Question

McLaurin series problem?

Answer 1

Write a second degree polynomial equation of McLaurin for the function \[f(x)=e ^{-x}(cosx-senx)\]and determine if in the range I=(0,1/2) the McLaurin polynomial equation rounds up or down and define the error.

Answer 2

*define the error when you substitute f(x) in the final polynomial equation in the given range.

Answer 3

let f = ab
f' = a'b + ab'
f'' = a''b + a'b' + a'b' + ab''
= a''b + 2a'b' + ab''

a = e^-x
a' = -e^-x
a'' = e^-x

b= .... etc

Answer 4

hmmm, im wondering if the error is the area between f(x) and M(x), integration

Answer 5

or is error, the amount of the remainder of the M(x) that is not worked out .... that seems more like it

Answer 6

wait a little bit

Answer 7

So I should put it like this? b= cosx-sinx, b'=-sinx-cosx and b''=-cosx+sinx

Answer 8

yes, that would be fine

Answer 9

By knowing this how do I proceed in order to find the error?

Answer 10

hopefully, there is a pattern that develops

f(0) = ab = 1(1) = 1
f'(0) = a'b + ab' = -1(-1) + 1(1) = 2
f''(0) = a''b + 2a'b' + ab'' = 1(1)+ 2(-1)(-1) + 1(-1) = 1+2-1 = 2

1 + 2x +2x^2/2!

if we knew a few more f's we might see a pattern develop

Answer 11

pfft, f' = 0, f''(0) = -2

those were a pain to keep track of

so, 1 - x^2 seems to be the Mac for it

Answer 12

f(x) = 1 - x^2 + E, for some E that we did not compute

f(x) - 1 + x^2 = E is the function of E that we get at best

Answer 13

let me see

Answer 14

further thought, it asks us to determine an interval x = 0 to 1/2

test the error function at x=1/2

Answer 15

Ok, but where did you take 1-x^2 from? Shouldn't it be sin(0)=0 and cos(0)=0?

Answer 16

cos0 = 1

Answer 17

but b''=-cox+sinx which is -1+0

Answer 18

* -cosx

Answer 19

yeah ..... i was doing b = c+s instead of c-s :)

Answer 20

c-s = 1
-s-c = -1
-c+s = -1

f = 1(1) = 1
f' = -1(1) + 1(-1) = -2
f' = 1(1) +2(-1)(-1)+ 1(-1) = 2

1 -2x +x^2 .... maybe?

Answer 21

yeah, that is soo much better

http://www.wolframalpha.com/input/?i=y%3De%5E%28-x%29%28cos%28x%29-sin%28x%29%29%2C+y%3D1-2x%2Bx%5E2

Answer 22

ook

Answer 23

i was simply making up my own problem lol

Answer 24

xD don't worry

Answer 25

once we get to that point, the Mac series is supposed to be equal to the original function, at least on some interval of convergence.

assuming that it converges between 0 and 1/2 ... we need to assess the "rest" of the series that we are omitting by stopping at the x^2 term

f(x) = Mac = 1 - 2x + x^2 + E, for some E that we ignored by stopping short, solving for E we get

f(x) - [1 -2x +x^2] = E, solve for x=1/2 to determine the error

if its positive, the error is rounds up; if negative it is rounding down

Answer 26

On my book it says that in order to find the error you should apply this equation:\[f(x)=T _{n,x _{0}}(x)+\frac{ f ^{n+1} (c)}{ (n+1)! }(x-x _{0})^{n+1}\]

Answer 27

ah ok

Answer 28

class is starting soon, so i have to be getting ready. good luck ;)

Answer 29

okky, thanksss

Answer 30

bye :)