Question

Couple of series that I cannot figure out. Please help me... Help this poor sleep deprive girl.

1.) Find the MacLaurin series.
f(x) = 1/(3-x^2)

Answer 1

1.)f(x) = \[\frac{ 1 }{ 1-3x^2 }\]
\[ans: \sum_{k=0}^{\infty} (2)^n(x)^2n \]
\[-\sqrt(2)/2 <x<\sqrt(2)/2\]

2.) \[\lim_{x \rightarrow 0} \frac{e^x-1-x}{x^2}\]

3)\[\int\limits_{}^{} e^(x^2)\]
ans:\[\sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{(n+1)!}\]

Answer 2

I hate to do this with a youtube video, but this one is probably exactly what you want! http://www.youtube.com/watch?v=9EGDZVcl_g8

Answer 3

if that doesn't make things a lot better for you, i'll try and walk you through it, but it is bedtime!

Answer 4

Hopefully it is. I gave up that strategy long ago.. Can't sleep when midterm is in 9 hours.

Answer 5

Thanks for the video though!

Answer 6

im on spring break! but mine are all next week

Answer 7

Answered #2 yes!

Answer 8

do u know taylor series

Answer 9

mclauren is taylor when x=0

Answer 10

:) good job!

Answer 11

Yeah, ok but I feel like I get lost in the process.

Answer 12

I stay lost in all the processes of cal 2 :( i haven't even started series yet. i have just been watching the videos to prepare as much as i can.

Answer 13

This \[\int\limits_{}^{}[1- \frac{x^2}{1!} + \frac{x^4}{2!}- \frac{x^6}{3!}+....] \] is the same as answer for number 3 right?

Answer 14

Like it's basically saying the same thing but in summation

Answer 15

the integral here means something else

Answer 16

summation only is right

Answer 17

Because if not I got this using the integral \[\sum_{n=1}^{\infty} (-1)^n \frac{(x^{2n+1})}{(2n+1)(n!)}\]

Answer 18

wha did u do

Answer 19

I took the integral by using the series of e^x\[e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\]

Answer 20

ok yes

Answer 21

wasn't it e^-x^2

Answer 22

it says e^x^2 is that what u want now

Answer 23

just plugged in after which gave me \[\sum_{n=0}^{\infty} (-1)^n \frac{(x^{2n})}{n!}\]

Answer 24

took the integral of that. What I want is the series of e^-x^2 which is what I did by manipulating the e^x series.

Answer 25

ok just saying because the question doesn't have a - in there for me

Answer 26

write a couple exapansion out u will see a simple pattern

Answer 27

since x=0 it keeps simplifying

Answer 28

Like I said I get lost in the process but I'll try not to lose my way haha

Answer 29

ok well to understand mclaruen series and taylor series

Answer 30

its like approximating the next value given the slope accereleration jerk and so on

Answer 31

if u have velocity
u know that x=integral v dt = v*t
if u have acceleration
x=integral integral a dt = 1/2 a t^2.. ignore other terms
jerk
x= .... = 1/3! j * t^3

Answer 32

thats how you forming this taylor approximation

Answer 33

\[
\frac {1}{1-u}= \sum_{n=0}^\infty u^n, \quad -1< u <1\\
\frac {1}{1-3x^2}= \sum_{n=0}^\infty (3 x^2)^n, \quad 3 x^2 <1\\

\]

Answer 34

oooh!! Okay got it. I did not see that for some reason, maybe because it's 3 in the morning here. Thanks!

Answer 35

YW

Answer 36

\[
e^x = 1 + x + \frac {x^2}{2!} + \frac {x^3}{3!} + \cdots \\
e^x - 1 - x =\frac {x^2}{2!} + \frac {x^3}{3!} + \cdots \\
\frac{e^x - 1 - x}{x^2} =\frac {1}{2!} + \frac {x}{3!} + \cdots \\
\lim_{x\to 0}\frac{e^x - 1 - x}{x^2}=\lim_{x\to 0}\frac {1}{2!} + \frac {x}{3!} + \cdots=\frac 12 \\

\]

Answer 37

\[
e^u=\sum_{n=0}^\infty \frac {u^n}{n!}\\
e^{x^2}=\sum_{n=0}^\infty \frac {(x^2)^n}{n!}=\sum_{n=0}^\infty \frac {x^{2n}}{n!}\\
\int e^{x^2}dx=\sum_{n=0}^\infty \int \frac {x^{2n}}{n!}dx=\sum_{n=0}^\infty
\frac {x^{2n+1}}{(2n+1)n!}
\\

\]

Answer 38

Thanks! That was reassuring and helpful! Good night!

Answer 39

Good morning for me.

Answer 40

just do the division ...

1/3 + 1/9 x^2 + ...
-------------------------
3-x^2 ) 1
-(1 - 1/3 x^2)
--------------
1/3 x^2
-(1/3 x^2 -1/9 x^4)
------------------
1/9 x^4
............

\[\sum\frac{x^{2n}}{3^n}\]
maybe ....

Answer 41

to clean it up some ...\[\sum_0^{\infty}~\frac{x^{2n}}{3^{n+1}}\]
but thats just a thought