# Expanding $\frac{1}{1-x}$

Is there any kind of expansion of $$f(x)=\frac{1}{1-x}$$, possibly with polynomials, such that with only a few terms I can represent with an error smaller than $$10\%$$ the function over the interval $$[0, 1)$$?

I understand that for the Taylor polynomial, if I expand around $$0$$, I need many terms as the derivative of the function $$f$$ increases quickly as $$x \rightarrow 1$$.

Edit:

My goal is:The fact is that the function for my particular problem is $$\frac{1}{1-x_{l_1} x_{l_2}}$$ where $$x_{l_1}, x_{l_2}$$ are two functions in the fourier space and $$l_2 = L-l_1$$. And I would like to apply the convolution theorem using an expansion. The problem is that $$0 can be near $$0.999$$ and I have to use a lot of terms.

So if I have: $$\int dl_1 \frac{1}{1-x_{l_1} x_{l_2}}$$ this would be $$\sum_n \int dl_1 P_n(x_{l_1},x_{l_2})$$ for some polynomials, such that for each polynomial I can apply the convolution theorem, i.e. $$F[F^{-1}[]_{l_1} \cdot F^{-1}[]_{l_2}]$$, with $$F$$ fourier transform and $$F^{-1}$$ its inverse.

• Why don't you want to use the function itself? It is pretty easy to evaluate. – MachineLearner Mar 13 '19 at 21:40
• The fact is that the function for my particular problem is $\frac{1}{1-x_{l_1} x_{l_2}}$ where $x_{l_1}, x_{l_2}$ are two functions in the fourier space and $l_2 = L-l_1$. And I would like to apply the convolution theorem using an expansion. The problem is that $0<x_{l_1} x_{l_2}<1$ can be near $0.999$ and I have to use a lot of terms. – Saladino Mar 13 '19 at 21:46

You can sample points $$(x_i, f(x_i))$$ in the region from $$[0,1)$$ with a higher density in towards the $$x\to 1$$.

Then use a polynomial regression for the dataset $$\mathcal{D}=\{(x_1, f(x_1)),\ldots,(x_N, f(x_N))\}$$ with $$N$$ data points. The problem with polynomials will be that the error is unbounded for $$x\to 1$$.

You could start by fitting a quadratic polynomial with the regression equation

$$f(x_i)=w_0+w_1x_i+w_2x_i^2+\varepsilon_i$$

• Isn't this computationally expensive? I would have an N degree polynomial, say if I had 1000 points it would be a nightmare. I was looking for something like some simple polynomial with a small order to handle this and with some particular coefficients(that could depend on some interpolation..). (btw good idea) – Saladino Mar 13 '19 at 21:43
• You can use recursive least square if you really want to get more data points. but for $10$ $\%$ from $0,0.999$ a few observations should be enough and $1000$ would be total overkill. – MachineLearner Mar 13 '19 at 21:47
• Can I use the convolution theorem with this type of expansion? (if you can see my edit aboe in the main post) – Saladino Mar 13 '19 at 21:49
• Ok, thanks @MachineLearner I will try it and let you know! My goal is to be able to apply this for a convolution – Saladino Mar 13 '19 at 21:50
• I think convolutions should not be problematic as you will obtain a polynomial as you wanted. – MachineLearner Mar 13 '19 at 21:50

You can't find a polynomial which could estimate f uniformly. Because f is unbounded on [0,1) and polynomials are bounded on any bounded domain. So the error always will be unbounded.

• yes, any taylor polynomial or polynomial can not do the job (but the taylor series can) – Saladino Mar 14 '19 at 0:20