Finding the Maclaurin Series Expansion of (1+x)ln(1+x)

[sooyong94]

Homework Statement

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).

Homework Equations

Product rule, Maclaurin series

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. :/

 

[Ray Vickson]

Homework Statement

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).

Homework Equations

Product rule, Maclaurin series

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. :/

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.

 

[Chestermiller]

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

 

[sooyong94]

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

The largest factor is 2? :/

 

[sooyong94]

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.

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

 

[Chestermiller]

the largest factor is 2? :/

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

 

[ehild]

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

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

ehild

 

[Ray Vickson]

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

Certainly that is the way I would do it.

 

[Chestermiller]

Certainly that is the way I would do it.

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):
$$(-1)^r\frac{(r-2)!x^r}{r!}$$

 

[HallsofIvy]

$x+\frac{x^{2}}{2} -\frac{x^{3}}{6} + \frac{x^{4}}{12} - \frac{x^{5}}{20}$

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

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).

 

[sooyong94]

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).

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

 

[Chestermiller]

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

No. It is r(r-1).

 

[sooyong94]

No. It is r(r-1).

OK, I see about that. :P