Numerical Analysis COT4501 Fall 2007 Midterm I

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					                                  Numerical Analysis
                                            COT4501
                                             Fall 2007
                                            Midterm I


                                          October 9, 2007


1. Taylor series approximation [35 points overall]
[10 points] What is the second order Taylor series approximation p2 (x) and the remainder
R2 (x) for f (x) = x log x when the approximation is carried out at x0 = 1 and in the interval
[1, 2]? [Here log(x) denotes the natural logarithm.]
[25 points]

  1. [5 points] Write down the nth term in the Taylor series approximation of f (x) = x log x
     when the approximation is carried out at an arbitrary x0 and is valid in the interval
     [x0 , 2x0 ].

  2. [10 points] Bound the nth term in the Taylor series approximation using the fact that
     x ≥ x0 for all points in the interval [x0 , 2x0 ]. Let us denote the bound by g (n) (x, x0 ).
                                ∞                                                          (x−x0 )2
  3. [10 points] Sum up         n=2 g
                                      (n)
                                          (x, x0 )   to show that f (x) − p1 (x) ≤        2(2x0 −x)
                                                                                                      or in other
                                  x                           (x−x0 )2
     words, show that       x log x0 − x + x0        ≤       2(2x0 −x)
                                                                         for all x ∈ [x0 , 2x0 ]. [Hint: Use a
     geometric series summation. Here p1 (x) is the rst order Taylor series approximation
     of f (x) = x log x.]

2. Derivative approximation [35 points overall]
                                                     f (x+ h )−f (x− h )
[10 points] Is the approximation f (x) ≈                   2
                                                              h
                                                                     2
                                                                           a valid rst derivative approxima-
tion? Discuss.
[25 points] Begin with f (x) ≈ Af (x + ah) + Bf (x + bh).

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   1. [10 points] Expand f (x + ah) and f (x + bh) up to second order using a Taylor series
       approximation.

   2. [15 points] Construct an approximation to f (x) by ensuring that i) the term involving
       f (x) is zero, ii) the term involving f (x) is zero, and that iii) the coecient of f (x)
       is one. Write down the resulting constraints involving A, B, a, b, and h. Pick a set of
       possible values for (A, B, a, b) that satisfy the constraints.

3. Linear Interpolation [30 points overall]
[10 points] Rolle's Theorem: Given a function f (x) which is continuous and dierentiable
in an interval [a, b] and with f (a) = f (b) = 0, show that there exists a point ξ ∈ [a, b] such
that f (ξ) = 0.
[20 points] Extension of the linear interpolation formula to quadratic interpolation. Given
three points (x0 , f (x0 )), (x1 , f (x1 )), and (x2 , f (x2 )):

   1. [10 points] Construct a single second order polynomial approximation p2 (x) (by ex-
       tending the linear approximation formula) of the form p2 (x) = ax2 + bx + c such that
       p2 (x0 ) = f (x0 ), p2 (x1 ) = f (x1 ) and p2 (x2 ) = f (x2 ). [Hint: We are looking for a single
       quadratic polynomial here. Do not combine two piecewise linear approximations.]

   2. [10 points] What are the values of a, b, and c?




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                                          List of Useful Formulae
                                                            (x−x0 )i f (i) (x0 )   (x−x0 )(n+1) f (n+1) (ξ[x0 ,x] )
Taylor series approximation: f (x) =                    n
                                                        i=0            i!
                                                                                 +             (n+1)!
                                                                                                                    with ξ[x0 ,x]
                                                                            h(n+1) f (n+1) (ξ
                                                                                             [x,h] )
                                                            hi f (i) (x)
                                                       n
in the interval [x0 , x]. Also, f (x + h) =            i=0       i!
                                                                          +        (n+1)!
                                                                                                     where ξ[x,h] depends
on both x and h.
Euler's method: Approximation of y = f (t, y ) is yn+1 = yn + hf (tn , yn ) with y0 = y(t0 ).
Dierence Approximations of Derivatives: f (x) = f (x+h)−f (x) − 1 hf (ξx,h ), f (x) =
                                                             h          2
f (x+h)−2f (x)+f (x−h)        1 2
         h2
                         −   12
                                hf   (ξx,h ).
Linear Interpolation formula: p1 (x) = x11−x0 f (x0 ) + xx−x00 f (x1 ).
                                       x −x
                                                          1 −x
                                  n
Bisection method: |α − xn | ≤ 1 (b − a) with the midpoint of the interval chosen at
                               2
each step.
Newton's method: Update rule for nding a root to f (x) = 0 is xn+1 = xn − f (xn )) with
                                                                           f
                                                                             (xn
initial condition x0 .
                                                                                    √
Quadratic formula: Solution of ax2 + bx + c = 0 is x = −b± 2a −4ac .
                                                            b                           2



Geometric series summation: ∞ arn = 1−r where r < 1.
                                 n=0
                                            a

Mean Value Theorem: Let f be a given function continuous on [a, b] and dierentiable
                                                                                      f (b)−f (a)
on (a, b). Then there exists a point ξ ∈ [a, b] such that f (ξ) =                         b−a
                                                                                                  .
Derivatives:       d
                  dx
                     x log x   = (1 + log x),    d
                                                dx
                                                             1
                                                     log x = x ,     d 1
                                                                    dx x
                                                                               1
                                                                           = − x2 ,    d 1
                                                                                      dx xn
                                                                                                    n
                                                                                              = − xn+1 .




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