Question

I need an easier way to solve this problem than what my professor taught me please!! Exam tomorrow!!!

How many terms of the Taylor series for ln(1+x) centered at x=0 do you need to estimate the value of ln(1.4) to three decimal points.

I know ln(1+x)=x-x^2/2+x^3/3-...
ln(1+.4)
abs of En(.4)
but thats all I know and understand

Answer 1

OH and f(.04)=ln(1.4)

Answer 2

if i am not mistaken, find the term that give a number less than \(.001\)
since the terms alternate, once you are there, the estimate is good to that many decimal places

Answer 3

wait what?

Answer 4

https://www.khanacademy.org/math/calculus/sequences_series_approx_calc/maclaurin_taylor/v/maclauren-and-taylor-series-intuition

Answer 5

DUde you can get a ti 89 titanium like mines and it will do it ;)

Answer 6

the series alternates, and the terms get smaller and smaller, so the error can be no more than the last term

Answer 7

im studying for my exam and were not allowed to use any calculators during exams or quizzes so i dont use mine anymore to practice not using it

Answer 8

\(\frac{.4^3}{3}=.021...\) so you need to go further out

Answer 9

if it's centered at 0 is it now the maclurian series?

Answer 10

my professor has been using these weird formulas calculating the error and he's not explaining anything and the exam is tomorrow so I'm SOL

Answer 11

hmm it is numerical exercise, can't think of a way to do it without checking the numbers.
i mean if it was \(\ln(1.1)\) then the answer would be different
if it was \(\ln(1)\) then the answer would be 1 term

Answer 12

what i need is steps to finding this problem just so i can get by with my exam please

Answer 13

https://www.khanacademy.org/math/calculus/sequences_series_approx_calc/maclaurin_taylor/v/maclauren-and-taylor-series-intuition

Watch the vids in this series. ANd just plug in values :D

Answer 14

this is using error not taylor polynomials

Answer 15

try this