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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 $0<x_{l_1} x_{l_2}<1$ 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.

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  • $\begingroup$ Why don't you want to use the function itself? It is pretty easy to evaluate. $\endgroup$ – MachineLearner Mar 13 '19 at 21:40
  • $\begingroup$ 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. $\endgroup$ – Saladino Mar 13 '19 at 21:46
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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

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

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  • $\begingroup$ 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) $\endgroup$ – Saladino Mar 13 '19 at 21:43
  • $\begingroup$ 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. $\endgroup$ – MachineLearner Mar 13 '19 at 21:47
  • $\begingroup$ Can I use the convolution theorem with this type of expansion? (if you can see my edit aboe in the main post) $\endgroup$ – Saladino Mar 13 '19 at 21:49
  • $\begingroup$ Ok, thanks @MachineLearner I will try it and let you know! My goal is to be able to apply this for a convolution $\endgroup$ – Saladino Mar 13 '19 at 21:50
  • $\begingroup$ I think convolutions should not be problematic as you will obtain a polynomial as you wanted. $\endgroup$ – MachineLearner Mar 13 '19 at 21:50
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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.

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

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