# Maclaurin series expansion

1. Jan 24, 2014

### sooyong94

1. The problem statement, all variables and given/known data
Given that $f(x)=(1+x) ln (1+x)$.
(a) Find the fifth derivative of f(x),
(b) Hence, show that the series expansion of f(x) is given by
$x+\frac{x^{2}}{2} -\frac{x^{3}}{6} + \frac{x^{4}}{12} - \frac{x^{5}}{20}$

(c) Find, in terms of r, an expression for the rth term, (r>=2) of the Maclaurin expansion for f(x).

2. Relevant equations
Product rule, Maclaurin series

3. The attempt at a solution

For (a) I have used product rule and simplified the answer as $-6(1+x)^{-4}$

For part (b), I just have to plug in 0 into f(x) and up to the fifth derivative, right?

Part (c)... Now I'm stuck. I know the sign alternates each other, so I have to use the term
$(-1)^{r}$ for that. The trouble is dealing with the denominator. It looks like a series, though the denominator looks like a series, but it doesn't look like an arithmetic nor geometric series. :/

2. Jan 24, 2014

### Ray Vickson

There should be a "..." after the degree-5 polynomial you wrote above. Also, in TeX/LaTeX you should use "\ln" intead of "ln", as it looks much nicer; compare $\ln(1+x)$ with $ln(1+x)$. (The same goes for "lim" and all the trig functions and their inverses.)

Finding the first few terms by differentiation is OK, but is not only way to find the nth term. Can you see an easier way? Think about how f(x) is constructed.

3. Jan 24, 2014

### Staff: Mentor

Look at the two largest integer factors of each of the denominators (for the terms with r≥2). This should allow you to see the pattern.

Chet

4. Jan 24, 2014

### sooyong94

The largest factor is 2? :/

5. Jan 24, 2014

### sooyong94

:thinking: Maybe I can use the standard series of $\ln(1+x)$, that is $(-1)^{r+1} \frac{x^{r}}(r}$ ?

6. Jan 24, 2014

### Staff: Mentor

6 = (3)(2) = 3!/1!
12 = (4)(3) = 4!/2!
20 = (5)(4) = 5!/3!

7. Jan 24, 2014

### ehild

Yes!!! multiply it with (1+x) what do you get?

ehild

8. Jan 24, 2014

### Ray Vickson

Certainly that is the way I would do it.

9. Jan 24, 2014

### Staff: Mentor

Not me. I would just use the simple induction method I suggested in #3 and #6, particularly since I already have the results from (a) and (b):
$$(-1)^r\frac{(r-2)!x^r}{r!}$$

10. Jan 24, 2014

### HallsofIvy

Yes, the terms alternate in sign, so that would be $(-1)^r$. For the coefficients note that, for r= 2, the coefficient is 2= 2(1), for r= 3 it is 6= 3(2), for r= 4 it is 12= 4(3), and for r= 5 it is 20= 5(4).

11. Jan 24, 2014

### sooyong94

So that actually looks like $r(r+1)$?

12. Jan 24, 2014

### Staff: Mentor

No. It is r(r-1).

13. Jan 24, 2014

### sooyong94

OK, I see about that. :P