によって Michael Collins 12年前.
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x^n
(-1)^nx^(2n)/(2n)!
(-1)^nx^(2n+1)/(2n+1)!
x^n/n!
put whatever into the known power series
null
do it like you do a definite integral
integrate each of the terms you get from the series
Divide
use long division
bottom into top
Multiply
multiply the 2 series term by term
gives you the series
f(x)=(cosx)^.5
just substitute the x^.5 into the known series
f(x)=(1+x)^k
(k(k-1)...(k-n+1)x^n)/n!
WRITE OUT
f^n(c)/n!*(x-c)^n
f(x)=lnx
use f(x)=1/x
or, more easily, integrate the series itself, using rules from 8.8
solve for C by letting x=center
integrate the terms of 1/x and you'll get lnx + c
use partial fractions to get 2 power series
make sure signs are correct, could be + or -
pull out common term
if adding 2 together
IOC = intersection of the two
if (-2,2) and (-1,1), IOC=(-1,1)
if 1/x, 1/(1-(-x+1)
cannot change the original form
must get it to this form, must have 1 on bottom and minus the x
a/(1-r)
goes to: Ear^n
integral
put to n+1 and divide by n+1
only integrate one with x in term
add constant, C
f'(x)
bring down n, put to n-1
only take derivative on n power with x in term
Radius
limit infinity, R=0
limit is 0, all reals: R=infinity
(-2,2): R=2
IOC
test endpoints for ( or [
use ratio test
Remember to not use (-1)^n
if limit = infinity converges at center
if left with |x-c| set < 1
if limit = 0 then converge for all reals
sin(.1)
subtract the 2
get approximation
get real value
R(x)=[f^(n+1)(z)/(n+1)!](x-c)^(n+1)
error = |R(x)|
f(x)=P(x)+R(x)
use series ln(1+x)
so plugin .1 for the series
x=.1
center at 0
f(c)+f'(c)(x-c) +...+f^(n)(c)/n!*(x-c)^n
center at some number, c
