# Thread: Estimating integral using Taylor Series, extra term

1. ## Estimating integral using Taylor Series, extra term

Hi

I have a question to find the Taylor Series for y(x)=ln(1+x) up to and including terms of 0(x^5).

I found that series to be: x - 1/2x^(2) + 1/3x^(3) - 1/4x^(4) + 1/5x^(5)...which may or may not be correct!

The next part of the question asks:

"Use the series to estimate the value of the integral from 1 to 0 of ln(1+x)/3x dx" (I am sorry I don't know how to type the integral sign properly).

How do I deal with the extra x term in the second part of the question ie the /3x. I know to find the integral of the series to integrate each term of the polynomial and then calculate for 1 to 0 however how do I "use the series" being without the extra term to calculate the integral asked for.

Apologies if I am incorrect in anything I have worked out.

Any help would be appreciated.

Kind regards
Beetle

2. ## Re: Estimating integral using Taylor Series, extra term

Okay, what is the Taylor's series for ln(1+ x)? What do you get if you divide that by 3x and then integrate?

3. ## Re: Estimating integral using Taylor Series, extra term

Do you mean divide each term of the polynomial "f(x) = x - 1/2x^(2) + 1/3x^(3) - 1/4x^(4) + 1/5x^(5)" by 3x then integrate ie x/3x - 1/2x^(2)/3x...then integrate each term. Sorry not sure I understand your answer.

Kind regards

Yes.

5. ## Re: Estimating integral using Taylor Series, extra term

Does that then make the polynomial: 1/3 - 3/2x + x^(2) - 3/4x^(3) + 3/5x^(4)- I think I have done this all wrong...

6. ## Re: Estimating integral using Taylor Series, extra term

Am I right in thinking that then works out as: x/3 - x^(2)/12 + x^(3)/27 - x^(4)/48 + x^(5)/75 or have I made some stupid mistake. Just wanting to check I have understood how to solve this.

Kind regards
Beetle

7. ## Re: Estimating integral using Taylor Series, extra term

$\ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dfrac{x^4}{4} + \dfrac{x^5}{5} - \, ...$

$\dfrac{\ln(1+x)}{3x} = \dfrac{1}{3} \cdot \dfrac{\ln(1+x)}{x} = \dfrac{1}{3} \left(1 - \dfrac{x}{2} + \dfrac{x^2}{3} - \dfrac{x^3}{4} + \dfrac{x^4}{5} - \, ... \right)$

$\displaystyle \int_0^1 \dfrac{\ln(1+x)}{3x} \, dx \approx \dfrac{1}{3} \int_0^1 \left(1 - \dfrac{x}{2} + \dfrac{x^2}{3} - \dfrac{x^3}{4} + \dfrac{x^4}{5}\right) \, dx$