Show that the Maclaurin Polynomials for f(x) = ln (1+x) are . . .

Question

$$T_n(x) = x - \frac{x^2}{2} + \frac{x^3}{3} + \dotsb + \frac{(-1)^{n-1} x^n}{n}$$

I of course understand how to show that it equals the first two terms, but so far have been unable to prove that it equals the 'general term' or the last term.

Answer 1

$$ f(x) = \sum_{n=0}^{\infty} \frac{f^{n}(x_0)}{n!} (x-x_0)^{n} $$

$$ \frac{d}{dx}\ln(1+x) = \frac{1}{1+x} = (1+x)^{-1} \rightarrow \frac{d^{n}}{dx^{n}} \ln(1+x)= (-1)^{n-1}(n-1)!(1+x)^{-n} \\ \frac{d^n}{dx^n}\ln(1+x) |_{x=0}= (-1)^{n-1}(n-1)! $$

$$ \ln(1+x) = \sum_{n=0}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n!} (x)^{n} = \sum_{n=0}^{\infty} \frac{(-1)^{n-1}}{n} (x)^{n} $$

Answer 2

If $f(x)=ln(x+1)$ then the derivatives are:

\begin{equation} f^{(1)}=\frac{1}{x+1}\\f^{(2)}=\frac{-1}{(x+1)^2}\\f^{(3)}=\frac{(-1)(-2)}{(x+1)^3}=\frac{2}{(x+1)^3}\\f^{(4)}=\frac{2(-3)}{(x+1)^4}=\frac{-6}{(x+1)^4}\\f^{(5)}=\frac{2*3*4}{(x+1)^5} \end{equation}

and so on...We can see that n-derivative will be:

\begin{equation} f^{(n)}=(-1)^{(n-1)}\frac{(n-1)!}{(x+1)^n} \end{equation}

If you calculate the derivatives in $x=0$ and multiply for $x^n$ you obtain the formula.

Answer 3

Hint: the general derivative of $\ln(1+x)$ is $$f^{(n)}=(-1)^{n-1}\frac{(n-1)!}{(1+x)^n}\quad{\text n>0}$$ at $x=0$, the derivative equation becomes $$f^{(n)}=(-1)^{n-1}\frac{(n-1)!}{(1+0)^n}=(-1)^{n-1}(n-1)!$$

Answer 4

We use $\ln(1+x) =\int_0^x \frac{dt}{1+x} $.

We have $\sum_{k=0}^n t^k =\frac{1-t^{n+1}}{1-t} $. Putting $-t$ for $t$, $\sum_{k=0}^n (-1)^kt^k =\frac{1-(-1)^{n+1}t^{n+1}}{1+t} $ so that $\frac{1}{1+t} =\sum_{k=0}^n (-1)^kt^k+\frac{(-1)^{n+1}t^{n+1}}{1+t} $.

Integrating from $0$ to $x$,

$\begin{array}\\ \ln(1+x) &=\int_0^x \frac{dt}{1+t}\\ &=\int_0^x dt\left(\sum_{k=0}^n (-1)^kt^k+\frac{(-1)^{n+1}t^{n+1}}{1+t}\right)\\ &=\sum_{k=0}^n(-1)^k\int_0^x t^k dt+\int_0^x dt\frac{(-1)^{n+1}t^{n+1}}{1+t}\\ &=\sum_{k=0}^n(-1)^k\frac{x^{k+1}}{k+1}+\int_0^x dt\frac{(-1)^{n+1}t^{n+1}}{1+t}\\ &=\sum_{k=1}^{n+1}\frac{(-1)^{k-1}x^{k}}{k}+e_n(x) \qquad\text{where } e_n(x) =(-1)^{n+1}\int_0^x dt\frac{t^{n+1}}{1+t}\\ \end{array} $

This is the MacLaurin series for $\ln(1+x)$ with an explicit error term which is nice).

To show that $e_n(x) \to 0$ as $n \to \infty$ for $0 < x < 1$, note that

$\begin{array}\\ |e_n(x)| &=\big|\int_0^x dt\frac{t^{n+1}}{1+t}\big|\\ &<\big|\int_0^x t^{n+1}dt|\\ &=\dfrac{x^{n+2}}{n+2}\\ &< \frac1{n+2}\\ & \to 0\\ \end{array} $

As is often the case, my answer is not original. It is taken from "100 Great Problems of Elementary Mathematics" by Heinrich Dörrie, available for less than $15 US from Amazon.

I have enjoyed and recommend this book.