In Exercises 31-32, approximate the value of the given definite integral by
using the first 4 nonzero terms of the integrand's Taylor series.This discussion will cover only #31 that is \[ \begin{aligned} \int^{\sqrt \pi}_{0} sin(x^2) \end{aligned} \]
First, let’s compute the Taylor series for \(sin(x^2)\). We know that the Taylor series of \(sin(x)\) is
\[ \begin{aligned} sin(x) = \sum^{\infty}_{n=0} (-1)^n x^{2n+1}/(2n+1)! \\ = x - x^3/3! + x^5/5! - x^7/10! + ... \end{aligned} \]
\[ \begin{aligned} sin(x^2) = \sum^{\infty}_{n=0} (-1)^n x^{2(2n+1)}/(2n+1)! \\ = \sum^{\infty}_{n=0} (-1)^n x^{4n+2}/(2n+1)! \\ = x^2 - x^6/3! + x^{10}/5! - x^{14}/7! + ... \end{aligned} \]
\[ \begin{aligned} \int^{\sqrt \pi}_{0} sin(x^2) \,dx= \int^{\sqrt \pi}_{0} \sum^{\infty}_{n=0} (-1)^n x^{4n+2}/(2n+1)! \,dx\\ = \int^{\sqrt \pi}_{0}x^2 - x^6/3! + x^{10}/5! - x^{14}/7! + ... \,dx \\ = \big[x^3/3 - x^7/(7*3!) + x^{11}/(11*5!) - x^{15}/(15*7!) + ...\big]^{\sqrt \pi}_0 \\ = (\sqrt \pi)^3/3 - (\sqrt \pi)^7/(7*3!) + (\sqrt \pi)^{11}/(11*5!) - (\sqrt \pi)^{15}/(15*7!)+ ... \end{aligned} \]
let’s try to estimate the numerical approximation of the first 4 terms above to compute \(\int^{\sqrt \pi}_{0} sin(x^2)\)
terms <- 0:3 # number of nonzero terms to use
# Function to compute the approximate definite integral of sin(x^2) from a non-zero
# upper limit to 0 (lower limit) using Taylor Series
integral_sin_x_squared_func <- function(x, n){
numerator <- x^(4*n+3)
denominator <- (4*n+3) * factorial(2*n+1)
return(sum(((-1)^n) * (numerator / denominator)))
}
sin_x_squared_approx <- integral_sin_x_squared_func(pi^0.5, terms)
sin_x_squared_approx## [1] 0.8877069The above value is the approximation of the integral using the first 4 terms in the Taylor Series of \(sin(x^2)\)
For n terms, the approximate solution for the integral is
\[ \begin{aligned} \sum^{\infty}_{n=0} (-1)^n \sqrt \pi^{4n+3}/[(4n+3) (2n+1)!] \end{aligned} \]