# Homework Help: Series expansion help

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1. Jan 20, 2015

### sooyong94

1. The problem statement, all variables and given/known data
Show that the first non-zero coefficient in the expansion of
$e^{-x}-\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}$
in ascending powers of x is that of x^5

2. Relevant equations
Series expansion, logarithmic series

3. The attempt at a solution
Let $y=e^{-x}-\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}$

${\ln y={\ln \left[\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}\right]\ }\ }$

${\ln y\ }={\ln \left(1-x\right)\ }-\frac{1}{2}{\ln \left(1-x^2\right)-\frac{1}{3}{\ln \left(1-x^3\right)\ }\ }$

${\ln y\ }=\left(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}\right)-\frac{1}{2}\left(-x^2-\frac{x^4}{2}\right)-\frac{1}{3}\left(-x^3\right)$

${\ln y\ }=-x-\frac{x^5}{5}$

$y=e^{-x}\cdot e^{-\frac{x^5}{5}}$

$ye^x=e^{-\frac{x^5}{5}}$

$\left[\frac{1-x}{{\left(1-x^2\right)}^{\frac{1}{2}}{\left(1-x^3\right)}^{\frac{1}{3}}}\right]e^x=e^{-\frac{x^5}{5}}$

I know I'm incredibly close to the answer, however it seems like I can't solve it.

2. Jan 20, 2015

### RUber

You seem to have left off the e^(-x) term in your first step with the log.

3. Jan 21, 2015

### sooyong94

I'm aware of this, however the reason why I do this is to find the series expansion of the (1-x)/((1-x^2)^(1/2) (1-x^3)^(1/3)) first, then use it to subtract from the series expansion of e^-x.

4. Jan 21, 2015

### RUber

Then perhaps you simply mislabeled your y.
If $y = \frac{ 1-x }{ (1-x^2)^{1/2}(1-x^3)^{1/3}}$ then I agree that you might write
$y = e^{-x - x^5/5}$
Your expression then is $e^{-x} - e^{-x - x^5/5}$
Take a look at the series representations of those two exponentials. The result should be clear.

5. Jan 22, 2015

### sooyong94

How do you get e^-x -e^(-x -x^5 /5) ?

6. Jan 22, 2015

### RUber

The way you expanded ln(y) was for the fraction part without the $e^{-x}$. So $e^{-x} - e^{ln y}$ should be your expression.
As you said in post #3, you wanted to subtract the fractional part from the exponential part.

7. Jan 22, 2015

### sooyong94

Wasn't if I use e^(ln y), the expression would change to e^(-x -x^5 /5) ?

8. Jan 22, 2015

### RUber

Back to my original point that in your first post you left $e^{-x}$ off when you took the log of y. So using the definition of y which I gave in post 4, you have the expression $e^{-x}-y$. You did all the work to express $y = e^{-x -x^5/5}$. Now you just have to subtract that from $e^{-x}$.

9. Jan 22, 2015

### sooyong94

Now I have e^-x -e^(-x-x^5/5) Should I expand both as a series?

10. Jan 23, 2015

### RUber

If you do, your answer will be evident.