The tutor looks at forming a Taylor polynomial with the example of square root 31.

A transcendental function is one there is no operation for. Rather, it’s represented by a series of expressions. Square root and sin are two examples.

The Taylor polynomial for a function is defined as

P(x)=f(c) + f'(c)(x-c) + f”(c)(x-c)²/2! + f”'(c)(x-c)³/3! + ….

For presentation purposes, we note that square root c = c^0.5. In general,

Following the form of the Taylor polynomial gives, for square root,

P(x) = (c)^0.5 + 0.5c^-0.5(x-c) -0.25c^-1.5(x-c)²/2! + 0.375c^-2.5(x-c)³/3! – 0.9375c^-3.5(x-c)⁴/4! + ….

The meaning of c

In the Taylor polynomial above, c is an “anchor value” at which you already know the output. Preferably it’s the closest value [to the one being evaluated] for which the exact answer is known.

Example: Evaluate square root 31 using a Taylor polynomial.

Solution: closest to 31 is 36, so c=36. Then

P(31) = 36^0.5 + 0.5(36)^-0.5(31-36) – 0.25(36)^-1.5(31-36)²/2! + 0.375(36)^-2.5(31-36)³/3! – 0.9375(36)^-3.5(31-36)⁴/4! + ….

which becomes

6 + (0.5/6)(-5) – (0.25/216)(-5)²/2! + (0.375/7776)(-5)³/3! – (0.9375/279936)(-5)⁴/4! + ….

and then

6 – 0.416666667 – 0.014467592 – 0.00100469393 – 0.00008721301476

=5.567773835

According to the calculator,

31^0.5 = 5.567765363

The difference between the values is 0.00000947215. Perhaps each term in the series gives an additional decimal place of accuracy.

HTH:)

Source:

Larson, Roland E. and Robert P. Hostetler. Calculus, 3rd ed. Toronto: DC Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.