what is the properties of single function

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What is the properties of single function?

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Jan
Jan 2018-6-30
编辑:Jan 2018-6-30
Dou you mean the function called single?
help single
doc single
It converts a number to a floating point value of type single, which stores about 8 digits. double 's store about 15 digits. While single uses 4 bytes per number, double needs 8 bytes.
  36 个评论
marwan mokbil
marwan mokbil 2018-7-15
Hi Walter, is it easy find the function of rational function (end beheivet function) if we know enough points of (x,y) or we will face the same problem as we faced before.
Walter Roberson
Walter Roberson 2018-7-16
Sorry, I cannot seem to find information about beheivet ?
For rational functions, it would depend upon what the individual terms were permitted to be.
If I understand correctly, there are signal processing tools to find ratios of polynomials approximating arbitrary signals. I would need to think more about the theoretical case.
If you were to use a finite set of points to find polynomials f(x), g(x) such that y = f(x)/g(x), then take any xn such that xn is not in the original x; then yn = (f(x)*(x-xn))/(g(x)*(x-xn)) is the same as f(x)/g(x) except at x = xn, where it is discontinuous. But since xn was an arbitrary point not present on input, we cannot know that the "real" behaviour at xn is not in fact discontinuous. So there are an infinite number of variant functions yn that describe the system as accurately as f(x)/g(x) does, so the ratio of polynomials cannot be unique.
Also, consider any finite subset of the original points, x0..xm with y0..ym, excluding at least one point, (xn, yn). We hypothesize that there is a unique y1(x) = f1(x)/g1(x) describing the subset.
Now, does that hypothesized unique y1(x) = f1(x)/g1(x) definitely predict that y1(xn) will be yn ? No, it cannot, because (xn, yn) could be arbitrary. So for each (xn, yn) not in the original set x0..xm with y0..ym, of points used to construct y1(x) = f1(x)/g1(x), there must be a distinct y2(x) = f2(x)/g2(x) that fully fits the fuller set of points x0..xn with y0..yn that includes (xn, yn) . But each of those infinite number of y2(x) must agree with y1(x) for each x in x0..xm, so each of those infinite f2(x)/g2(x) are also valid but different functions for f1(x)/g1(x), which contradicts the hypothesis that f1(x)/g1(x) can be found uniquely from x0..xm with y0..ym. Therefore it is not possible to find a unique ratio of polynomials to fit the points.

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