Question

f(x) = 1/(1-x^2) through x^8 term is = X^8+x^6+x^4+x^2+1, use partial expansion to in power of x to come up with the value of derivative f'(0),f''(0),f'''(0)...f''''''''(0)

Answer 1

what do you mean by "through x^8 term" ?
f(x) = 1/(1-x^2) correct ?

Answer 2

in 8 times power

Answer 3

grind it til you find it
\[f'(x)=8x^7+6x^5+4x^3+2x\implies f'(0)=0\]

Answer 4

\[f''(0)=2\] etc

Answer 5

so f'''(0)=0 and f''''(0)=4 ?

Answer 6

\(f'''(0)=0\) but i don't think \(f^{(4)}(0)=4\)

Answer 7

why? i though in power of 4 times, the number is 4?

Answer 8

i think it is 24 but it is late and i am doing it in my head
try and see what you get

Answer 9

how did you get the 24? how did you calculate

Answer 10

it is from the term that started out at \(x^4\)
when you take successive derivatives you get \(4x^3, 12x^2, 24x, 24\) aka 4!

Answer 11

so \(f^{(4)}(0)=4!\) which i think is where this is headed
and of course \(f^{(5)}(0)=0\)

Answer 12

want to guess at \(f^{(6)}(0)\) ?

Answer 13

is 6!?

Answer 14

you found the sound

Answer 15

it is only apply for this formula or it's universal?

Answer 16

of course you can check by differentiating, but i am sure that is what you will get

Answer 17

well you have coefficients of 1 for each term, so that makes it easier
also this is the derivative at zero, so we have to be careful when generalizing
if the expansion was different, you would get different answers, but there would certainly be an \(n!\) involved, as you can see

Answer 18

thank you!

Answer 19

you are being set up to learn about taylor polynomials, where the coefficients look like
\[\frac{f^{(n)}(0)}{n!}\]
yw

Answer 20

if a is not 0, just an "a", where we should put (x-a), after n! or other place?