#  >> Continuing Education >> Other Continuing Education

How to Solve Infinite Series Trigonometry

Infinite series are powerful mathematical tools in the world advanced mathematics. They are used to analyze limits, convergence, differential equations, numerical analysis, estimating the behavior of functions and trigonometry functions. The infinite series equation for the trig function sine(x) is SUM [(-1)^n/(2n + 1)!(x^(2n +1))] where the value inside the bracket change for each value of n and the SUM means the individual values of n are added together to create the series. The symbol "!" in the formula represent factorial where 3! is 6 (3 x 2 x 1) and 5! is 120 (5 x 4 x 3 x 2 x 1).

Instructions

  1. Solve the infinite series for a sine function using the formula SUM [(-1)^n/(2n + 1)!(x^(2n +1))]

    • 1

      Choose the number of times you want to expand the infinite series. As an example, choose 4 times for n = 0, 1, 2 and 3

    • 2

      Calculate the individual values of the series for each value of n using the formula: (-1)^n/(2n + 1)!(x^(2n +1)). For example:

      For n = 0

      [(-1)^n/(2n + 1)!(x^(2n +1))] = SUM [(-1)^0/(2(0) + 1)!(x^(2(0) +1))] = 1/1 * x = x.

      For n = 1

      [(-1)^n/(2n + 1)!(x^(2n +1))] = [(-1)^1/(2(1) + 1)!(x^(2(1) +1))] = -1/(3!)x^3 = -X^3/3! = -x^3/6.

      For n = 2

      [(-1)^n/(2n + 1)!(x^(2n +1))] = SUM [(-1)^2/(2(2) + 1)!(x^(2(2) +1))] = 1/(5!)x^5 = x^5/5! = x^5/120

      For n = 3

      [(-1)^n/(2n + 1)!(x^(2n +1))] = SUM [(-1)^3/(2(3) + 1)!(x^(2(3) +1))] = -1/(7!)x^7 = -x^7/7! = -x^7/5040

    • 3

      Add the individual values together. In this example:

      sinx = SUM [(-1)^n/(2n + 1)!(x^(2n +1))] = x - x^3/6 + x^5/120 - x^7/5040 ........

Related articles
Other Continuing Education
Latest Articles
Learnify Hub © www.0685.com All Rights Reserved