Question

find the maclaurin series

Answer 1

know of generalized binomial theorem?

Answer 2

actually i was given this formula
\[\sum_{n=0}^{\infty} \frac{ f ^{(n)}(0) }{ n! }(x)^n= f(0)+f'(0)+...+\frac{ f ^{(n)} (0)}{ n! }(x)^n\]

Answer 3

Looks like you should start taking derivatives then, don't you think?

Answer 4

http://mathworld.wolfram.com/MaclaurinSeries.html

Answer 5

Yep the exact same formula

Answer 6

well, i know how to plug in values but idk how to form to the nth term

Answer 7

Just differentiate it for a few times and you should be able to see the pattern

Answer 8

Start writing down terms but whenever you multiply it's generally helpful to keep them factored.

So you should write 24 as really 2*3*4.

Answer 9

use it then ...

Answer 10

since you have a factor (1-x) ... the pattern shouldn't be that difficult.

Answer 11

f'(x)= 2/(1-x)^3, f''(x)=6/(1-x)^4

Answer 12

set x=0

Answer 13

okay hold on

Answer 14

Yep. Don't forget your negative exponent though.

f'''=2*3*4(1-x)^-5, f''''=2*3*4*5(1-x)^-6

Answer 15

ah, i think i got the pattern from the above form.. hmm so i just need to plug tht in to the formula?

Answer 16

\[f ^{(n)}(x)=\frac{ 1\times2\times3\times4...n }{ (1-x)^{n+2} }\] ???

Answer 17

how do i simplify the numerator?

Answer 18

have you done binomial series yet?

Answer 19

i certainly did when i was in highschool, but now im in college doing calculus 2, so we were only provided mclauren and taylor series formula

Answer 20

oh ok. then just use Talyor series

Answer 21

hmm, you mean transform it into factorial?

Answer 22

yeah

Answer 23

isit (n+1)!?

Answer 24

wait what?

Answer 25

Just stick with Mclaurin Series

Answer 26

http://www.wolframalpha.com/input/?i=D%5B1%2F%28x-1%29%5E2%2C+%7Bx%2C+n%7D%5D

Answer 27

\[\sum_{n=0}^{\infty} 1 +2x + \frac{ 6 }{ 2! }x^2+...\]

Answer 28

http://www.wolframalpha.com/input/?i=Expand+1%2F%281-x%29%5E2+at+x%3D0

Answer 29

looks like you are doing okay.

Answer 30

hmm.. but i need tofind the nth term, i'm not exactly sure but let me try

Answer 31

n-term ... isn't it obvious, you will get (n+1)! from f^n(0) ... and you have n! down there.

Answer 32

there are tons of other ways, best you use geometric series expansion.
\[ \frac{1}{1-x} = 1 + x + x^2 + \dots \]
just differentiate both sides.

Answer 33

\[\sum_{n=0}^{\infty} \frac{ x ^{(n+1)} }{ (n+1)! }\] is this correct? haha

Answer 34

no ...
\[ \sum_{n=0}^\infty \frac{(n+1)!}{n!}x^n\]

Answer 35

hahah, oh ya that makes sense! hahah, i got it from earlier but i was abit confused.. anyway to simplify that it would be\[\sum_{n=0}^{\infty} (n+1)x^n\] ????

Answer 36

yes ... just take derivative on both sides

\[ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \dots \]

Answer 37

*differentiate on both sides.

Answer 38

why do we need that? arent we done ? haha

Answer 39

oh ya need to find the radius of convergence

Answer 40

i think i've got it already, r: 1, since using the ratio test we get |x| so -1<x<1.

Answer 41

thank you! :D

Answer 42

okay ... 1 works