Taylor Polynomial Series

In mathematics, the Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is named after the English mathematician Brook Taylor. If the series is centered at zero, the series is also called a Maclaurin series, named after the Scottish mathematician Colin Maclaurin. It is common practice to use a finite number of terms of the series to approximate a function. The Taylor series may be regarded as the limit of the Taylor polynomials.  In the article we shall discuss about Taylor polynomial series.(Source: Wikipedia)

Taylor polynomial series:

Find the Taylor series for sin x.

    Sin x = `sum_(0)^oo` `((-1)^k)/((2k + 1)!)` x^2k+1

The remainder term is not expressible in any simple way but can be estimated by using the Lagrange's form of the remainder. The coefficients

   `((-1)^k)/((2k + 1)!)`

are easily verified by calculating successive derivatives of f(x) = sin x and using the formulas

 a_k=`(f^(k)(0))/(k!)`

To check convergence of the series, apply Lagrange's form for R_a(x); For each x`in` R. there exists Z such that

R_n(x) = `(f^(n + 1)(z))/((n + 1)!)` xⁿ⁺¹

Now |fⁿ⁺¹(z)| equals either |cos z| or |sin z| So, in either case,|fⁿ⁺¹ (z)|`<=`1 ,and

       |R_n(x) |`<=` | x |ⁿ⁺¹ /(n+1)!

Since | x |ⁿ⁺¹ /(n+1)!`->`0 as n `|->` `oo`  for all x`in` R, we can see that the remainder term |R_n(x)|`|->` 0 as n `|->` `oo`

 for all x`in` R. Thus the series representation is completely justified for all real x.

  Observe that our estimate for |R_n (x)|,

                                                    |R_n (x)|`<=`|x|^_n+1/(n+1)!

gives also a sense of the rate of convergence of the series for fixed x, for example, for | x|`<=` 1, we find

                                                    |R_n (x)|`<=`1/(n+1)!

Thus, if we want to calculate sin x on (-1, 1) to within .01, we need take only the first five terms of the series (n = 4) to achieve that degree of accuracy.

 Had we used the integral form for R_n (x) we would have obtained a similar estimate.

Sample problem for Taylor polynomial series:

Pro:  Evaluate the definite integral `int_0^1` Sin (x) dx

Sol:  The integrand has no anti derivative expressible in the terms of familiar functions. Howebver, we know how to find its Taylor series. we know that

      Sin t = t -`(t^3)/(3!)` + `(t^5)/(5!)` - `(t^7)/(7!)` + ----

Now if we substitute t = x, we have

     Sin (x) = X - `(x^3)/(3!)` + `(x^5)/(5!)` - `(x^7)/(7!)` + ----

In spite of the fact that we cannot antidifferentiate the function, we can antidifferentiate the Taylor series:

    `int_0^1` Sin (x) dx = `int_0^1` (X - `(x^3)/(3!)` + `(x^5)/(5!)` - `(x^7)/(7!)` + ----) dx

                            =(`(x^2)/(2)` -`(x^)/(4*3!)` + `(x^6)/(6*5!)`-`(x ^ 7)/(8*7!)`+ -----) |₀¹

                            = (`(x^2)/(2)` -`(x^)/(4*3!)` + `(x^6)/(6*5!)`+`(x^7)/(8*7)` + ---)

Notice that this is an alternating series so we know that it converges. if we add up the first four terms, the pattern becomes ckear: the series converges to 0.2871