Polynomials and Taylor Series:

1
Polynomials and Taylor Series How Functional is
your Function ?
Earth's Interior temperature profile
How well would a linear equation fit this curve ?
2
Polynomials and Taylor Series How Functional is
your Function ?
Borehole - geothermal profiles

• Measured in the Field

• Theoretical

3
What is the Difference Between a Differential
and a Derivative ?
A little bit, rise/run, which is which ?
A differential is an infinitesimally small amount
a little bit, dx.
4
What is the Difference Between a Differential
and a Derivative ?
A little bit, rise/run, which is which ?
5
The Derivative

• The derivative of f with respect to x
• is the ratio of differentials .
• A small change in x will produce a small change
  in f .

• We can separate the differentials,

df f '(x) dx
Similar to the equation for a line.....
6
The Derivative

• In formal calculus the derivative is written,

• The definitions for differentials and
  derivatives only hold
• In the limit of small ?x going to 0.
• If the limit is removed, the equalities are only
  approximate

??????????f f '(x) dx ?or?????f f (x
?x) - f (x)
So how close is approximate ?
The Taylor Series has the answer...
7
Imagine that you are on a hillside...

• Hillside slope is upward to the east
• No slope in N-S direction
• Topo contours run N-S only
• We are interested in elevation, h
• h depends only on x, giving h(x)

• Suppose Jessica is sitting at x 1200 m
• Her elevation is 1125 m
• The hillside slope is 0.2 (or 20 grade or
  11o)

• What would be her elevation if she got up and
  walked to x 1300 m ?

8
Or in Mathematical Language...

• If a 1200
• h(a) 1125
• h '(x) 0.2
• What is h(x) ?

• If you know the elevation and slope at a point,
  a ,
• How can you calculate your elevation at a nearby
  point, x ?

9
In Formal Calculus...

• We need to solve for h(x)
• Do this removing the limit stuff...

h(x) h(a) (x-a) h'(a)

• Now just substitute in the values

h(1300) 1125 100 0.2
1145 m

• This is called projecting the hillside upward
• (assuming a constant slope) a straight line

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But what if the slope changes ?

• If the slope changes as Jessica moves uphill
• Then other answers are possible for h(1300)

• Here are 3 other functions all satisfying the
  conditions
• h(1300) 1125 and h'(1300) 0.2

h 309 1.16x - 0.0004x2 h -555
2.6x - 0.001x2 h -3435 7.4x -
0.003x2
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But what if the slope changes ?
h 309 1.16x - 0.0004x2 h -555
2.6x - 0.001x2 h -3435 7.4x -
0.003x2

• Are all these functions a straight line ?
• How do the slopes change with distance ?

• The slopes decrease with increasing distance, x

• In this case, the projected value for h(1300)
  will be smaller
• Than in the equation for a line with a constant
  slope

• The departure from the projected value increases
  with
• The magnitude of the coefficient of the 3rd
  term for x2

12
Polynomials are curves with changing slopes
h 1125 0.2x
h 309 1.16x - 0.0004x2

• The slope of the line and polynomial curve both
  pass through
• The point for h 1125 at x 1200.

13
Polynomials are curves with changing slopes
Slope 0.2

• The slope of the line and polynomial curve all
  pass through
• The point for h 1125 at x 1200.

14
Parabolas
h 309 1.16x - 0.0004x2

• These parabolas are concave downward
• An infinite number of parabolas can be drawn
• To fit the initial conditions
• We can represent these possibilities by a second
  order polynomial

h(x) co c1x c2 x2
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Polynomials
h(x) co c1x c2 x2

• Second Order refers to the highest power of
  the variable x
• The coefficients, ci, determine the shape and
  location
• The first coefficient, co, gives vertical
  position
• The second coefficient, c1, gives the slope at
  the y intercept
• The third coefficient, c2, determines how much
  it curves

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Polynomial Coefficients
h(x) co c1x c2 x2

• Each of the coefficients, co , c1 , c2 can be
  determined
• by differentiation and use of the initial
  conditions
• The derivative h '(x) is

h'(x) c1 2c2 x

• The second derivative h ''(x) is

h''(x) 2c2

• We already know h(a) at x a 1125 (at
  x 1200) ,

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Polynomial Coefficients

• Plug in h(x) 1125 at x 1200) into first
  and second deriv's
• And solve for c1

h(x) co c1x c2 x2
h'(x) c1 2c2 x
c1 h'(a) - 2ac2
This gives

• Plug into first equation and solve for co

co h(a) a h'(a) a2c2
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Polynomial Coefficients

• We know from the second derivative that

c2 h''(a) / 2

• Plug this into equation for co and you can get
  all 3 coefficients

• This brings us back to the original polynomial

h -555 2.6x - 0.001x2

• Message the coefficients are interrelated
• If you change one, the others will adjust to
  keep the parabola
• Passing through the original point

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Polynomial and Taylor Series

• Polynomial functions can be represented by Taylor
  Series

First order function
h(x) h(a) (x-a) h'(a)
h(x) h(a) (x-a) h'(a) (x-a)2/2 h''(a)
Second order function
Third order cubic function
h(x) h(a) (x-a) h'(a) (x-a)2/2! h''(a)
(x-a)3/3! h'''(a)
These Taylor Series can also be used for error
propagation