Something weird about a math function

**Mapisto** · 02-02-2012, 01:40 PM

Hi,

I've been asked to write a program that calculates cos(x) value with it's Taylor equivalent.
I've done it in 2 ways: 1) a loop 2)Recursion.

The thing is, it gives me the right values for every X between 0 and 2pi (using radians), as long as I'm giving it the "n" value of 10. If I input let's say n=100, it gives me "NaN".
What's more weird, if I stay with n=10, but give it a value of X=3 (pi), it gives me a wrong answer :|

You can find the Taylor equivalent of cos(x) here: Taylor series - Wikipedia, the free encyclopedia

And here's my program:

Java Code:

import java.util.Scanner;

public class RecursionEx2 {

	
	public static void main(String[] args) {
		Scanner input= new Scanner(System.in);
		
		System.out.print("Please enter your n: ");
		int n= input.nextInt();
		System.out.print("Please enter your X: ");
		double X= input.nextDouble();
		loop(n,X);
		System.out.println("Rec: "+rec(n,X));
	
	}
	public static void loop (int n, double X){
		double res=0;
		for(int i=0; i<=n; i++){
			res+=((Math.pow(-1, i)*Math.pow(X, 2*i))/(factorial(2*i)));	
		}
		System.out.println("Loop: "+res);
		
	}
	
	
	public static double rec (int n, double X){
		if (n==0) return (Math.pow(-1, n)*Math.pow(X, 2*n)/factorial(2*n));
		else return  ((Math.pow(-1, n)*Math.pow(X, 2*n)/factorial(2*n))+rec(n-1,X));
	}
	
	public static int factorial (int n){
		
		if(n>1)return n*factorial(n-1);	
		else return 1;
		
	}
	
}

Thanks :)

**JosAH** · 02-02-2012, 01:57 PM

Your factorial method overflows for (reasonably) large values of n.

kind regards,

Jos

**Mapisto** · 02-02-2012, 02:05 PM

Oh.. right!
Forgot how fast factorials grow :|
But that explains only the n problem.

What about X when it's larger than 2 (pi)?

Thanks!

**JosAH** · 02-02-2012, 02:14 PM

Originally Posted by Mapisto

Oh.. right!
Forgot how fast factorials grow :|
But that explains only the n problem.

What about X when it's larger than 2 (pi)?

Thanks!

A Taylor expansion around zero (a McLaurin expansion actually) is quite inaccurate the further away you approximate a function value. Have a look at this link and pay special attention to the figure on the right side of the page. It is a Taylor expansion of the sine function. See how inaccurate it is around 2*pi.

kind regards,

Jos

**Mapisto** · 02-02-2012, 03:32 PM

I really needed this calculous refreshment :|
I get it now :)

Thanks man!

**JosAH** · 02-02-2012, 04:34 PM

Originally Posted by Mapisto

I really needed this calculous refreshment :|
I get it now :)

Thanks man!

You're welcome of course; you could always try the Taylow expansion around pi if you want to know a value (a bit) larger than that; you do need to know the derivatives around pi (instead of zero). See the link I posted in a previous reply.

kind regards,

Jos