8.7. Taylor and Maclaurin Series http://www.ck12.org
Then
f 1 ′(x) =
( 2
x^3)·e−x (^12) x 6 = 0
0 x= 0
It can be verified thatf 1 ′( 0 ) = 0 ,f 1 ′′( 0 ) = 0 ,f 1 ′′′( 0 ) = 0 ,....
So the Maclaurin series is 0, clearly difierent fromfexcept atx=0.
Nevertheless, here is a positive result.
TheoremIffhas a power series representation atx=x 0 , i.e.
f(x) =∑∞n= 0 an(x−x 0 )nfor|x−x 0 |<Rc, then the coefficients are given byan=f(n)n(!x^0 ).
So any power series representation atx=x 0 has the form:
f(x) =
∞
n∑= 0 f
(n)(x 0 )
n! (x−x^0 )
n
Exercises
- Find the higher order derivatives of the functionf 1 (x)above thus recursively showingf 1 (n)( 0 ) =0 forn≥ 0
- Verify the Theorem using term-by-term differentiation.
Taylors Formula with Remainder, Remainder Estimation, Truncation Error
Recall the remainderRn(x)of thenth−degree Taylor polynomial atx=x 0 is given byRn(x) =f(x)−Tn(x).
Theorem(Convergence of Taylor series)
If limn→∞Rn(x) =0 for|x−x 0 |<Rc, thenfis equal to its Taylor Series on the interval|x−x 0 |<Rc.
The above condition limn→∞Rn(x) =0 could be achieved through the following bound.
Theorem(Remainder Estimation)
If|f(n+^1 )(x)|≤Mfor|x−x 0 |≤r, then we have the following bound forRn(x):
|Rn(x)|≤(n+M 1 )!|x−x 0 |n+^1 for|x−x 0 |≤r.
Example 1The functionexis equal to its Maclaurin series for allx. Proof. Letf(x) =ex. We need to find the above
bound onRn(x).
If|x|≤r,f(n)(x) =ex≤erforn≥0 and the remainder estimation gives|Rn(x)|≤(n+er 1 )!|x|n+^1 for|x|≤r.
Since limn→∞(n+er 1 )!|x|n+^1 =erlimn→∞(|nx|+n 1 +)^1 != 0 ,limn→∞|Rn(x)|=0 by the squeeze Theorem. So limn→∞Rn(x) =0.
Henceexis equal to its Maclaurin series∑∞n= 0 xnn!for allx.
Example 2(Truncation Error) What is the truncation error of approximatingf(x) =√ 1 +xby its third-degree
Maclaurin polynomial in for|x|≤ 0 .1.
Solution.