chapter 9 (Power series) Flashcards
general form of a power series
sigma(k to inf) ck(x-a)^k
ck in power series
the coeffiecients
a in a power series
the offset of the center
power series
an infinite series whoos terms include powers of a variable
linear approximation of f(x) at a
L(x) = f(a) +f’(a)(x-a)
quadratic approximation of f(x) at a
Q(x) = f(a) +f’(a)(x-a) + f’‘(a)/2(x-a)^2
the nth order taylor polynomial
sigma(k to n) ck(x-a)^k
ck in a Taylor polynomial
kth derivative of f(a)/k!
remainder for an nth order polynomial with M
Rn <= M((|x-a|^n+1)
remainder formula
f^n+1(c)/(n+1)! (x-a)^n+1
differences between geometric series as a series and as a power series
basically the r in a/1-r becomes an x
so the power series is 1/1-x
power geometric series
1/1-x = sigma x^k
find a power series representation for x^4/1-x using a geometric power series
sigma(k to inf) (x^4)(x^k)
deriving or intergrating geometric power series to find power series to match ones
basically take the derivative or integral of a power series if the derivative or integral gets you closer to what you need to approximate it to
when you derive or integrate the non sigma side, what must be done to the sigma side
the same exact process
interval of convergence
this is the interval that the power series is still converghent uppon
how to find the interval of convergent
use the root or ratio test.
root test
p= lim of kthsqrt(ak)
ratio test
l= lim of a(k+1)/ak
if the root or ratio limit is 0
the radius is inf
the interval is also inf
if the root or ratio limit is inf
the radius is 0
the interval does not exist. diverges everywhere
if the root or ratio limit is anything else
put it in absolute values and solve for x. then find the radius and the 2 potential endpoints
what to do with two endpoints for the interval of convergence
test each with different series tests. if they converge then the endpoint is inclusive. if they diverge the endpoint is noninclusive
makalaureans series
a taylor series that has an a of 0
finding the taylor series for a function
make the collums
- the kth derivative
- the kth derivative evaluated at a
- f^k(a)/K!
- that multiplied by (x-a)^k
find the damn pattern
how to get a taylor series to alternate values and skip odd values used for cosine
x must be raised to the 2k
it must be 2k!
and k must start at 0
how to get a taylor series to alternate values and skip even values
used for sin
x must be raised to the 2k+1
it must be (2k+1)!
and k must start at 0
taylor series for 1/1-x
sigma (k=0 to inf) x^k
taylor series for e^x
sigma (k=0 to inf) x^k/k!
taylor series for ln(1+x)
sigma (k=1 to inf) ((-1^k+1)x^k/k for -1
taylor series for sinx
sigma (k=0 to inf) ((-1^k)(x^2k+1)/(2k+1)!
taylor series for cosx
sigma (k=0 to inf) ((-1^k)(x^2k)/(2k)!
approximate
use the nth order taylor polynomial.
plug in a value close to what you are approxximating
calc
estimate the remainder
find what f^n+1 is bound by
find M then plug in with the given n to the < one
