LAGRANGE REMAINDER OR ERROR BOUND
Like alternating series, there is a way to tell how accurately your Taylor polynomial approximates the actual
function value: you use something called the Lagrange remainder or Lagrange error bound.
Lagrange Remainder: If you use a Taylor polynomial of degree n centered about c to approximate the value
x, then the actual function value falls within the error bound
R
n(x) = f(n+1) (z) (x – c)n+1
(n + 1)! , where z is some number between x and c.
Translation: Similar to alternating series, the error bound is given by the next term in the series, n + 1. the only
tricky part is that you evaluate f(n+1) , the (n + 1)th derivative, at z, not c. z is the number that makes f(n+1)(z)
as large as it can be. This error bound is supposed to tell you how far off you are from the real number, so we
want to assume the worst. We want the error bound to represent the largest possible error. In practice, picking
z is pretty easy.
Example 1:
Approximate cos (.1) using a fourth-degree Maclaurin polynomial, and find the associated Lagrange
remainder (error bound).
Solution:
Since the 4th degree Taylor polynomial for cos x = 1 – x2
2! + x4
4! , then
cos (.1) ≈ 1 – (.1)2
2! + (.1)4
4! ≈ .99500416667
Now, the associated Lagrange remainder after n = 4 (denoted R4(x)) is
R
4(x) = f(5)(z) (x – c)5
5! .
The fifth derivative of cos x is –sin z. Now, plug in x = .1 and c = 0 to get
R
4(.1) = (– sin z)(.1)5
5! .
We need – sin z to be as large as possible. The largest value of – sin z is 1. By assuming
– sin z is the largest possible value, we are creating the largest possible error; so, plug in 1 for
– sin z. The actual remainder will be less that this largest possible value.
R
4 (.1) < (1) (.1)5
5! = .15
5! = .0000000833
Therefore, our approximation of .99500416667 is off by less than .0000000833.
