Question

Medal to person who can help me.....
Use a graphical method to determine the approximate interval for which the second order Taylor polynomial for ln(1+x) at x=0 approximates ln(1+x) with an absolute error of no more than 0.04... I have no idea where to start. My book does a bad job at explaining series problems--especially this special case.

Answer 1

@jim_thompson5910

Answer 2

@timo86m

Answer 3

@ganeshie8

Answer 4

@Luigi0210

Answer 5

@dumbcow

Answer 6

Sorry, only in calc I.

Answer 7

Oh well. Thank you very much regardless.

Answer 8

First you have to find the second order taylor polynomial for ln(1+x) centered at x = 0

Answer 9

any ideas on how to do that?

Answer 10

well its derivative right? Well then that entails yo uto find derivative of ln(1+x) which is 1/1+x. Now second order, so find 2nd derivative so now its -(1+x)^-2. Right?

Answer 11

so far, so good

Answer 12

now how do you use those derivatives to find the taylor polynomial?

Answer 13

I'm pretty sure its plug in zero for x since that is what you are subsitituting for. Plus I use McLauren Series cause its centered at x=0..... So the taylor series would be....\[\ln(1+x)=1+-.5x\] Right?

Answer 14

Just for 2nd order thats why I went only to two right?

Answer 15

when you plug x = 0 into the first derivative, you get 1/(1+0) = 1/1 = 1

You divide this by 1! = 1 to get 1*1 = 1

Then you multiply that result by (x-0)^1 = x^1 = x

So the first term of this taylor polynomial is x

Answer 16

for the second derivative, you plug in x = 0 to get

-1/(1+x)^2 = -1/(1+0)^2 = -1

Then you divide this by 2! = 2*1 = 2 to get 1/2

and then you multiply by (x-0)^2 = x^2

The second term is (-1/2)x^2

Answer 17

So overall, the 2nd order taylor polynomial is x - (1/2)x^2

Answer 18

call that T(x), so T(x) = x - (1/2)x^2

Answer 19

I do not mean to insult you at all in any manner but are you sure?

Answer 20

yes, I'm sure

Answer 21

have a look at this page if you're not sure

http://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx

Answer 22

ok cool. now what?

Answer 23

f(x) is the original function, so f(x) = ln(1+x)

the difference in the two functions is f(x) - T(x) = ln(1+x) - (x - (1/2)x^2) = ln(1+x) - x + (1/2)x^2

Answer 24

Why are we subtracting the Taylor polynomial by the function?

Answer 25

If T(x) perfectly matched up with f(x), then f(x) - T(x) would always be zero

but that's not the case. There's error. So that's why f(x) - T(x) is for the most part nonzero

Answer 26

we subtract to calculate the error E(x)

E(x) = |f(x) - T(x)|

and we have those absolute value bars in there to make sure the difference is always positive

Answer 27

oh ok so now the error is ln(1+x) + x + (1/2)x^2 ? Now do I substitute the .04 error in the equation? If so this makes sense haha

Answer 28

the error is E(x) = |f(x) - T(x)| which is the same as saying

E(x) = ln(1+x) + x + (1/2)x^2

we then plug in E(x) = 0.04 since we want the error to be 0.04. Solving for x will give you the values of x that make E(x) = 0.04 true

Answer 29

sorry I meant to say

E(x) = |ln(1+x) + x + (1/2)x^2|

Answer 30

How do we get the upper and lower limit? The answers given in the multiple choice consists of an inequality involving an upper and lower limit...?

Answer 31

tell me what you get when you solve |ln(1+x) + x + (1/2)x^2| = 0.04

Answer 32

you can do so graphically by plotting

y1 = |ln(1+x) + x + (1/2)x^2|
y2 = 0.04

Answer 33

or you can plot y1 = |ln(1+x) + x + (1/2)x^2|-0.04 and find the roots

Answer 34

Oh Ok I see. I understand now. Thank you veyr much. You just solve with one equaling 0.04 and the other -0.04 Am I correct? From there you have the limits. Cool Thank you!!!

Answer 35

well if you don't use absolute values, then yes, you'll use 0.04 and -0.04

Answer 36

it will turn out that everything between those two solutions will have the error smaller than 0.04

anything outside that interval will have E(x) > 0.04