Question

he function f(x)=ln(9−x) is represented as a power series

Answer 1

\[f(x)=\sum_{n=0}^{oo}c_{n}x^n\]

find c0, c1,c2 etc.

Answer 2

@jim_thompson5910

Answer 3

check out this page

https://www.math.ucdavis.edu/~thomases/W11_16C1_lec_3_4&3_7_11.pdf

specifically what I'm attaching as an image

Answer 4

ok thanks, the book i have is more confusing then this

Answer 5

np

Answer 6

how do i solve for c?

Answer 7

@jim_thompson5910

Answer 8

c is where you center the power series at

Answer 9

so if you center it at say x = 0, then c = 0

or if you center it at x = 1, then c = 1

Answer 10

im trying to find f'(c)(x-c)

so i get 1/9(x-0) = x/9

but what is x?

Answer 11

can I see the full problem as a screenshot?

Answer 12

on that pdf, the first term is f(c)

since the first term you have is ln(9) and f(c) = ln(9), this means

f(x) = ln(9-x)
f(c) = ln(9-c)
ln(9) = ln(9-c)
9 = 9-c
c = 0

Answer 13

so this power series is centered at x = 0

Answer 14

yes i got that far

Answer 15

but the next one c1

Answer 16

c1 is not = 0

Answer 17

c1 = f ' (c)
c1 = -(9-c)^(-1)
c1 = -(9-0)^(-1)
c1 = -1/9

Answer 18

f ' (x) = -(9-x)^(-1)

Answer 19

c2 = f '' (c)/2

c3 = f ''' (c)/6

etc etc

look at page 5 of that pdf

Answer 20

how did you get negative?

Answer 21

the derivative of ln(x) is 1/x

so the derivative of ln(9-x) is 1/(9-x) times the derivative of the inside terms. Don't forget the chain rule

Answer 22

1/(9-x) * derivative of (9-x) = (9-x)^(-1) * (-1) = -(9-x)^(-1)

Answer 23

ah ok

Answer 24

so x and c are the same

Answer 25

in most of these problems or?

Answer 26

the same? what do you mean

Answer 27

x is a variable, c is a constant

Answer 28

c=0

f(0)=f'(0)(x-0)?

Answer 29

so it looks like you're computing the first term

Answer 30

yes

Answer 31

-1/9(x)

Answer 32

ah ok so -1/9 is just the c part

Answer 33

yes the coefficient to the (x-c)^n term

Answer 34

alright i should be able to get the rest thanks alot man.

Answer 35

life saver

Answer 36

glad to be of help