# 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.

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