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

In summary: So the final answer is##(-1)^{r+1} \frac{x^{r}}{r(r-1)}##or##(-1)^{r+1} \frac{x^{r}}{r^{2}-r}## ?OK, I see about that. :P So the final answer is##(-1)^{r+1} \frac{x^{r}}{r(r-1)}##or##(-1)^{r+1} \frac{x^{r}}{r^{2}-r}## ?That is certainly OK, but it is also possible to combine the two terms in the denominator into r(r-1) instead of r^2-r
  • #1
sooyong94
173
2

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. :/
 
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  • #2
sooyong94 said:

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.
 
  • #3
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
Chestermiller said:
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? :/
 
  • #5
Ray Vickson said:
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}## ?
 
  • #6
sooyong94 said:
the largest factor is 2? :/

6 = (3)(2) = 3!/1!
12 = (4)(3) = 4!/2!
20 = (5)(4) = 5!/3!
 
  • #7
sooyong94 said:
: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
 
  • #8
sooyong94 said:
: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.
 
  • #9
Ray Vickson said:
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):
[tex](-1)^r\frac{(r-2)!x^r}{r!}[/tex]
 
  • #10
sooyong94 said:
##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 [itex](-1)^r[/itex]. 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
HallsofIvy said:
Yes, the terms alternate in sign, so that would be [itex](-1)^r[/itex]. 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)##?
 
  • #12
sooyong94 said:
So that actually looks like ##r(r+1)##?
No. It is r(r-1).
 
  • #13
Chestermiller said:
No. It is r(r-1).

OK, I see about that. :P
 

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

1. What is a Maclaurin series expansion?

A Maclaurin series expansion is a mathematical tool used to represent a function as a sum of infinitely many terms. It is named after Scottish mathematician Colin Maclaurin and is similar to a Taylor series expansion, but specifically centered at the point x=0. This type of expansion is particularly useful for approximating functions and solving differential equations.

2. How is a Maclaurin series expansion calculated?

A Maclaurin series expansion is calculated by finding the derivatives of a function at x=0 and using those values to construct the series. The general formula for a Maclaurin series is f(x) = f(0) + f'(0)x + (f''(0)x^2)/2! + (f'''(0)x^3)/3! + ... + (f^(n)(0)x^n)/n! where f(0), f'(0), f''(0), etc. are the derivatives of the function evaluated at x=0.

3. What is the purpose of a Maclaurin series expansion?

A Maclaurin series expansion can be used to approximate the value of a function at any given point, as long as the function is differentiable at that point. It is also commonly used in calculus to solve differential equations and in physics to model natural phenomena.

4. Are there any limitations to using a Maclaurin series expansion?

Yes, there are some limitations to using a Maclaurin series expansion. The series may only converge for certain values of x, and it may not provide an accurate approximation for the function outside of the radius of convergence. Additionally, the function must be differentiable at x=0 for a Maclaurin series to exist.

5. Can a Maclaurin series be used to find the value of a function at x=0?

Yes, if a Maclaurin series exists for a given function, then the value of the function at x=0 is simply the first term in the series. This is because when x=0, all of the higher-order terms in the series become 0, leaving only the first term. However, it is important to note that this is only true if the function is differentiable at x=0.

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