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Lagrange Form of the Remainder (also called Lagrange Error Bound or Taylor's Theorem ... introduced his formula for generating polynomial terms in 1712!
Typology: Exercises
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Lagrange Form of the Remainder (also called Lagrange Error Bound or Taylor’s Theorem Remainder)
Suppose we didn’t have a calculator, but we were interested in the value of sin1. How could we approximate its value?
Eyeball it on the graph?
Compare it to
sin 3 2
Tangent line approximation? Euler’s method? Taylor polynomial of degree 1?
Suppose we choose the third or fifth option, after all, polynomials are relatively easy to evaluate. Is there a way, depending on the degree of the polynomial we decide to use, to determine how accurate our approximation is? That is, do we know what remains from our approximation to get the actual value?
YES! Yes, we do... kind of.
When a Taylor polynomial, Tn (^) x , centered at x c is used to approximate a function, f (^) x , at a value
x a near the center, we can express our result as follows:
If: Function = Polynomial Approximation + Remainder,
Then:
Remainder = Function – Polynomial Approximation
Written mathematically, if:
f a ( ) Tn ( ) a Rn ( ) a
Then: Rn ( ) a f a ( ) Pn ( ) a
We define the remainder, Rn ( ) a , to be:
1 1
n n n n
f z R a T a f a a c n
Taylor’s Theorem
Historically, the remainder was not due to Taylor but to a French mathematician, Joseph Louis Lagrange (1736-1813), the MVT guy.
Historical Note: Up to the time of Mr. Lagrange, there seemed to have been very little interest in determining the error between exact and approximate values, but rather, there was a shared enthusiasm of “good feeling” for the rapidity of approximations and the goodness of the values they were finding.
Langrange wanted to quantify this “good feeling,” and first published his estimation of the remainder for the Taylor series in his Théorie des fonctions analytiques in 1797. Remember that Brook Taylor introduced his formula for generating polynomial terms in 1712!
Langrange’s efforts, thus ended the 85 year Era of “Good Feelings,” (not to be confused with the period in American History that coincided with Monroe’s presidency.)
For this reason, Rn ( ) a is called the Lagrange form of the remainder.
When applying Taylor’s Formula, we would not expect to be able to find the exact value of z. Rather, we
are merely interested in a safe upper bound (maximum value) for ^ 1 ( ) n f z from which we will be able to
tell how large the remainder Rn ( ) a is.
If a function f is differentiable through order n 1 in an interval containing the center x c , then for each x a in the interval, there exists a number x z between a and c such that
n n n
f c f c f a f c f c a c a c a c R a n
where the remainder (^) Rn ( ) a is given by
1 1
n n n
f z R a a c n
, called the Lagrange Remainder
(or Lagrange Error Bound).
a hmm, hmm, hmm, hmm.
Example 2: (Calculator Permitted) Let’s return to our original quandary—approximating sin1with an idea of how accurate our approximation would be.
(a) Find the fifth-degree Maclaurin polynomial for sin x. Then use your polynomial to approximate sin1, and use Taylor’s Theorem to find the maximum error for your approximation. Give five decimal places.
(b) Use your answer from (a) to find an interval (^) a b , such that a sin1 b.
(c) Could sin1equal 0.87? Why or why not?
Example 3: (No Calculator)
(a) Write the fourth-degree Maclaurin polynomial for f ( ) x ex. Then use your polynomial to
approximate e , and find a Lagrange error bound for the maximum error when x 0.5.
(b) Use your answer from (a) to find an interval (^) a b , such that a e b.
Example 4: (Calculator Permitted) The function f has derivatives of all orders for all real numbers x. Assume that f (2) 6 , f (2)^ 4 , f (2)^ 7 , f (2)^ 8.
(a) Write the third-degree Taylor polynomial for f about x 2 , and use it to approximate f (2.3). Give three decimal places.
(b) The fourth derivative of f satisfies the inequality f ^4 ^ ( ) x 9 for all x in the closed interval (^) 2, 2.3.
Use the Lagrange error bound on the approximation of f (2.3)found in part (a) to find an interval a b , ^ such that^ a^ ^ f^ (2.3) b. Give three decimal places.
(c) Based on the information above, could f (2.3)equal 6.992? Explain why or why not.