Calc 1 Lecture NotesSection 3.1Page 1 of 9

Section 3.1: Linear Approximations and Newton’s Method

Big Idea #1 (of 2): If you look at a small enough region of a function near a given point, the function can be approximated pretty well by its tangent line at that point.

Here is a graph of the function y = f(x) = x3 –3x2 + 2x + 1 and its tangent line at the point (2,1). The equation of the tangent line is y = 2x – 3.

Here is the same function and tangent line in the range of –1.9x < 2.1. Note how close they are.

In fact, the percent error between the function’s y-value and the tangent line’s y-value at the left endpoint of the interval [1.9, 2.1] is:

,

while the percent error between the function’s y-value and the tangent line’s y-value at the right endpoint of the interval [1.9, 2.1] is

.

Advantages / Uses of linear approximations:

  1. Many times, errors of the size above are acceptable in exchange for being able to approximate y values of the function with the simple equation of the tangent line. (The closer you are to the point, the better the approximation.)
  2. Linear approximations are useful if we want to calculate values of a function that a calculator can’t handle with its limited number of digits.
  3. Linear approximations can be used to interpolate between data points.

Equation for the tangent line (linear approximation) of a function f(x) at x = x0:

y = L(x) = f(x0) + f(x0)(x – x0)

This comes from the point slope form for the tangent line at the point (x0, y0) = (x0, f(x0)) that has slope m = f(x0):

Practice:

  1. Find a linear approximation to f(x) = sin(x) near x = π/3.
  1. Find a linear approximation to f(x) = sin(x) near x = 0. Use the linear approximation to obtain an approximate value for x = 0.000 000 000 289 6, and compare to your calculator’s answer.
  1. How close to 0 do the x values need to be in order for the linear approximation of sin(x) to be accurate to within 10%?
  1. Find an approximate value for (0.999 999 999 9)½. Compare this to what your calculator returns as an answer.

Note: A good choice for x0 should meet the following two criteria:

  • f(x0) and f(x0) should be easy to evaluate exactly
  • x0 should be close to the number you are trying to plug into the function
  1. Use a linear approximation to calculate an approximate value of .
  1. Use the fact that and to approximate cos(8º) using linear interpolation.

Newton’s Method

Big idea:Newton’s Method is a procedure for finding a numerical approximation for a zero of a function. The method involves finding the tangent line to a function near the zero and then finding where the tangent line crosses the x axis, and then repeating this procedure until a desired precision is achieved. This method is usually much faster than the method of bisections. The general iterative formula for successively better approximations to a zero that results from this method is: .

Example: Let’s try to find the zero near x = 1 of the function f(x) = x4 – 6x2 + 3, whose graph is shown below.

  • How do we know there is a zero is in the interval [0,1]?
  • Show that the four zeros are given exactly by . Use this expression to write down the value of the zero closest to x = 1 to as many digits as your calculator shows.

  1. The first step of Newton’s Method is to make a rough guess for the zero. We’ll call our initial guess x0 and choose x0 = 1. Why choose x0 = 1?
  1. The second step of Newton’s Method is to find the tangent line to the function for the initial guess. Do this now.
  1. The third step of Newton’s Method is to compute where the tangent line crosses the x axis. Do this now. This crossing point should be a much better approximation of the zero of the function than your initial guess x0. If it is not, then you made a bad initial guess. A picture of the situation through this third step is shown below. It is customary to name this next, better approximation x1. You should find that x1 = 0.75.

Just for comparison, what would the method of bisections give us for a first estimate of the zero after starting from the interval [0, 1]?

The rest of Newton’s Method involves repeating the first three steps, except that instead of making a new guess for the zero each time, you use the zero you got from the tangent line you just found. Also, each time you find a new crossing point for the tangent line, it should get labeled with a sequential subscript. The picture below shows the result of applying Newton’s Method a second time. Here we see x1 = 0.74198717, which is pretty darn close to the actual answer.

Just for comparison, what would the method of bisections give us for a second estimate of the zero?

You might have noticed that it gets to be a pain to write down all those calculator digits…

So, let’s re-do the second step above without relying on the numbers:

This representation should help you see the general formula for Newton’s Method, which is called the Newton-Raphson Formula:

In summary, the steps of Newton’s Method are:

1)Make an initial guess x0 “close” to the real zero.

2)Find the equation of the tangent line for that x0:

L(x) = f(x0) + f ’(x0)(x – x0)

3)Calculate where the tangent line crosses the x axis:

L(x) = f(x0) + f ’(x0)(x – x0) = 0 

4)Call this intersection point x1. Go back to step 2 using x1 in place of x0.

5)Repeat to find x2, x3, etc. until you are “close enough.”

To use the Newton-Raphson Formula efficiently, you can program the formula into your graphing calculator and never have to write down a single number until the final answer. See the separate handout with today’s notes, and the image below.

Practice:

  1. What is the zero of f(x) = x3 + 2?
  1. Where do the graphs of y = cot x and y = x intersect?
  1. Approximate .
  1. Derive an iterative formula for computing the square root of any number.
  1. Derive an iterative formula for computing the nth root of any number.