SERIESSUM

Sums the first terms of a power series.

Syntax:

SERIESSUM(x, n, m, ar)

A power series may be represented as:

$f(x)=a_0{x}^n+a_1{x}^{n+m}+a_2{x}^{n+2m}+a_3{x}^{n+3m}+\cdots$

The SERIESSUM function calculates the sum of the first terms of such a series, where:

x is the variable,

n is the power of x for the first term,

m is the increment by which the power of x increases with each term, and

ar refers to a range containing the a coefficients of the terms to be included.

Example:

The following power series may be used to express the mathematical constant e raised to a power:

$e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$

When x = 1, summing the terms in the series will approximate e. To sum the first 5 terms using SERIESSUM we should set:

x = 1

n = 0

m = 1

ar to B1:B5, where B1, B2, B3, B4, B5 contain respectively:

          = 1/FACT(0), = 1/FACT(1), = 1/FACT(2), = 1/FACT(3), = 1/FACT(4)

Now, using these values in the SERIESSUM function:

SERIESSUM(1, 0, 1, B1:B5)

returns 2.70833333333333, an approximation of e ( = 2.71828182845904...). Using more terms would give a closer approximation.

Using the SERIESSUM function with these values:

SERIESSUM(x, n, m, coefficients)

SERIESSUM(1, 0, 1, {1, 1, 0.5, 0.166667, 0.041667, 0.008333})

The function calculates the sum as follows:

$(1\times 1^0)+(1\times 1^1)+(0.5\times 1^2)+(0.166667\times 1^3)+(0.041667\times 1^4)+(0.008333\times 1^5)$

$=1+1+0.5+0.166667+0.041667+0.008333$

$=2.716667$

The result is an approximation of $e^1$, and as more terms are included, the approximation becomes more accurate.