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

[xWaffle]

Homework Statement

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).

Homework Equations

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

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

 

[clamtrox]

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).

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:
$$f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{1}{2} f''(x_0)(x-x_0)^2 + ...$$

 

[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.

 

[xWaffle]

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:
$$f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{1}{2} f''(x_0)(x-x_0)^2 + ...$$

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

 

[clamtrox]

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

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

 

[xWaffle]

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

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

 

[clamtrox]

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

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