) tangent line x-intercept y = (x 0.5) y = (x 0.379) y = (x 0.353)

18 f (x) = x 7 + 3x 3 + 7x 2 1 f (x) = 7x 6 + 9x x i x i f (x i ) f (x i ) tangent line x-intercept y = (x 0.5) y = (x 0.379) y = (x 0.353) y = (x 0.352) 0.352

19 Newton s Method Step 1: Pick a place to start. Call it x 0.

20 Newton s Method Step 1: Pick a place to start. Call it x 0. Step 2: The tangent line at x 0 is y = f (x 0 ) + f (x 0 ) (x x 0 ). To find where this intersects the x-axis, solve 0 = f (x 0 ) + f (x 0 ) (x x 0 ) to get x = x 0 f (x 0) f (x 0 ). This value is your x 1.

21 Newton s Method Step 1: Pick a place to start. Call it x 0. Step 2: The tangent line at x 0 is y = f (x 0 ) + f (x 0 ) (x x 0 ). To find where this intersects the x-axis, solve 0 = f (x 0 ) + f (x 0 ) (x x 0 ) to get x = x 0 f (x 0) f (x 0 ). This value is your x 1. Step 3: Repeat with your new x-value. In general, the next value is x i+1 = x i f (x i) f (x i )

22 Newton s Method Step 1: Pick a place to start. Call it x 0. Step 2: The tangent line at x 0 is y = f (x 0 ) + f (x 0 ) (x x 0 ). To find where this intersects the x-axis, solve 0 = f (x 0 ) + f (x 0 ) (x x 0 ) to get x = x 0 f (x 0) f (x 0 ). This value is your x 1. Step 3: Repeat with your new x-value. In general, the next value is x i+1 = x i f (x i) f (x i ) Step 4: Keep going until your x i s stabilize. What they stabilize to is an approximation of your root!

23 Caution! Bad places to pick: Critical points! (where f (x)=0) xi Tangent line has no x-intercept!

24 Caution! Bad places to pick: Critical points! (where f (x)=0) xi Tangent line has no x-intercept! Even near critical points, the algorithm goes much slower. Just stay away!

25 You try: Approximate a root of f (x) = x 2 x 1 near x 0 = x f (x) = i x i f (x i ) f (x i ) x i+1 = x i f (x i ) f (x i )

26 You try: Approximate a root of f (x) = x 2 x 1 near x 0 = x f (x) = 2x 1 i x i f (x i ) f (x i ) x i+1 = x i f (x i ) f (x i ) / /3 1/9 7/3 34/

27 Back to the example: f (x) = x 7 + 3x 3 + 7x 2 1 f (x) = 7x 6 + 9x x

28 Back to the example: f (x) = x 7 + 3x 3 + 7x 2 1 f (x) = 7x 6 + 9x x r

29 Back to the example: f (x) = x 7 + 3x 3 + 7x 2 1 f (x) = 7x 6 + 9x x r 1 r 2 r

30 Back to the example: f (x) = x 7 + 3x 3 + 7x 2 1 f (x) = 7x 6 + 9x x r r r

31 Linear approximations of functions Goal: approximate functions

32 Linear approximations of functions Goal: approximate functions Example: approximate

33 Linear approximations of functions Goal: approximate functions Example: approximate y = (x 1)

34 Linear approximations of functions Goal: approximate functions Example: approximate y = (x 1) (2 1) = 1.5

35 Linear approximations of functions Goal: approximate functions Example: approximate y = (x 1) (2 1) = 1.5 ( 2 = )

36 Linear approximations If f (x) is differentiable at a, then the tangent line to f (x) at x = a is y = f (a) + f (a) (x a). For values of x near a, then f (x) f (a) + f (a) (x a). This is the linear approximation of f about x = a. We usually call the line L(x).

37 Approximate 5:

38 Approximate 5: Our last approximation told us This isn t great... (3 2 = 9) 1 5 L(5) = 1 + (5 1) = 3 2

39 Approximate 5: Our last approximation told us This isn t great... (3 2 = 9) 1 5 L(5) = 1 + (5 1) = 3 2 Better: Use the linear approximation about x = 4!

40 Approximate 5: Our last approximation told us This isn t great... (3 2 = 9) 1 5 L(5) = 1 + (5 1) = 3 2 Better: Use the linear approximation about x = 4! The tangent line is so Better! ( = ) L(x) = (x 4) L(5) = 2 + (5 4) =

41 Even better approximations... The linear approximation is the line which satisfies L(a) = f (a) + f (a)(a a) = f (a) and L (a) = d ( f (a) + f (a)(x a) ) = f (a) dx

42 Even better approximations... The linear approximation is the line which satisfies L(a) = f (a) + f (a)(a a) = f (a) and L (a) = d dx ( f (a) + f (a)(x a) ) = f (a) A better approximation might be a quadratic polynomial p 2 (x) which also satisfies p 2 (a) = f (a): p 2 (x) = f (a) + f (a)(x a) f (a)(x a) 2

43 Even better approximations... The linear approximation is the line which satisfies L(a) = f (a) + f (a)(a a) = f (a) and L (a) = d dx ( f (a) + f (a)(x a) ) = f (a) A better approximation might be a quadratic polynomial p 2 (x) which also satisfies p 2 (a) = f (a): p 2 (x) = f (a) + f (a)(x a) f (a)(x a) 2 or a cubic polynomial p 3 (x) which also satisfies p (3) 3 (a) = f (3) (a): p 3 (x) = f (a) + f (a)(x a) f (a)(x a) f (3) (a)(x a) 3

44 Even better approximations... The linear approximation is the line which satisfies L(a) = f (a) + f (a)(a a) = f (a) and L (a) = d dx ( f (a) + f (a)(x a) ) = f (a) A better approximation might be a quadratic polynomial p 2 (x) which also satisfies p 2 (a) = f (a): p 2 (x) = f (a) + f (a)(x a) f (a)(x a) 2 or a cubic polynomial p 3 (x) which also satisfies p (3) 3 (a) = f (3) (a): p 3 (x) = f (a) + f (a)(x a) f (a)(x a) f (3) (a)(x a) 3 and so on... These approximations are called Taylor polynomials (read 2.14)

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