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I have this function, $e^{-x}$ bounded between 0 and 1500 and I have an approximation by Taylor Series of the same function bounded between 0 and 0.5. I would like to express my function $e^{-x}$ bounded between 0 and 15oo in terms of the last approximation I described, what is bounded between 0 and 0.5. Does anyone know how I can do it? Thank you.

NOTE: This question is related with Development of a hardware arquitecture for a particular algorithm.

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Where is the Taylor series for (0,1500) centered? If at zero, it is the same Taylor series as for (0,0.5). But you really don't want a Taylor series for (0,1500)-it would take an enormous number of terms for any reasonable accuracy. Numerical Recipes (chapters 5 and 6) and all other numerical analysis books have suggestions for better models. Abramowitz and Stegun has a polynomial fit over (0,1) that is much more accurate than you need. Given what you had in the previous question, you can't represent $\exp(-x)$ for $x$ much over 11 because you will underflow.

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If I were in his shoes, I'd do scaling and squaring: for positive argument, keep dividing by 2 until it's small enough, evaluate a Padé approximant of appropriate order, and square the result of that an appropriate number of times, and then reciprocate. –  J. M. Dec 24 '10 at 16:33
    
(Answer to "Where is the Taylor series for (0,1500) centered?"): I was thinking about two possibilities. One, to use the maclaurin series, the other, would be to chose a middle point in the bound, for the bound (0,1500), I would chose as middle point, 750. –  Peterstone Dec 30 '10 at 8:31
    
@Peterstone: Those are both possibilities. The middle point will give better accuracy, as the ends of the interval are not so far away. But can you fit exp(-1500) in your data type without underflow? Even exp(-15)? –  Ross Millikan Dec 30 '10 at 13:47
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