MacLaurin Series for f(x)=ln|1+x^3|

In summary, the conversation is about finding the Maclaurin series representation for f(x)=ln|1+x^3|. The attempt at a solution involves finding the derivatives of ln(1+x^3) and using the geometric series to find the Maclaurin series. The conversation also mentions finding the radius of convergence and the number of terms needed to approximate the integral of ln|1+x^3| from 0 to 2.
  • #1
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Homework Statement


Find the MacLaurin series representation for f(x)=ln|1+x^3|

Homework Equations


1/(1-x) = [tex]\sum[/tex]x^n = 1+x+x^2+x^3+... |x|<1

The Attempt at a Solution


right.
so maclaurin series by default means it expands as a taylor series where x=0
f(0)= ln|1+x^3| = 0
f'(0)= 3(0)^2/(1+0^3)^1 = 0/1 = 0
f''(0)= -3(0)^2/(1+0^3)^2 = -0/1 = 0
f'''(0)= 6(0)^6-42(0)^3+6/(1+0^3)^3 = 6
so on and so forth

it's taken me a few hours but so far, i can't seem to use that relevant equation to find the maclaurin series.

i may have gotten kinda close, b/c I've found so far that
fn(0)=(very complex numerator that kind of looks like a quadratic or polynomial formula with switching signs)/(1+x3)n

i've done upto 9th derivative(and further more) and found that this goes something like:
0,0,0,6,0,0,-360,0,0,120960,0,0,-119750400,0,0,261534973600
and that the series the pretty much follows (-1)^(n-1)*3*(n-1)!, where n starts at 0, with 2 blanks in between.

i know I'm missing something, and this isn't that hard.
please help.

--
then I'm supposed to find radius of convergence, which i think i can probably get using by ratio test..
then need to find out how many terms are needed to appx it within 0.0001...
then find out why appx'ing integral of ln|1+x^3| from 0 to 2 by using the series representation is wrong.
but i think i can figure those out once the first part is done.
 
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  • #2
Find the Maclaurin series for the derivative of ln(1+x^3), by using the geometric series.
 
  • #3
Or find the Maclaurin series for ln(1+t) and then let t=x^3.
 

Related to MacLaurin Series for f(x)=ln|1+x^3|

What is a MacLaurin series?

A MacLaurin series is a type of Taylor series which is a mathematical representation of a function as an infinite sum of terms. It is centered at x=0 and is used to approximate the value of a function at a given point.

How is the MacLaurin series for f(x)=ln|1+x^3| derived?

The MacLaurin series for f(x)=ln|1+x^3| is derived by using the Taylor series formula and finding the derivatives of the function at x=0. This results in a simplified form of the series that can be used to approximate the value of the function at any point within its radius of convergence.

What is the radius of convergence of the MacLaurin series for f(x)=ln|1+x^3|?

The radius of convergence of the MacLaurin series for f(x)=ln|1+x^3| is 1. This means that the series will only converge for values of x within the interval (-1,1).

How accurate is the MacLaurin series for f(x)=ln|1+x^3|?

The accuracy of the MacLaurin series for f(x)=ln|1+x^3| depends on the value of x and the number of terms used in the series. Generally, the more terms used, the more accurate the approximation will be. However, for values of x near the endpoints of the interval, the accuracy may decrease.

What is the significance of the MacLaurin series for f(x)=ln|1+x^3|?

The MacLaurin series for f(x)=ln|1+x^3| is significant because it allows us to approximate the value of the natural logarithm function at any point within its radius of convergence without having to use a calculator. This can be useful in many applications, especially in fields such as physics, engineering, and economics.

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