Example:
Approximate ln (1 + x) by a polynomial centered
around x = 0 (c = 0):
P(0) = ln(1 + x) at x = 0 P(x) = ln (1 + x) so P(0) = 0
P '(0) = d (ln(1 + x)) at x =
0
P'(x) = 1/(1 + x) so P'(0) = 1
dx
P ''(0) = d2 (ln(1 + x))
at x = 0
P''(x) = -1/(1 + x)2 so P''(0) = -1
dx2
P'''(0)= d3 (ln(1
+ x)) at x = 0
P'''(x) = 2/(1+x)3 so P'''(0) = 2
dx3
P(4)(0)= d4
(ln(1 + x)) at x = 0
P(4)(x) = -6/(1 + x)4 so P(4)(0)
= -6
dx4
P(5)(0)= d5 (ln(1
+ x)) at x = 0
P(5)(x) = 24/(1 + x)5 so P(4)(0)
= 24
dx5
Taylor Series: x - x2/2! + 2x3/3! - 6x4/4! + 24x5/5! + ... + (-1)n-1xn/n! + ... , for n > 1 (The first term is 0 and is not included) Working out the factorials and reducing the fractions produces
x - x2/2 + x3/3 - x4/4 + x5/5
+ ... + (-1)n-1xn/n! + ... , for
n > 1
Common MacLaurin series that should be memorized.
¥
1 =
1 + x + x2 + ... + xn + ... = å
xn , for |x| < 1
1 - x
n=0
¥
1 =
1 - x + x2 + ... + (-x)n + ... =
å (-1)n xn
, for |x| < 1
1 + x
n=0
¥
ex = 1 + x + x2/2!
+ ... + xn/n! + ... = å
xn/n! , for all real x
n=0
¥
sin x = x - x3/3! + x5/5!
-... + (-1)n x2n+1/(2n+1)! + ... =
å (-1)n x2n+1/(2n+1)!
, for all real x
n=0
¥
cos x = 1 - x2/2! + x4/4!
-... + (-1)n x2n/(2n)! + ... =
å (-1)n x2n/(2n)!
, for all real x
n=0
¥
ln (1 + x) = x - x2/2
+ x3/3 - ... + (-1)n-1xn/n + ...
= å (-1)n-1xn/n
, for -1 < x < 1
n=1
¥
arctan x = x - x3/3 + x5/5 -... +
(-1)n x2n+1/(2n+1) + ... = å
(-1)n x2n+1/(2n+1) , for
|x| < 1
n=0
On their intervals of convergences, Taylor series can be added, subtracted, and multiplied by constants and powers of x, and the resulting series will also be Taylor series.
To find a series for y = cos 2x, simply substitute 2x in for each x in the Taylor series for cos x.
A third order polynomial approximating a function means a polynomial
of degree 3. Notice, this is not always the same as
writing three terms of a series approximating a function.
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