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Using Newton's Method to Solve an Implicit Equation

Date: 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/ 
Associated Topics:
College Analysis
College Calculus
College Trigonometry
High School Analysis
High School Calculus
High School Trigonometry

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