The polynomials that we found above using the method
of successive derivatives are called
Taylor polynomials.
What about higher degree Taylor Polynomials?
Explore the animations for Taylor Polynomials.
http://www.cs.gasou.edu/faculty/demos/TaylorPolynomials/gallery/
Changing the point of tangency
In the previous examples we found Taylor polynomials
were the "point of tangency" to the given
function is found when x = 0. This is referred to as
"expanding the polynomial about x = 0". Actually,
we can expand about any point on the curve. If x =
a
is the abcissa of the "point of tangency"
and the function is repeatablely differentiable at
this point,
then the resulting Taylor polynomial is said to be centered at a.
How do we expand a Taylor polynomial about a point x
= a? Think about shifting a
function.
We proceed with an example. Suppose we want to
find the cubic Taylor polynomial approximation to the
function
expanded about some point
x =
a,
then we need to find the constants
to
such that
:
Notice that the polynomial is written in powers of x - a.
Problem: Find the cubic Taylor polynomial for 
expanded
about the point a = 1.
To find
. substitute x = 1 into Eq. 2).
A cubic polynomial
is completely defined by the four coefficients
,
,
and
.
Problem:
Find constants
,
, 
and 
such that
thus,
.
Step 2. Calculate
the values of
,
and
using
(Hint: To take successive derivatives use the operator
)
Change the value for n from 1 to 3.
Find the 3th degree Taylor polynomial T6(x) for
expanded about the point x =1.
The first insertion point is the center of the expansion.
Type x <ctrl>= 1
The second insertion point determines the number of
terms in the expansion.
Compare this with what you found in the previous exercise.
This may also be written in terms of thew powers of
x:
1) Given the nth-degree Taylor polynomial approximation
for the function f(x),
show that
,
,
and that in general
.
2) Find the
approximation for
about
.
Write your answer in both of the forms:
3) Find the
approximation for
about
.
Check by graphing, include the graph of the point
of tangency.
Use
to estimate
.
_________________________________________________________
Step 2. Calculate
,
and
.
Change the value for n from 1 to 3.
Check by graphing. Include the graph of the tangent
point.
The Algebraic Representation
Given a function
f(x),
that is at least n-times differentiable when
x = a,
the nth-degree Taylor polynomial,
is written as:
Your job is to see that Eq. 3) is exactly a symbolic
representation of the method of successive differentiation,
and that denotes exactly what we have already done
in the previous examples!
A little note of cheer, Mathcad can actually
calculate 
using Eq. 3).
Calculating Taylor Polynomials Using
Mathcad
Mathcad's symbolic menu has a Expand in Series Keyword
that may be used to calculate
Since
, we have
.
Step 2: Find b as follows.
P'(0) = f '(0) (
The slopes are equal)
The requirement that at the point (0, 1), the derivative
of the cosine equals the derivative of the polynomial,
simply says that at this point they have the same slope.
This is written as:
Take the derivatives to get:
At the "point of tangency", (0, 1),
.
Since the
sin(0) = 0
we find:
or
.
Now our polynomial looks like
Finally, we find a value for a.
Step 3: Find a
as follows. P2
''(0) = f ''(0) (
Second derivatives are also equal)
Next we require that the second derivatives of
both functions are also equal at the "point of
tangency", (0, 1). Thus:
Again, we let
and
solve for
a.
We have found the approximating polynomial,
.
Note: in this lab bold equal is displayed as equal.
____________________________________
Problem: How to Calculate Values for Transcendental
Functions
Since calculators and computers are so common today,
the problem of evaluating a transcendental function
like e2.3
or asec(1.57) is relatively easy. However not too long
ago these calculations were either done by hand or
by looking up the result in a book of tables. The need
to evaluate transcendental functions led to the development
of several techniques that were based on ideas from
calculus. Linear approximation, or tangent line approximation,
is an example of one such method.
Look at the
linear
approximation to the
cosine function at
shown in the graph
below
.
If we want to find approximate values for the cosine
function by using the line, we are forced to use values
only
near the point of tangency. So why use a straight line?
Instead, why not use a curve as an approximation function.
But what function should we choose for our curve?
And, what advantages do we gain by such an approximation
in the first place?
Suppose we try a simple polynomial, a quadratic.
Polynomials are desirable because they are formed only
using the operations of elementary arithmetic, addition,
subtraction and multiplication, that are easily done
by hand.
Now we have the problem of finding a quadratic
that somehow fits the cosine function. How do we select
a quadratic? First, we require that it pass throught
the "
point of tangency
" when x = 0. Any other requirements that you might
add? You might remember that every quadratic
is completely determined by the coefficients a, b
and c.
But how do we find the coefficients, a, b and c? You
may also recall that every quadratic polynomial is
determined by three points, just as a line is determined
by two points.
Website Exploration:
Every Quadratic Polynomial is Determined by Three Points
Find the equation of the quadratic that passes through
three given points.
http://www.openteach.com/javaapplets/parabola.html
While there are many possible methods for finding
a, b and c such that
we will use the requirement that the slope of the quadratic
must equal the slope of the cosine at the "point
of tangency". In fact, why not make all of the
derivatives of the quadratic equal to those of the
cosine?
Let's call this the
Method of Successive Differentiations.
The Method of Successive Differentiations
The basic idea for the method is that the polynomial
and the function are the equal at a "point of
tangency", in this case (0, 1), and also the
derivatives at this point are equal. Let's see how
this helps us to find the a, b and c values.
Step 1: Find c as follows.
P2
(0) = f(0) (
This is the point of tangency)
If
, then the equation Eq. 1) becomes
.
Since the
, we solve to get
.
Use Trace to find the intersection points of the two
graphs.
The interval is for the linear approximation is:
?
The interval for the quadratic approximation is: ?
Solution 2: A Numerical Method Using the root( ) function.
The right endpoint of the interval:
(Change the value from x = 1 to x = -1 to find the
left endpoint.)
What comparisons can you make about the linear and quadratic
approximations?
Let's try our method of succesive differentiations
using a cubic polynomial instead of a quadratic.
Find a Cubic
approximation
to 
at the point
(0, 0)
How well does our quadratic approximate the cosine function?
a) Compare the graphs of
and
.
Graph both functions in the same window.
b) Compare the numerical values of
and
.
Make a table to compare the Quadratic approximation
with the exact value of
.
Step 1) Define the range variable x over the interval
[-1, 1].
Step 2) Type x = , cos(x) = , P.2(x) = , cos(x) - P.2(x)
=
Table 1: The Numerical Comparison of cos(x) and
.
Notice the absolute errors are symmetric about the point
of tangency.
The polynomial
is
symmetric about the
y-axis
, can you tell why from looking at the Table above,
or from the equation?
Finding the Interval of Approximation with a Specified
Accuracy
As we look at the graph and table above, we can
see that our polynomial approximates the cosine function
only for a limited interval. To confidently use an
approximation, we must know that given a certain requirement
on the number of correct decimal places or equivalently
the size of acceptable error, what is the interval of approximation
. This "goodness of fit" is often described
as the interval for which the function and the approximation
have an absolute error less than some specified amount.
Find the Interval of Approximation
Find the intervals of approximation for the linear
approximation y = 1 and P2(x) for the cosine function
with an absolute error less than 0.01?
. 
Eq. 1)
?
