Writing an Arcsin Function in C++Date: 04/17/2002 at 19:17:56 From: Matt Subject: Writing an arcsin function in C++ Hey, I'm taking a computer science class, and I'm stuck on some homework. I have to write a function that computes the arcsin of an input. The book gives this formula: arcsin x = x + (x^3)/3! + x^5 * (3^2)/5! + x^7 * 3^2 * (5^2)/7! + x^9 * 3^2 * 5^2 * (7^2)/9! +... The problem is that I have to do this until a newterm is < 10^-6. What I'm stuck on is the formula. I have to break down the formula above, and write the function. It's Computer Science 111, so I'm just learning C++. The most advanced thing I can use right now is loops, so I was planning on breaking the formula down and using loops to repeat until that 10^-6 marker, but I can't seem to break it down properly. I see a pattern, but the beginning is throwing me. I'd appreciate any help... Thanks a lot. :) Matt Date: 04/18/2002 at 17:52:15 From: Doctor Jeremiah Subject: Re: Writing an arcsin function in C++ Hey Matt, Cool assignment. There are two ways: loops and function calls (the function calls can be used recursively). Since you know about loops and calling functions we could use either, so I will use both. Say we have a function called newterm() that makes a term given the term number: arcsin x = x^1 / 1! (term 1) + x^3 / 3! * (1)^2 (term 2) + x^5 / 5! * (1*3)^2 (term 3) + x^7 / 7! * (1*3*5)^2 (term 4) + x^9 / 9! * (1*3*5*7)^2 (term 5) + ... + x^(2k-1) / (2k-1)! * (product of odd numbers less than 2k-1)^2 (term k) The exponent is the kth odd number. Odd numbers have the form 2k-1 so when k = 5, for example, the exponent will be 2(5)-1 = 5. To calculate an exponent we multiply the base times itself a certain number of times. We will have a function like: double power(double base,long exponent) { // start with power=base and then // use the loop to multiply it the // required number of times double power=base; for(long i=1;i<exponent;i++) power*=base; return power; } The factorial is the product of all the values less than a number and greater than one. As a recursive function we could do something like: long factorial(long value) { // stop when we get to 1 if (value < 2) return 1; // otherwise multiply by the number // that is one lower return value * factorial(value - 1); } So putting in factorial(3) would return 3*2*1 = 6 The product of odd numbers is just as easy: long oddfactorial(long value) { // stop when we get to 1 if (value < 2) return 1; // otherwise multiply by the number // that is two lower (next odd number) return value * oddfactorial(value - 2); } So putting in oddfactorial(7) would return 7*5*3*1 = 105 Then our function newterm() would look like: double newterm(double value,long term) { long oddnum = 2*term-1; // the reason we use oddfactorial(oddnum-2) // is because we don't want to include the // odd number itself, just the ones below it return power(value,oddnum) / factorial(oddnum) * oddfactorial(oddnum-2) * oddfactorial(oddnum-2); } Then the main function must add the terms together. A pretty simple while loop. The while loop just adds the term and checks it see how small it is. Let me know if you need more info on what I did here. If you aren't allowed to use recursion you will have to change factorial and oddfactorial to use loops. If you need help with that let me know. - Doctor Jeremiah, The Math Forum http://mathforum.org/dr.math/ |
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