Question

http://i1106.photobucket.com/albums/h362/sona441/Screenshot2014-04-07031223_zpsf598c20b.png

Answer 1

??

Answer 2

is it not readable?

Answer 3

http://i1106.photobucket.com/albums/h362/sona441/Screenshot2014-04-07032428_zpsc352624a.png

Answer 4

The limit could be a lot of things. You need to tell us what x is approaching.

I'll assume you mean
lim (as x goes to 0) ((cosx -1)/x^2)

Multiply the top and bottom by cosx +1
(cosx -1)/x^2
= (cosx -1)(cosx +1) / [x^2 (cosx +1)]
= ((cosx)^2 -1) / [x^2 (cosx +1)]
= -(1 -(cosx)^2) / [x^2 (cosx +1)]
= -(sinx)^2 / [x^2 (cosx +1)]
= -[(sinx)/x]^2 * 1/(cosx +1)
= -1(1)^2 *1/(1+1) as x goes to 0
= -1/2

Answer 5

@jinniekim12 you there

Answer 6

sorry

Answer 7

????

Answer 8

are you responding to part a?
how would i find the general term of the taylor

Answer 9

I just explained it you try to understand it

Answer 10

i can find the nonzero terms of the taylor expansion, but im having trouble finding the general term of the taylor for (cosx-1)/x^2

Answer 11

first do you know taylar series

Answer 12

mhm and i know the first three will be -1/2 + x^2/24 - x^4/720

Answer 13

TRY TO UNDERSTAND WHAT I HAVE DONE HERE

I'll assume you mean lim (as x goes to 0) ((cosx -1)/x^2) Multiply the top and bottom by cosx +1 (cosx -1)/x^2 = (cosx -1)(cosx +1) / [x^2 (cosx +1)] = ((cosx)^2 -1) / [x^2 (cosx +1)] = -(1 -(cosx)^2) / [x^2 (cosx +1)] = -(sinx)^2 / [x^2 (cosx +1)] = -[(sinx)/x]^2 * 1/(cosx +1) = -1(1)^2 *1/(1+1) as x goes to 0 = -1/2

Answer 14

did you not just find the first term of the taylor expansion where if n=0 of the sum it will be -1/2
i need \[\sum_{n=0}^{\infty} \frac{ (-1)^n (x ^{2n}) }{ (2n)! }\] something like this

Answer 15

did you understood it first

Answer 16

kind of, I can barely read it but yes, i see how -1/2 is correct

Answer 17

understand it first and then ask

Answer 18

i want to know how i can get from cosx = \[\sum_{n=0}^{\infty} (-1)^n x^(2n) / (2n)!\] to (cosx-1)/x^2

Answer 19

∑ got this

Answer 20

yes

Answer 21

what goes next to it though?

Answer 22

you know x right i mean uou found it

Answer 23

yes

Answer 24

so just try to concentrate and solve my teachwer says eat some suagr before doing msath helped me it can help you

Answer 25

it's 4 am, i don't need sugar, I need this done,
is there any way to incorporate the -1 and /x^2 to the general term....you put up a message earlier but deleted it, i wanted that

Answer 26

do u hv a exam or this ur homework

Answer 27

it's homework but like a test: it's absolutely critical that i get this question correct

Answer 28

when is your school what time

Answer 29

today school starts at 9, i need to leave by 8:30 am and its 4 am and i haven't gotten any sleep yet, also i'm in 11th grade of highschool

Answer 30

I HIGHLY RECOMMED TO SLEEP ATLEAST FOR 2 HOURS EATING SOMETHING LIKE FRENCH OMELLET AND THEM DO IT WILL HELP A LOT

Answer 31

i will, after i finish this problem [; not the eating part though, I'll probably just throw it up

Answer 32

ARE U DOING ENGNEERING

Answer 33

I'M IN HIGH SCHOOL engineering - college
and i probably won't in college

Answer 34

so, the problem [;

Answer 35

no did to get mood off just ask ur teacher he/she may help you

Answer 36

it's due tomorrow, and he can't help us, it's like a test

Answer 37

\[\sum_{n=1}^{\infty} \frac{ (-1)^n x ^{2n-2} }{ (2n)! }\] = \[\frac{ \cos(x)-1 }{ x^2 }\] ?

Answer 38

you know cos??????????????
and x ??????????????????????
and x square????????????????????????

Answer 39

what do you mean know? can you tell me if the statement i just typed up there is true

Answer 40

first find the value of x

Answer 41

i don't need to find the value of x

Answer 42

okay so first find x square and root it to ge x

Answer 43

I don't need any of that

Answer 44

x is every number

Answer 45

k just relax and try to get motivated to do the sum

Answer 46

I'm sorry, but I think you've been wasting my time. Bye now.

Answer 47

no not wasting your time the thing is you are not understanding it jusy ask you teacher to reexplain you taylars serires

Answer 48

@mathmale

Answer 49

talk to him