Taylor’s Polynomials & LaGrange Error Review

Maclaurin Series: (generated by f at ) If we want to center the series (and it’s graph) at some point other than zero, we get the Taylor Series: Taylor Series: (generated by f at )

The more terms we add, the better our approximation. Hint: On the TI-89, the factorial symbol is:

example: Rather than start from scratch, we can use the function that we already know:

When referring to Taylor polynomials, we can talk about number of terms, order or degree. This is a polynomial in 3 terms. It is a 4th order Taylor polynomial, because it was found using the 4th derivative. It is also a 4th degree polynomial, because x is raised to the 4th power. The 3rd order polynomial for is , but it is degree 2. The x3 term drops out when using the third derivative. A recent AP exam required the student to know the difference between order and degree. This is also the 2nd order polynomial.

There are some Maclaurin series that occur often enough that they should be memorized. They are on your formula sheet, but today we are going to look at where they come from. Maclaurin Series: (generated by f at )

List the function and its derivatives.

a = 1 and r = x. List the function and its derivatives. Evaluate column one for x = 0. This is a geometric series with a = 1 and r = x.

We could generate this same series for with polynomial long division:

This is a geometric series with a = 1 and r = -x.

We wouldn’t expect to use the previous two series to evaluate the functions, since we can evaluate the functions directly. They do help to explain where the formula for the sum of an infinite geometric comes from. We will find other uses for these series, as well. A more impressive use of Taylor series is to evaluate transcendental functions.

Both sides are even functions. Cos (0) = 1 for both sides.

Both sides are odd functions. Sin (0) = 0 for both sides.

If we start with this function: and substitute for , we get: This is a geometric series with a = 1 and r = -x2. If we integrate both sides: This looks the same as the series for sin (x), but without the factorials.

p

Finding Truncation Error in a Taylor Polynomial Graph the function y1 = ln (1 + x) and it’s corresponding Taylor Polynomial

Finding Truncation Error in a Taylor Polynomial Graph the function y1 = ln (1 + x) and it’s corresponding Taylor Polynomial

Finding Truncation Error in a Taylor Polynomial To find the error between the functions, graph y3 = abs (y2 – y1).

Finding Truncation Error in a Taylor Polynomial To find the error between the functions, graph y3 = abs (y2 – y1). Where is the error the smallest?

Use Table to see where truncation error is least on (-1,1)

Finding Error in a Taylor Polynomial Graph the function y1 = sin x and it’s corresponding Taylor Polynomial and find the interval in which the Taylor polynomial is accurate to the thousandths place.

Finding Error in a Taylor Polynomial The third order Taylor Polynomial for y = sin x is accurate to the thousandths place on the interval (-.65, .65). Graph y3 = abs (y1(x) – y2(x)), and y4 = .001.

Taylor series are used to estimate the value of functions (at least theoretically - now days we can usually use the calculator or computer to calculate directly.) An estimate is only useful if we have an idea of how accurate the estimate is. When we use part of a Taylor series to estimate the value of a function, the end of the series that we do not use is called the remainder. If we know the size of the remainder, then we know how close our estimate is.

For a geometric series, this is easy: ex. 2: Use to approximate over . Since the truncated part of the series is: , the truncation error is , which is . When you “truncate” a number, you drop off the end. Of course this is also trivial, because we have a formula that allows us to calculate the sum of a geometric series directly.

Taylor’s Theorem with Remainder If f has derivatives of all orders in an open interval I containing a, then for each positive integer n and for each x in I: Lagrange Form of the Remainder Remainder after partial sum Sn where c is between a and x.

Lagrange Form of the Remainder Remainder after partial sum Sn where c is between a and x. This is also called the remainder of order n or the error term. Note that this looks just like the next term in the series, but “a” has been replaced by the number “c” in . This seems kind of vague, since we don’t know the value of c, but we can sometimes find a maximum value for .

Lagrange Form of the Remainder If M is the maximum value of on the interval between a and x, then: Taylor’s Inequality Note that this is not the formula that is in our book. It is from another textbook. This is called Taylor’s Inequality.

so the series converges. Prove that , which is the Taylor series for sin x, converges for all real x. ex. 2: Since the maximum value of sin x or any of it’s derivatives is 1, for all real x, M = 1. Taylor’s Inequality so the series converges.

Find the Lagrange Error Bound when is used to approximate and . ex. 5: On the interval , decreases, so its maximum value occurs at the left end-point. Remainder after 2nd order term

Find the Lagrange Error Bound when is used to approximate and . ex. 5: Taylor’s Inequality On the interval , decreases, so its maximum value occurs at the left end-point. error Error is less than error bound. Lagrange Error Bound

Example using Taylor’s Theorem with Remainder For approximately what values of x can you replace sin x by with an error magnitude no greater than 1 x 10-3 ?

Taylor’s Theorem with Remainder Taylor’s Theorem with Remainder works well with the functions y = sin x and y = cos x because |f (n+1)(c)| ≤ 1. Note: Factorial growth in the denominator is larger than the exponential growth in the numerator. Since | Rn(x)|  0 then the Taylor Series converges as n →