Transcript Taylor Series

Warm Up
1)
2)
Taylor Series
AP Questions & Error Revisited
Alternating Series Remainder
If an alternating series converges then the
absolute value of the remainder Rn involved in
approximating the sum S by Sn is less than (or
equal to) the first neglected term.
That is, |S – Sn| = |Rn| < an+1
The difference
between the “real”
sum and the
approximate sum.
the next term
in the series
x
1 3 1 5 1 7
Given:  e dt  x  x  x  x  ...
3
10
42
0
t 2
Use the first two terms of the series to estimate the value of
1
2
e
t 2
dt
0
Explain why the estimate differs from the actual value by less
than 1/200.
Since the series is alternating
|R2(1/2)| < |a3(1/2)| =
  
1
10
1 5
2
1
1
 320
 200
1 4 1 6 1 8
Given: ln( x  1)  x  x  x  x  ...
2
3
4
2
2
Determine a rational number A such that
 4   1100
A  ln 5
Since the series is alternating, |A – ln(5/4)| = Rn(1/2) < An+1
The first term of the series to be less than 1/100 is the third term
1
1 6
1
3
2
192
  

So A is the sum of the first two terms: 7/32
2004 form B
Let f be a function having derivatives of all orders for all real
numbers. The third degree Taylor polynomial for f about x = 2 is
given by
T ( x)  7  9( x  2) 2  3( x  2)3
a) Find f(2) and f”(2).
f(2) = 7 and f”(2) = -18
b) Is there enough information given to determine whether f has a
critical point at x = 2? If not, explain why not. If so, determine
whether f(2) is a relative maximum, a relative minimum, or
neither, and justify your answer.
Let f be a function having derivatives of all orders for all real
numbers. The third degree Taylor polynomial for f about x = 2 is
given by T ( x)  7  9( x  2) 2  3( x  2)3
c) Use T(x) to find an approximation for f(0). Is there enough
information given to determine whether f has a critical point at
x = 0? If not, explain why not. If so, determine whether f(0) is a
relative maximum, a relative minimum, or neither, and justify your
answer.
d) The fourth derivative of f satisfies the inequality |f 4(x)| <6 for all x
in the interval [0,2]. Use the Lagrange error bound on the
approximation to f(0) found in part c to explain why f(0) is
negative.