Newton s Method and Linear Approximations 10/19/2011

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Newton s Method and Linear Approximations 10/19/2011

Curves are tricky. Lines aren t. Newton s Method and Linear Approximations 10/19/2011

Newton s Method Goal: Where is f (x) =0? f (x) =x 7 +3x 3 +7x 2 1 2-1 -0.5 0.5-2

Newton s Method Goal: Where is f (x) =0? f (x) =x 7 +3x 3 +7x 2 1 2-1 -0.5 0.5 x0-2

Newton s Method Goal: Where is f (x) =0? f (x) =x 7 +3x 3 +7x 2 1 2-1 -0.5 0.5 x0-2

Newton s Method Goal: Where is f (x) =0? f (x) =x 7 +3x 3 +7x 2 1 0.3 0.4 0.5 x0

Newton s Method Goal: Where is f (x) =0? f (x) =x 7 +3x 3 +7x 2 1 0.3 0.4 0.5 x1 x0

Newton s Method Goal: Where is f (x) =0? f (x) =x 7 +3x 3 +7x 2 1 0.3 0.4 0.5 x1 x0

Newton s Method Goal: Where is f (x) =0? f (x) =x 7 +3x 3 +7x 2 1 0.36 0.38 x1

Newton s Method Goal: Where is f (x) =0? f (x) =x 7 +3x 3 +7x 2 1 x2 0.36 0.38 x1

f (x) =x 7 +3x 3 +7x 2 1 f (x) =7x 6 +9x 2 + 14x i x i f (x i ) f (x i ) tangent line x-intercept 0 0.5 1 2 3

f (x) =x 7 +3x 3 +7x 2 1 f (x) =7x 6 +9x 2 + 14x i x i f (x i ) f (x i ) tangent line x-intercept 0 0.5 1.133 9.359 y = 1.133 + 9.359(x 0.5) 0.379 1 2 3

f (x) =x 7 +3x 3 +7x 2 1 f (x) =7x 6 +9x 2 + 14x i x i f (x i ) f (x i ) tangent line x-intercept 0 0.5 1.133 9.359 y = 1.133 + 9.359(x 0.5) 0.379 1 0.379 2 3

f (x) =x 7 +3x 3 +7x 2 1 f (x) =7x 6 +9x 2 + 14x i x i f (x i ) f (x i ) tangent line x-intercept 0 0.5 1.133 9.359 y = 1.133 + 9.359(x 0.5) 0.379 1 0.379 0.170 6.619 y = 0.170 + 6.619(x 0.379) 0.353 2 3

f (x) =x 7 +3x 3 +7x 2 1 f (x) =7x 6 +9x 2 + 14x i x i f (x i ) f (x i ) tangent line x-intercept 0 0.5 1.133 9.359 y = 1.133 + 9.359(x 0.5) 0.379 1 0.379 0.170 6.619 y = 0.170 + 6.619(x 0.379) 0.353 2 0.353 3

f (x) =x 7 +3x 3 +7x 2 1 f (x) =7x 6 +9x 2 + 14x i x i f (x i ) f (x i ) tangent line x-intercept 0 0.5 1.133 9.359 y = 1.133 + 9.359(x 0.5) 0.379 1 0.379 0.170 6.619 y = 0.170 + 6.619(x 0.379) 0.353 2 0.353 0.007 6.084 y = 0.007 + 6.084(x 0.353) 0.352 3

f (x) =x 7 +3x 3 +7x 2 1 f (x) =7x 6 +9x 2 + 14x i x i f (x i ) f (x i ) tangent line x-intercept 0 0.5 1.133 9.359 y = 1.133 + 9.359(x 0.5) 0.379 1 0.379 0.170 6.619 y = 0.170 + 6.619(x 0.379) 0.353 2 0.353 0.007 6.084 y = 0.007 + 6.084(x 0.353) 0.352 3 0.352

f (x) =x 7 +3x 3 +7x 2 1 f (x) =7x 6 +9x 2 + 14x i x i f (x i ) f (x i ) tangent line x-intercept 0 0.5 1.133 9.359 y = 1.133 + 9.359(x 0.5) 0.379 1 0.379 0.170 6.619 y = 0.170 + 6.619(x 0.379) 0.353 2 0.353 0.007 6.084 y = 0.007 + 6.084(x 0.353) 0.352 3 0.352 0.00001 6.060 y = 0.00001 + 6.060(x 0.352) 0.352

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

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.

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 )

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!

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

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!

You try: Approximate a root of f (x) =x 2 x 1 near x 0 = 1. 0.5-0.5 0.5 1 1.5 x0-0.5-1 f (x) = i x i f (x i ) f (x i ) x i+1 = x i f (x i ) f (x i ) 0 1 1 2

Back to the example: 2-1 -0.5 0.5-2

Back to the example: 2-1 -0.5 0.5-2.352 r 3 0.352

Back to the example: 2-1 -0.5 0.5-2.352 r 1 r 2 r 3 0.352

Back to the example: 2-1 -0.5 0.5-2.352 r 1 1.217 r 2 0.418 r 3 0.352

Linear approximations Goal: approximate functions

Linear approximations Goal: approximate functions Example: approximate 2 1 1 2 3

Linear approximations Goal: approximate functions Example: approximate 2 1 1 2 3 y =1+ 1 2 (x 1)

Linear approximations Goal: approximate functions Example: approximate 2 1 1 2 3 y =1+ 1 2 (x 1) 2 1+ 1 2 (2 1) = 1.5

Linear approximations Goal: approximate functions Example: approximate 2 1 1 2 3 y =1+ 1 2 (x 1) 2 1+ 1 2 (2 1) = 1.5 ( 2=1.414...)

Linear approximations If f (x) isdifferentiable at a, then the tangent line to f (x) atx = a is y = f (a)+f (a) (x a). For values of xneara,then f (x) f (a)+f (a) (x a). This is the linear approximation of f about x = a. Weusuallycall the line L(x).

Approximate 5:

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

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!

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

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)+ 1 2 f (a)(x a) 2

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)+ 1 2 f (a)(x a) 2 or a cubic polynomial p 3 (x) whichalsosatisfiesp (3) 3 (a) =f (3) (a): p 3 (x) =f (a)+f (a)(x a)+ 1 2 f (a)(x a) 2 + 1 2 3 f (3) (a)(x a) 3

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)+ 1 2 f (a)(x a) 2 or a cubic polynomial p 3 (x) whichalsosatisfiesp (3) 3 (a) =f (3) (a): p 3 (x) =f (a)+f (a)(x a)+ 1 2 f (a)(x a) 2 + 1 2 3 f (3) (a)(x a) 3 and so on... These approximations are called Taylor polynomials (read 2.14)

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