Question

let \[F(r) = \int\limits_{0}^{x}e ^{-t^2}dt\]
a) find Maclorian Series (i.e Taylor Series about 0) of F(r). (You may use Maclorian Series of well-known functions)

b) approximate F(1/2) with error less then 0.00001

Answer 1

ahh finally someone :D

Answer 2

i got something but i have go afk for like 20-30 mins ill say what i found :D when i back

Answer 3

So it goes 1-t^2 +t^4/2 -t^6/6 etc

1,3/4,13/16,51/64 so

maybe u can use Taylor's Theorem/Remainder to calculate error bounds until u are within your limit...

Answer 4

solve my question mybe ...

Answer 5

I don't do numbers..:-)

Answer 6

well :D let me write what i got

Answer 7

\[e^t = \sum_{n=0}^{\infty}(t^n / n!) \nu t \epsilon iR\]

Answer 8

What's that?

Answer 9

and hence; \[e ^{-t^2} = \sum_{n=0}^{\infty}((-ı^n)t ^{2n})/n!\]

Answer 10

\[\int\limits_{0}^{x}e^{-t^2}dt = \sum_{n=0}^{infity}((-1)^n x^{2n+1}) / ((2n+2)n!) = F(x)\]

Answer 11

I already did that above (and F(1/2))

Answer 12

\[\nu x \epsilon ı\mathbb{R}\]

Answer 13

What's that say?

Answer 14

thats what i got :D

Answer 15

dunno :P just calculated what i see there o.O

Answer 16

So it goes 1-t^2 +t^4/2 -t^6/6 etc Is the expansion

1,3/4,13/16,51/64 Is F(1/2)

Now u have to test error.

Answer 17

not calculate we can say order

Answer 18

yeap, but i dunno how to test error -.-'

Answer 19

It's Taylor's Theorem..if remainder = f(x) -p(x) then

remainder = f^(n+1)*c/(n+1)! * x^(n+1) and u have to calculate function and each partial to test for your limit...

Answer 20

http://en.wikipedia.org/wiki/Taylor%27s_Theorem

Answer 21

o.O
hmm

Answer 22

Don't look at me:-)

Answer 23

lol :) really good stuff

Answer 24

You are calculating an "error" against a function which is only an approximation to begin with, lol.

Answer 25

nono i mean the theorem ^.^

Answer 26

\[f(1/2) \approx (1/2) - (1/24) + (1/320) - (1/5376)\]

Answer 27

?? is this ture? :D

Answer 28

Here

http://www.wolframalpha.com/input/?i=series+e^-x^2

Answer 29

hmm