# Multiplying power series - Maclaurin series

• Mar 7th 2009, 05:56 AM
triggy
Multiplying power series - Maclaurin series
Hi, I've run into some trouble trying to answer this power series question that asks:

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for the given function.

$e^{7x}ln(1-x/5)$

This is my work so far (all sums are n=0 to infinity):
$e^x$ = $\sum$ $\frac{x^n}{n!}$

$e^{7x}$ = $\sum$ $\frac{7^nx^n}{n!}$

and
$
ln(1+x)$
= $\sum$ $(-1)^{n-1}\frac{x^n}{n}$

$ln(1-x/5)$ = $\sum$ $(-1)^{n-1}(-1)^n\frac{5^{-n}x^n}{n}$
$ln(1-x/5)$ = $\sum$ $(-1)^{2n-1}\frac{x^n}{n5^n}$

To find the first 3 non-zero terms, I think I'm supposed to do something like the following:

n=3 $\longrightarrow$ $a_0b_3+a_1b_2+a_2b_1+a_3b_0$
n=2 $\longrightarrow$ $a_0b_2+a_1b_1+a_2b_0$
n=1 $\longrightarrow$ $a_0b_1+a_1b_0$
n=0 $\longrightarrow$ $a_0+b_0$

What I got from doing that was:
( $(-1)^{2n-1}$ becomes -1 for all n)

n=3
$-\frac{(7x)^0}{0!}*\frac{x^3}{3(5^3)}-\frac{(7x)^1}{1!}*\frac{x^2}{2(5^2)}-\frac{(7x)^2}{2!}*\frac{x^1}{1(5^1)}-\frac{(7x)^3}{3!}*\frac{x^0}{0(5^0)}$
= $-\frac{x^3}{375}-\frac{7x^3}{50}-\frac{49x^3}{10}-\frac{343x^3}{0}$
= $-\frac{3781x^3}{375}-\frac{343x^3}{0}$

n=2
$-\frac{(7x)^0}{0!}*\frac{x^2}{2(5^2)}-\frac{(7x)^1}{1!}*\frac{x^1}{1(5^1)}-\frac{(7x)^2}{2!}*\frac{x^0}{0(5^0)}$
= $-\frac{x^2}{50}-\frac{7x^2}{5}-\frac{49x^2}{0}$
= $-\frac{71x^2}{50}-\frac{49x^2}{0}$

n=1
$-\frac{(7x)^0}{0!}*\frac{x^1}{1(5^1)}-\frac{(7x)^1}{1!}*\frac{x^0}{0(5^0)}$
= $-\frac{x}{5}-\frac{7x}{0}$

n=0
$-\frac{(7x)^0}{0!}-\frac{x^0}{0(5^0)}$
$-1-\frac{1}{0}$

How do I get my 3 terms from the above? I have a lot of fractions where I'm dividing by 0.
Edit: I just realized that I changed $e^{7x}$ into a power series, would the Maclaurin series for that function be the same?
• Mar 7th 2009, 06:59 AM
Moo
Hello,

I haven't checked your formulae, it looks long :D

However, note that the power series of the exponential starts at n=0, and the power series of the logarithm stars at n=1.
This should solve the problem of dividing by 0.

The MacLaurin series is a power series. The expansion of a function into a series is unique. So expanding a function as a power series is equivalent to getting the MacLaurin series. (I don't know if I'm clear here ? :s)

To simplify calculations, you can use the Cauchy product for series. But again, be careful that the series should start at the same point
• Mar 7th 2009, 09:43 AM
triggy
Alright, so if I changed the index for the logarithmic series to start at 1, the following:

n=3 http://www.mathhelpforum.com/math-he...a13771b7-1.gif http://www.mathhelpforum.com/math-he...8f499a51-1.gif
n=2 http://www.mathhelpforum.com/math-he...a13771b7-1.gif http://www.mathhelpforum.com/math-he...f7d2bba9-1.gif
n=1 http://www.mathhelpforum.com/math-he...a13771b7-1.gif http://www.mathhelpforum.com/math-he...85d83429-1.gif
n=0 http://www.mathhelpforum.com/math-he...a13771b7-1.gif http://www.mathhelpforum.com/math-he...6b9ac073-1.gif

..would change to:

n=3 $\longrightarrow$ $a_0b_3 + a_1b_2 + a_1b_1$
n=2 $\longrightarrow$ $a_0b_2 + a_1b_1$
n=1 $\longrightarrow$ $a_0b_1$

Would that be correct?
• Mar 7th 2009, 04:05 PM
triggy

I rechecked my work and fixed the power series for the ln function. I'll show my work for it.

$ln(1+x) = \sum (-1)^{n+1}\frac{x^{n+1}}{n+1}$

$ln(1-x/5) = \sum (-1)^{n+1}\frac{(-\frac{x}{5})^{n+1}}{(n+1)}$
$ln(1-x/5) = \sum (-1)^{n+1}(-1)^{n+1}\frac{x^{n+1}}{(n+1)5^{n+1}}$
$ln(1-x/5) = \sum (-1)^{2n+2}\frac{x^{n+1}}{(n+1)5^{n+1}}$
$ln(1-x/5) = \sum \frac{x^{n+1}}{(n+1)5^{n+1}}$

$a_n$ = $\sum \frac{(7x)^n}{n!}$
$b_n$ = $\sum \frac{x^{n+1}}{(n+1)5^{n+1}}$

With the n=3 $\longrightarrow a_0b_3 + a_1b_2 + ...$ etc. and same for n=2, n=1, n=0

I got:
$n=0 \longrightarrow 1+\frac{x}{5}$
$n=1 \longrightarrow \frac{71x^2}{50}$
$n=2 \longrightarrow \frac{3782x^3}{750}$
$n=3 \longrightarrow \frac{89568x^4}{7500}$

So are the first 3 non-zero terms of the series: $1, \frac{x}{5},$ and $\frac{71x^2}{50}$?
• Mar 7th 2009, 11:24 PM
Moo
There is a little problem in the Taylor series for the logarithm !

The formula is $\ln(1+x)=\sum_{n=1}^\infty (-1)^{n-1} \frac{x^n}{n}$
which is $\sum_{n=0}^\infty (-1)^{\color{red}n} \frac{x^{n+1}}{n+1}$

so $\ln(1-x/5)={\color{red}-} \sum_{n=0}^\infty \frac{\left(\tfrac x5\right)^{n+1}}{n+1}$

this last modification should give you the correct answers (Wink)

Edit : there is another problem, $c_0=a_0b_0=-\frac x5$. There is no 1 ;)