Difficult Maclaurin Series

[BurgerTown]

First time posting, so hopefully it works out and my code isn't too bad. I have been trying for a long time to find this Macluarin series, but to no avail. The function goes

f(x) = integral (1-ln(1+x))/x dx

. After I have the series, I need to use this to find a polynomial approx. to f(x) which is accurate to 3 decimal places on the interval [-1,1]. So if you could help me out and give me a walk through or something through the problem that would be great.

Thanks!

 

[soroban]

Hello, BurgerTown!

Welcome aboard!
I'll do the first part . . .

$\displaystyle \text{Find the Macluarin series for: }\;f(x) \:=\:\int \frac{1-\ln(1+x)}{x}\,dx$

$\displaystyle \text{Use this to find a polynomial approxomation to }f(x)$
. . $\displaystyle \text{ which is accurate to 3 decimal places on the interval }[\text{-}1,1].$

$\displaystyle \text{First, find (or recall) the Maclaurin series for }\ln(1+x):$

. . \(\displaystyle \ln(1+x) \;=\;x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \hdots\)


\(\displaystyle \text{Then: }\;1 - \ln(1+x) \;=\;1 - x + \frac{x^2}{2} - \frac{x^3}{3} + \frac{x^4}{4} - \hdots\)

\(\displaystyle \text{Hence: }\;\frac{1 - \ln(1+x)}{x} \;=\;\frac{1}{x} - \frac{x}{2} - \frac{x^2}{3} + \frac{x^3}{4} - \hdots\)


\(\displaystyle \text{Therefore: }\;f(x) \;=\;\int\left(\frac{1}{x} - \frac{x}{2} - \frac{x^2}{3} + \frac{x^3}{4} - \hdots\right)\,dx\)

. . . . . . . . . . . . \(\displaystyle =\;\ln x - x + \frac{x^2}{2^2} - \frac{x^3}{3^2} + \frac{x^4}{4^2} - \hdots\)

 

[galactus]

$\displaystyle \int_{-1}^{1}\frac{1-ln(1+x)}{x}dx$ is undefined on the interval [-1,1].

Mainly because $\displaystyle \int_{-1}^{1}\frac{1}{x}dx$ is undefined on the interval [-1,1].

See the ln(x) in Soroban's series?. It is undefined on the given interval.

But, if we wanted $\displaystyle \int_{-1}^{1}\frac{ln(1+x)}{x}dx$, then we could hammer out a solution using series and integration.

It works out to be $\displaystyle \frac{{\pi}^{2}}{4}$

But, then $\displaystyle \int_{-1}^{1}\frac{1}{x}dx=-{\pi}i$. In other words, undefined.

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The usefulness of ln(1+x) is limited because of its slow convergence and the restriction $\displaystyle -1< x\leq 1$

But, if we replace x with -x to get the series for ln(1-x), then subtract the two series we get:

$\displaystyle ln\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\frac{x^{7}}{7}+\cdot\cdot\cdot\right), \;\ -1\leq x<1$......[1]

The last series can be used to compute the natural log of any positive number y by letting:

$\displaystyle y=\frac{1+x}{1-x}$

$\displaystyle x=\frac{y-1}{y+1}$.............[2]

and noting that $\displaystyle -1<x<1$.

Let compute ln(2) by letting y=2 in [2]. This gives $\displaystyle x=1/3$.

Sub this into [1] and get $\displaystyle ln(2)=2\left(1/3+\frac{(1/3)^{3}}{3}+\frac{(1/3)^{5}}{5}+\frac{(1/7)^{7}}{7}+\cdot\cdot\cdot\right)$

This gives ln(2)=.6931 to four decimal places.

We could just let x=1 in the series for ln(1+x), but it converges too slow to be of any real computational value.

 

[BurgerTown]

Thank you so much to the two of you. It really helped me understand how to do Maclaurin Series.

 

[lookagain]

\(\displaystyle \ \ \text{Then: }\;1 - \ln(1+x) \;=\;1 - x + \frac{x^2}{2} - \frac{x^3}{3} + \frac{x^4}{4} - \hdots\)

\(\displaystyle 1) \ \ \text{Hence: }\;\frac{1 - \ln(1+x)}{x} \;=\;\frac{1}{x} - \frac{x}{2} - \frac{x^2}{3} + \frac{x^3}{4} - \hdots\)


\(\displaystyle 2) \ \ \text{Therefore: }\;f(x) \;=\;\int\left(\frac{1}{x} - \frac{x}{2} - \frac{x^2}{3} + \frac{x^3}{4} - \hdots\right)\,dx\)

. . . . . . . . . . . . \(\displaystyle 3) \ \ =\;\ln x - x + \frac{x^2}{2^2} - \frac{x^3}{3^2} + \frac{x^4}{4^2} - \hdots\)

Going back to this post, while continue having the post of galactus
in the front of the mind, make these changes:


\(\displaystyle 1) \ \ \text{Hence: }\;\frac{1 - \ln(1+x)}{x} \;=\;\frac{1}{x} -1 + \frac{x}{2} - \frac{x^2}{3} + \frac{x^3}{4} - \hdots\)


\(\displaystyle 2) \ \ \text{Therefore: }\;f(x) \;=\;\int\left(\frac{1}{x} - 1 + \frac{x}{2} - \frac{x^2}{3} + \frac{x^3}{4} - \hdots\right)\,dx\)

. . . . . . . . . . . . \(\displaystyle 3) \ \ =\;\ln|x| - x + \frac{x^2}{2^2} - \frac{x^3}{3^2} + \frac{x^4}{4^2} - \hdots \ + C\)