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- Representing functions as power series
- A list of common Maclaurin series
- Taylor Series

One of the major uses for Taylor Polynomials is using them as approximations for other, transcendental, functions. Another use is for approximating values for definite integrals, especially when the exact antiderivative of the function cannot be found. Here is a list of the three examples used here, if you wish to jump straight into one of them.

We cannot
find the value of exp(*x*) directly, except for a very few values
of
*x*. However,
we can create a table of values using Taylor polynomials as approximations:

.

.

for some *z* in [0,*x*]. If *x* is sufficiently small,
this gives a decent
error bound. For instance,

.

So, *** Error Below: it should be 6331/3840 instead of 6331/46080 ***

since exp(*x*) is an increasing function, 0 <= *z* <=
*x* <= 1/2, and

.

That tells us that *** Error Below: it should be 6331/3840 instead of 6331/46080 ***

or *** Error Below: it should be 6331/3840 instead of 6331/46080 ***

to at least three decimal places. Taking a larger-degree Taylor Polynomial
will make the approximation closer. For instance, the 10th degree polynomial
is off by at most *(e*^*z)*x*^10/10!, so for sqrt(*e*), that
makes the error less
than .5*10^-9, or good to 7decimal places.

Similarly, you can find values of trigonometric functions.
Say you wanted
to find
sin(0.1). Use a Taylor expansion of sin(*x*) with *a* close to
0.1 (say, *a*=0),
and find the 5th degree Taylor polynomial.

Since |cos(*z*)| <= 1, the remainder term can be bounded.

So, for *x*=0.1,

with an error of at most

,

or sin(0.1) = 0.09983341666... with an error of at most .139*10^-8, or good to seven decimal places.

Let's try a more complicated example. Suppose you needed to find

.

Since exp(*x*^2) doesn't have a nice antiderivative,
you
can't do the problem directly. Instead, use Taylor polynomials to find a
numerical approximation. Let's try a Taylor polynomial of degree 5 with
*a*=0:

, , , , , ,

(where *z* is between
0 and *x*)

So,

So,

with error

.

The error is

(with
*z* between 0 and
*x*)

,

so the answer .54479 is accurate to within .0006588, or at least to two
decimal places. Note that the inequality comes from the fact that
*f*^(6)(*x*) is increasing, and 0 <= *z* <= *x*
<= 1/2 for all *x* in [0,1/2].

- Return to the Power Series starting page
- Representing functions as power series
- A list of common Maclaurin series
- Taylor Series

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