Using Newton's Method to Solve an Implicit EquationDate: 10/14/2007 at 13:33:23 From: Vanessa Subject: solving sine theta/theta = 0.7031. How do you solve the equation sine theta/theta = 0.7031? The answer given for the value of theta was 80.3 degrees and I cannot seem to see how they did it. The text just said "solving numerically, the answer is 80.3 degrees." Thanks for your help! Date: 10/15/2007 at 23:05:33 From: Doctor Minter Subject: Re: solving sine theta/theta = 0.7031. Hi Vanessa, Solving equations like these is the basis for an entire branch of mathematics called numerical analysis. That's why your textbook just says "solving numerically," and leaves it at that. Usually, numerical analysis methods are taught in a numerical analysis (aka "numerics") course, so your textbook (calculus, perhaps?) does not go into how this type of equation is solved. Let me elaborate on what I mean by "this type" of equation. This equation is called "implicit." You cannot isolate the variable on one side of the equal sign in an implicit equation. From your problem, we have sin(t) / t = 0.7031 (where t is used instead of theta) so sin(t) = 0.7031 * t To invert the sine function, we would have to take the arcsine of both sides. That would give us t by itself on the left, but an arcsine function of t on the right. The only way to invert the arcsine function is to take the sine of both sides, and we're back where we started, and no better off. Thus, we cannot solve this equation for t. Because you cannot solve for t, this equation is implicit. For your information, there are a great many ways to find values of t that solve this equation. All of them are approximations. By applying the numerical methods, you can get an approximation that is as accurate as you like (that is, up to as many decimal places as desired), but the result will never be exact. Of course, there's not much difference in two numbers whose first several decimal places are equal, so at some point, we have to say "That's close enough." In my opinion (but it's just an opinion), the best way to solve this type of equation is called Newton's Method, after the same Isaac Newton that you hear about often in basic physics. Other methods that come to mind are the Bisection Method, the Secant Method, and the Fixed Point Algorithm. However, Newton's Method tends to be superior under the right conditions. I won't get too far into what the "right conditions" are, but suffice it to say that this problem definitely has all the right conditions to use Newton's Method. There is some very good information about how Newton's Method works on Wikipedia and Math World at the following links: Newton's Method http://en.wikipedia.org/wiki/Newton's_method http://mathworld.wolfram.com/NewtonsMethod.html The result of the derivation of Newton's Method is t_(n+1) = t_n - f(t_n)/f'(t_n) where t_(n+1) is your "next" approximation, t_n is your "current" approximation, f is the function that you're working with, and f' is that function's derivative. That is, f' = df/dt. The underscores denote subscripts if you're writing it on paper. Make a special note of this above equation because we'll be using it throughout. Let's start with the function f in this problem. What is this function? We have sin(t)/t = 0.7031 How do we get a function out of this? Newton's Method is also called a "root-finding" method, so we need to transform the above equation into something of the form f(t)=0. While it is perfectly legitimate to subtract 0.7031 from both sides to obtain sin(t)/t - 0.7031 = 0, we'll end up having to differentiate this function, so looking ahead, we should use sin(t) = 0.7031 * t sin(t) - 0.7031 * t = 0 where the left side is much easier to differentiate. Thus, we let f(t) = sin(t) - 0.7031 * t and we have our form f(t)=0. Now, we need our t_n. Since we have t_n and t_(n+1, the equation for Newton's Method will always have two t's, and their subscripts will be one number apart. So if we know t_0, we can find t_1, which we can use to find t_2, which we can use to find t_3, and so on. So all we need is t_0 and we can find the rest by re-using the formula for Newton's Method. For your information, this type of equation that we can keep reusing to find more values of t_n is called a "recursive relation." How do we find t_0 ? The answer is, "you don't." That's not to say that we don't NEED a t_0 because we certainly do. But there's no way to FIND t_0. What do we do? Just guess! That