Polynomials and Taylor Series:How Functional is your Function ?

Earth's Interior temperature profile

How well would a linear equation fit this curve ?

Polynomials and Taylor Series:How Functional is your Function ?

Borehole - geothermal profiles

Measured in the Field Theoretical

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.

What is the Difference Between a Differential and a Derivative ?

“A little bit”, “rise/run”, which is which ?

A derivative is a rate of change, , or f '(x)dfdx

a

e

c

d

b

How does the slope change ?

The Derivative

f '(x) = dfdx

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.....

The Derivative

f '(x) = lim f(x + x) - f(x)

xx 0

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

orf = f (x + x) - f (x)

So how close is approximate ? The Taylor Series has the answer...

Imagine that you are on a hillside...

E

h

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)

1200 m

1125

m

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

1300

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 ?

In Formal Calculus...

h '(x) = lim h(x ) - h(a)

x-ax-a) 0

Jessica now

Jessica moves uphill

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

If equation for astraight line, what isslope, and y intercept ?

But what if the slope changes ?

h = 309 + 1.16x - 0.0004x2

h = -555 + 2.6x - 0.001x2

h = -3435 + 7.4x - 0.003x2

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

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

Slope = 0.2

Polynomials – are curves with changing slopes

h = 309 + 1.16x - 0.0004x2h = 1125 + 0.2x

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

Polynomials – are curves with changing slopes

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

Slope = 0.2

Parabolas

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 = 309 + 1.16x - 0.0004x2

h(x) = co + c

1x + c

2 x2

Polynomials

h(x) = co + c

1x + c

2 x2

“Second Order” refers to the highest power of the variable x The coefficients, c

i, 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

The ratio of coefficients c1 and c

2 can give the maximum of a

concave down curve

Xsummit

= - c1 / 2c

2

Polynomial Coefficients

h(x) = co + c

1x + c

2 x2

Each of the coefficients, co , c

1 , c

2 can be determined

by differentiation and use of the initial conditions The derivative h '(x) is :

h'(x) = c1 + 2c

2 x

The second derivative h ''(x) is :

h''(x) = 2c2

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

Polynomial Coefficients

h(x) = co + c

1x + c

2 x2

h'(x) = c1 + 2c

2 x

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

1

c1

= h'(a) - 2ac2

This gives

Plug into first equation and solve for co

co = h(a) – a h'(a) + a2c

2

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

Polynomial and Taylor Series

Polynomial functions can be represented by Taylor Series

h(x) = h(a) + (x-a) h'(a)

First order function

h(x) = h(a) + (x-a) h'(a) + (x-a)2/2 h''(a)

Second order function

h(x) = h(a) + (x-a) h'(a) + (x-a)2/2! h''(a) + (x-a)3/3! h'''(a)

Third order cubic function

These Taylor Series can also be used for error propagation