Question

Interval of convergence

Answer 1

@hartnn

Answer 2

yes,i think...but not sure

Answer 3

u missed typed in wolfram!
http://www.wolframalpha.com/input/?i=series+k+%3D1+to+infiniti+%28%28-1%29%5E%28k%2B1%29%29%2Fk%29++%282x-1%29%5Ek&dataset=

Answer 4

Yeah. so do i just take the derivative of each term for b?

Answer 5

part B is correctly typed by them,as series goes till k=infinite
for f`(x),u just need 4 terms as asked,3 already given....
yup take derivative of each term,but only 4 terms as asked

Answer 6

so what i have down is correct right?

Answer 7

seems correct,all 4

Answer 8

how do i write part c?

Answer 9

ok,since u have alternate + and -1,
u will have (-1)^k
2(2x-1) will be there
1st term its power 0,2nd term its power 1 and so on....so power will be k
and summation can be written as
\[\sum_{0}^{\infty}(-1)^k2(2x-1)^k\]

Answer 10

alright thanks. how about part d?

Answer 11

put,x= 2/5 in above summation....but even i am thinking,how it can be simplified afterwards.....

Answer 12

Yeah, I am just unsure about how we would simplify it after that

Answer 13

or maybe u can put x=2/5 only in 1st 4 terms u found

Answer 14

http://www.wolframalpha.com/input/?i=series+k+%3D0+to+infiniti++%28-1%29%5Ek+2+%284%2F5-1%29%5Ek

Answer 15

hmm,so its 2.5.....

Answer 16

what u get when u put x=2/5 only in 1st 4 terms u found

Answer 17

I got it. so the the formula for geometric is a/(1-r) so you do 2/(1-1/5)

Answer 18

ok,but where does this come from? formula is correct...

Answer 19

so a is the coeffiecent and r is 1/5 because that is what is to the power of k

Answer 20

isn't it -1/5 ?

Answer 21

maybe but that would give you the wrong answer so i guess not

Answer 22

oh no! its not -1/5,i missed - sign which is outside
its 1/5,correct.....

Answer 23

i have another taylor questoin. should i post that here or a new question