# Maclaurin series question

• May 3rd 2010, 11:04 PM
Paymemoney
Maclaurin series question
Hi

I need help on the following question:
1) Obtain the Maclaurin Series for $e^x$ and $ln(1+x)$. Hence obtain by multiplication the Maclaurin series for $e^{x}ln(1+x)$ giving the first three terms.

I have worked out the series for the $e^{x}$ and $ln(1+x)$ by themselves.
I then tried to multiple the values together, but was unsuccessful.

$e^{x}$ series is:

$1 + x + \frac{x^2}{2} + \frac{x^3}{6}$

$ln(1+x)$ series is:

$ln(1) + x - \frac{x^2}{2} + \frac{2x^3}{6}$

P.S
• May 3rd 2010, 11:49 PM
matheagle
It's called the Cauchy product and it's not that clever.
Just start multiplying the two infinite sums.
Ln(1)=0, by the way.

$\biggl(a_1+a_2+\cdots \biggr) \biggl(b_1+b_2+\cdots \biggr)=a_1b_1+a_1b_2+a_2b_1+a_2b_2+ \cdots$
• May 3rd 2010, 11:57 PM
Tikoloshe
So if you want only the first three terms, it’s a simple case of multiplying multinomials at this point. Evaluate the multiplication:

$\left(1 + x + \frac{x^2}{2}\right)\left(x - \frac{x^2}{2} + \frac{x^3}{3}\right)$

and take the first three terms.

If, on the other hand, you want an expression for the Maclaurin series of $e^{x}\ln(1+x)$, you could multiply together the individual series:

$\left(\sum_{k=0}^\infty\frac{x^k}{k!}\right)\left( \sum_{j=1}^\infty\frac{(-1)^j}{j}x^j\right)=\sum_{j=1}^\infty\sum_{k=0}^\in fty\frac{(-1)^jx^{j+k}}{jk!}$
• May 4th 2010, 03:42 AM
Paymemoney
ok, i have multiplied the values and this is what i get, however its not what the book's answer says.

$x - \frac{x^2}{2} + \frac{x^3}{3}$ ,

$x^2 - \frac{x^3}{2} + \frac{x^4}{3}$ ,

$\frac{x^3}{2} - \frac{x^4}{4} + \frac{x^6}{6}$
• May 4th 2010, 06:50 AM
matheagle
Quote:

Originally Posted by Paymemoney
ok, i have multiplied the values and this is what i get, however its not what the book's answer says.

$x - \frac{x^2}{2} + \frac{x^3}{3}$ ,

$x^2 - \frac{x^3}{2} + \frac{x^4}{3}$ ,

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

add these now, it's only one series in the end
• May 5th 2010, 01:38 AM
Paymemoney
Quote:

Originally Posted by matheagle
add these now, it's only one series in the end

This is what i have:

$x - \frac{x^2}{2} + \frac{x^3}{2} + \frac{x^6}{6}$

$1 + x + \frac{x^2}{2} + \frac{x^3}{6} + ...... ,$
$x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + .......,$
$x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ......,$
• May 5th 2010, 10:22 AM
Tikoloshe
Quote:

Originally Posted by Paymemoney
$1 + x + \frac{x^2}{2} + \frac{x^3}{6} + ...... ,$
$x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + .......,$
$x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + ......,$

This is for all three answers:
$e^x=1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ldots$,
$\ln(1+x)=x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} +\ldots$, and
$e^x\ln(1+x)=x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} +\ldots$

Quote:

Originally Posted by Paymemoney
ok, i have multiplied the values and this is what i get, however its not what the book's answer says.

$x - \frac{x^2}{2} + \frac{x^3}{3}$ ,

$x^2 - \frac{x^3}{2} + \frac{x^4}{3}$ ,

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

I believe you have the last one wrong.
$\left(1 + x + \frac{x^2}{2}\right)\left(x - \frac{x^2}{2} + \frac{x^3}{3}\right)=\left(x-\frac{x^2}{2}+\frac{x^3}{3}\right)+\$ $\left(x^2-\frac{x^3}{2}+\frac{x^4}{3}\right)+\left(\frac{x^3 }{2}-\frac{x^4}{4}+\frac{x^5}{6}\right)$ $=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{12}+\fra c{x^5}{6}$

The first three terms here are of course all that’s correct.