CSE 541 Numerical Methods

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CSE 541Numerical Methods

• Taylors Series, Taylors Theorem, Mean-Value
  Theorem, Rolles Theorem

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Function Evaluation (1)

• Problem Compute f(x)
• Easy for well known functions
• For example, f (x) 1, f (x) x2

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Function Evaluation (2)

• Suppose f (x) is a black box
• Difficult and/or expensive to compute
• For example a complex function, an experiment, a
  procedure
• Limited information about the function
• Suppose we know information about the function at
  a particular domain location, x c
• f (c), f (c), f (c), f (c),

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Taylors Series

• Evaluate f (x) about a point c, called the point
  of expansion
• Observations
• An infinite series
• (x c) is the distance from the point of
  expansion
• F (x) is a polynomial!!!

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Taylors Theorem

• Where do we truncate this series?
• Taylors Theorem states the following

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Taylors Theorem

• Observations
• (x - c)(n 1) ? 0, if x - cltlt 1
• (x - c)(n 1) ? quickly, if x - cgtgt 1
• Higher-order derivatives may get smaller, e.g.
  smooth functions).

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Some Common Derivatives
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Some Resulting Series

• About c 0, called Maclaurin Series

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Some Resulting Series

• About c 0

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Books Introduction Example

• Eight terms for first series not even yielding a
  single significant digit.
• Only four for second serieswith
  foursignificantdigits.

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Mean-Value Theorem

• A special case of Taylors Theorem
• Take two locations x a and x b, where (a, b)
  is an interval.
• Assume f(x) is continuous and its first
  derivative exists everywhere within (a, b).
• Take Taylors Theorem where the point of
  expansion is a, x b, and n 0.

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Mean-Value Theorem

• There exists a location ? in the interval (a, b)
  where the first derivative is equal to the secant

secant
f(x)
a
b
?
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Rolles Theorem

• A special case of the Mean-Value Theorem
• If a and b are roots of function f (x), i.e. f
  (a) f (b) 0, then f (?)0 for some point ? in
  (a, b).
• What goes up, must come down

f(x)
f(?) 0
?
a
b
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Caveats

• For Taylors Series and Taylors Theorem to hold,
  the function and its derivatives must exist
  within the range you are trying to use it
• The function cant change its current value
  faster than the derivative will allow
• So, the function does not go to infinity, or have
  a discontinuity (implies f(x) does not exist),

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Implement a Fast sin()Application Riemann Sum

• const int Max_Iters 100,000,000
• float x -0.1
• float delta 0.2 / Max_Iters
• float Reimann_sum 0.0
• for (int i 0 i lt Max_Iters i)
• Reimann_Sum sinf(x)
• x delta
• Printf(Integral of sin(x) from 0.1-gt0.1 equals
  f\n, Reimann_Sumdelta )

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Implement a Fast sin()

• const int Max_Iters 100,000,000
• float x -0.1
• float delta 0.2 / Max_Iters
• float Reimann_sum 0.0
• for (int i 0 i lt Max_Iters i)
• Reimann_Sum my_sin(x) //my own sine func
• x delta
• Printf(Integral of sin(x) from 0.1-gt0.1 equals
  f\n, Reimann_Sumdelta )

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Version 1.0 Taylors

• my_sin( const float x )
• float x2 xx
• float x3 xx2
• return (x x3/6.0 x2 x3/120.0 )

Use three terms of the Taylors series expansion
for sin(x)
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Version 2.0 Horners Rule

• Static const float fac3inv 1.0 / 6.0f
• Static const float fac5inv 1.0 / 120.0f
• my_sin( const float x )
• float x2 xx
• return x(1.0 x2(fac3inv - x2fac5inv))

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Version 3.0 Inline code

• const int Max_Iters 100,000,000
• float x -0.1
• float delta 0.2 / Max_Iters
• float Reimann_sum 0.0
• for (i 0 i lt Max_Iters i)
• x2 xx
• Reimann_Sum x(1.0 x2(fac3inv -
  x2fac5inv)
• x delta
• Printf(Integral of sin(x) from 0.1-gt0.1 equals
  f\n, Reimann_Sumdelta )

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Timings

• Pentium III, 600MHz machine

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Observations

• Is the result correct?
• Why did we gain some accuracy with version 2.0?
• Is (0.1, 0.1) a fair range to consider?
• Is the original sinf() function optimized?
• How did we achieve our speed-ups?