may seem like a strange thing to do in math, but it's actually quite common in numerics and differential equations. But we shouldn't just shout out some random number. It's always better to have an "educated guess." How do we come up with an educated guess? Just think of a number that you think might come CLOSE to solving sin(t)/t = 0.7031 A good way to do so is to rewrite this equation as sin(t) = 0.7031 * t and try to come up with a value of t that makes the two sides equal. A graph is a very powerful tool in this case. Graph the function sin(t) and the function 0.7031 * t on the same axes. The intersection point is the value of t that we're ultimately looking for, because that's the value that solves the implicit equation. I want to take a second to tell you that this problem should be solved in RADIANS, and not DEGREES. Many of the mathematical formulas are designed for use in radians, and not degrees. Many methods will not work if degrees are used instead of radians, and this method that we're working with here is no exception. One radian is approximately 57.3 degrees. Thus, why don't we start with t_0 = 1 radian? Beforehand, we know that the answer is just over 80 degrees, so we expect the answer to be between 1 and 2 radians. It doesn't matter if t_0 is too high or too low. It's just a starting point to work with. Now, we need to differentiate f(t) because Newton's method involves the function f'(t). Basic differentiation rules will show that f'(t) = - cos(t) - 0.7031 so f'(t) = - (cos(t) + 0.7031) Now we have all of the pieces that we need. We can start with the Newton's Method formula with n=0. Thus, let t_0 = 1, and let f(t_0) = sin(t_0) - 0.7031 * t_0 = sin(1) - 0.7031 and f'(t_0) = - (cos(t_0) + 0.7031) = - (cos(1) + 0.7031) Putting these pieces into the formula gives us t_1 = 1 + (sin(1) - 0.7031*1)/(cos(1) + 0.7031) where the two minus signs (one from the formula, one from the denominator) have canceled. That's why there's a plus after the 1, when there was a minus before. Putting this into a calculator or computer shows that t_1 is approximately 1.111. The next step is to plug the number we just obtained BACK into the formula for Newton's Method, with n=1. Thus, t_2 = 1.111 + (sin(1.111) - 0.7031*1.111)/(cos(1.111) + 0.7031) Putting this into a calculator, we see that t_2 = 1.211. Now we plug 1.211 into the formula with n=2 to obtain t_3 = 1.291, plug that into the formula with n=3 to obtain t_4 = 1.346, and so on. The drawback to these "recursion relations" is that you can't skip ahead. Note that we needed t_1 to obtain t_2, and t_2 to obtain t_3, and so on. Nevertheless, recursion relations are powerful tools indeed, as long as you have the beginning terms to get you going. In this case, that was t_0, which, for obvious reasons, is sometimes called an "initial guess." If you keep plugging in these numbers, you'll soon notice that with each step, they don't change very much as you keep going. When you get to t_10 and t_11, you should notice that both are 1.402, out to the thousandths decimal place. Of course, t_10 and t_11 aren't EXACTLY equal, but they're very close, and recall that at some point, we have to say, "That's close enough." When the answers stop changing by very much, that means that you're very close to the actual value of t that solves the original equation. Thus, since t_10 and t_11 are so close, let's use t=1.402 as our approximation to the answer. Let's see if it works: sin(1.402)/1.402 = 0.7031 And it does! Now, we note that 1.402 is approximately 80.33 degrees, using 1.402 * 180 / pi. So we have arrived at the answer that your textbook gave, which is a very good sign. Please refer to the links above to see exactly how Newton's Method works, if you're interested. It's very clever. Please also note that sometimes Newton's Method DOES NOT WORK, and also that sometimes Newton's Method can be VERY SLOW, in that you may have to repeat the process hundreds or even thousands of times to get a good approximation. This problem, however, worked very nicely. Keep in mind that the closer your initial guess is to the right answer, the less times you have to apply the formula to get a good approximation. I hope this has helped. Please feel free to write again if you need further assistance, or if you have any other questions. Thanks for using Dr. Math! - Doctor Minter, The Math Forum http://mathforum.org/dr.math/ |
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