# Help with Maclaurin series of (1/x), (1/x^2), etc

1. Nov 1, 2012

### xWaffle

1. The problem statement, all variables and given/known data
I have the equation

$f(x) = \frac{\lambda^{2}}{ax^{2}}-\frac{\gamma ab}{x}$

What I am assigned to do is find a value of x at it's smallest, then approximate the value of the function when x - x(smallest) is much much greater than x(smallest).

2. Relevant equations

$f(x) = f(0) + f'(0)x + \frac{f''(0)x^{2}}{2!} + \frac{f^{3}(0)x^{3}}{3!} + \ldots$

3. The attempt at a solution

I re-wrote the equation to make it easier on the eyes and to help me see what exactly I'm supposed to do..

$f(x) = \frac{1}{x^{2}} \frac{\lambda^{2}}{a}- \frac{1}{x} \gamma ab$

From this I see that there may be a way to see when terms of the (1/x^2) become insignificant compared to the term with (1/x).

But how in the world do I expand the function with x in the denominator to show this? Am I approaching this wrong to begin with?

My idea was to find that first value of x, which I thought might be the 'a0' term of its Maclaurin Series (we are not dealing with Taylor Series about any points except the origin). But I can't find a Maclaurin series for a function where I need to plug in zero in the denominator.

Remember, the end goal is to approximate the original function when this "smallest significant x-value" is much much less than the value of the function. I think this can be re-written in a way to say, when $|x - x_{0}| << x_{0}$.

I hope I'm on the right track. If I'm not, then disregard the question about the Maclaurin series for now and help me get back on track.. Thanks

2. Nov 2, 2012

### clamtrox

that should be "... much much smaller than x(smallest)", I guess.

If that's the case, then you want to expand it into a series. But not a Maclaurin series, that's no use here. Instead use the general Taylor formula:
$$f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{1}{2} f''(x_0)(x-x_0)^2 + ...$$

3. Nov 2, 2012

### voko

The first part implies, I think, that yo find the value x were the function is minimal. This has nothing to do with Maclaurin or Taylor expansion.

The second part implies that you change the variable $z = \frac {1} {x - x_0}$, and express the function in terms of this variable, then obtain its linear approximation.

4. Nov 2, 2012

### xWaffle

But how am I supposed to find this value of x-zero to use in the series? I thought I had to use the series to find it.

Or do I take its limit as it approaches zero, or..? I'm still lost

5. Nov 2, 2012

### clamtrox

Oh, I thought it was the value where f(x) is the smallest. Maybe I misunderstood?

6. Nov 2, 2012

### xWaffle

Maybe I'm jumbling up notation.. The value I was calling x0 is where the function is a minimum value, and then I need to approximate the function for when |x - x0| is <<< than x0

I was assuming I needed to use some sort of series because it's included in the "series" part of the assignment

7. Nov 3, 2012

### clamtrox

You can find the minimum value easily by just requiring that f'(x0) = 0. Then expand the function in the vicinity of this point.