Example 2.4 (Taylor Series for log). Let f be the function f : x + logx. Then we know f'(x) x-1 f" (x) f (x) = = = (x-1)- = ⠀ f(n) (x) Let us find the Taylor approximations at x = 1. f(1) 0 ƒ' (1) 1 f" (1) = -1 f"" (1) = 2 = 2x-3 = = ⠀ -x-² (-1)-1(n-1)!x". f(n) (1) The nth Taylor formula at 1 is therefore (x-1)² log.x=0+ (x- -1)- 2 = (-1)"-¹ (n-1)!. (x-1)² 2 where t is a number between 1 and x. How to get it? +(−1)n-2 (x − 1)n−1 n-1 + (−1)n-1-n (x − 1)″ n +... +... + (−1)n-2 + (−1)n- Taylor's theorem: fibs = feas + f'cas(b-a) + f(a) (b-a) +...+ {n-19 (b-a)^²+ (n-1)! 3tela. b) a approximated value. 6: exact value. Approximation at 0 f(0) + f²₁o³x + £121² x ² + + +n! X".. Tip: Lagrange remainder is used to measure bound. error -1)"- n-1 n ¹1/(*7¹1) " n t r₁(x) (it) -Ent (6-a)² Lagrange remainder n! a=o, 1 b=x
Example 2.4 (Taylor Series for log). Let f be the function f : x + logx. Then we know f'(x) x-1 f" (x) f (x) = = = (x-1)- = ⠀ f(n) (x) Let us find the Taylor approximations at x = 1. f(1) 0 ƒ' (1) 1 f" (1) = -1 f"" (1) = 2 = 2x-3 = = ⠀ -x-² (-1)-1(n-1)!x". f(n) (1) The nth Taylor formula at 1 is therefore (x-1)² log.x=0+ (x- -1)- 2 = (-1)"-¹ (n-1)!. (x-1)² 2 where t is a number between 1 and x. How to get it? +(−1)n-2 (x − 1)n−1 n-1 + (−1)n-1-n (x − 1)″ n +... +... + (−1)n-2 + (−1)n- Taylor's theorem: fibs = feas + f'cas(b-a) + f(a) (b-a) +...+ {n-19 (b-a)^²+ (n-1)! 3tela. b) a approximated value. 6: exact value. Approximation at 0 f(0) + f²₁o³x + £121² x ² + + +n! X".. Tip: Lagrange remainder is used to measure bound. error -1)"- n-1 n ¹1/(*7¹1) " n t r₁(x) (it) -Ent (6-a)² Lagrange remainder n! a=o, 1 b=x
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Could u explain step by step that how to derive the taylor expansion of ln(x)? Thx!
PS: in this case, log(x) is the same ln(x).
:)
Transcribed Image Text:Example 2.4 (Taylor Series for log). Let f be the function f : x + logx. Then
we know
f'(x)
f" (x)
f"" (x)
=
log.x=0+ (x- -1)-
3tela.b). a
=
= (x-1)-
=
⠀
f(n) (x)
Let us find the Taylor approximations at x = 1.
f(1)
0
ƒ' (1)
1
f" (1)
= -1
f"" (1)
= 2
=
=
x-1
-x-²
2x-3
⠀
=
f(n) (1)
The nth Taylor formula at 1 is therefore
(x - 1)²
2
(-1)-1(n-1)!x".
=
where t is a number between 1 and x.
(-1)"-¹ (n-1)!.
(x-1)²
2
How to get it?
+ (−1)n-2 (x−1)n-1
n-1
+...
+ (−1)n-17-n (x−1)n
n
·+····
+ (−1)n-2
+ (−1)"-
n
-1²-¹²/ (+¹) "
n
t
Taylor's theorem: fibs = feas + f'cas(b-a) + f(a) (b-a) +...+ {n-19 (b-a)^²+
: approximated value. b: exact value.
Approximation at 01. f(0) + f²₁o³x + £121² x ² + + + ² x..
Tip: Lagrange remainder is used to measure
bound.
error
-1)n-1
n-1
r₁(x)
(it)
-Ent (6-a)² Lagrange remainder
n!
a=o,
b=x
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