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How to make Taylor expansion with fixed numerical precision?

asked 2019-12-25 05:37:23 +0200

John Bao gravatar image

updated 2019-12-25 05:40:29 +0200

Following command


shows result:

x |--> -0.4207354924039483*(x - 1.0)^2 + 0.5403023058681398*x + 0.30116867893975674

The precision of result has too many bits, what I want is 3 bits, such as:

x |--> -0.420*(x - 1.0)^2 + 0.540*x + 0.301

How to do it?

Thanks for your help.

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May be you could just use 1 instead of 1.0? As in taylor(f,x,1,2) this gives -1/2*(x - 1)^2*sin(1) + (x - 1)*cos(1) + sin(1)

Nasser gravatar imageNasser ( 2019-12-25 06:42:47 +0200 )edit

Thanks @Nasser. What I need is a numerical solution.

John Bao gravatar imageJohn Bao ( 2019-12-25 06:48:36 +0200 )edit

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answered 2019-12-25 22:35:40 +0200

dsejas gravatar image

updated 2019-12-25 23:08:34 +0200

Hello, @john-bao! You could change the polynomial ring of the Taylor Polynomial. For example,

f(x) = sin(x)
T(x) = taylor(f, x, 1.0, 2)
R = PolynomialRing(RealIntervalField(10), 'x')

This will give you something like

 -0.421?*x^2 + 1.39?*x - 0.120?

The ? sign indicates you have a loss of precision. I am working with 10 bits, because 3 gives you very poor precision. If you don't like the ?s, you can change the RealIntervalField(10) with Reals(10), for example. If you want to know how many bits you need in order to have 4 digits, for example, you can use the digits_to_bits function:

from sage.arith.numerical_approx import digits_to_bits
R = PolynomialRing(RealIntervalField(digits_to_bits(4)), 'x')

Here is an alternative approach (although a little slower). This allows you to fix the number of bits or the number of digits (I'll use digits). Define the function:

def poly_prec(p, digits=4):
    pol = 0
    for op in p.operands():
        if not op.is_constant():
            ex = op.operands()
            pol += ex[0] * N(ex[1], digits)
            pol += N(op, digits)
    return pol

I hope this helps!

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Asked: 2019-12-25 05:37:23 +0200

Seen: 283 times

Last updated: Dec 25 '19