(new-plot)

(message "
1. Use Newton's method to generate an approximation to the zeros of the
function

	f(x) = e^x - \frac{1}{x}

Note: f(x)=0 is a transcendental equation: it's solution cannot be obtained
analytically, only approximated. Numerical approximations are important!

	1) Write down

                 x_{n+1} = x_{n} - \frac{f(x_n)}{f^\prime(x_n)}

for this particular case.

	2) Generate at least 5 iterates (x_1 through x_6) using a
calculator or computer.
")
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(func f(x) exp x - 1 / x)
;; trying for initial values students tried:
(newton-root f 1 5)
(newton-root f -1 5)
(newton-root f 3 5)
(newton-root f .6 5)
(newton-root f .568 5)
(newton-root f 2 5)

(domain .1 1)
(newton f 2)
(new-plot)
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(message "
2. Try to solve the following using Newton's method, and report what
happens:

	f(x) = x^2 + 1, \hspace{1in} x_0 = 1

	ln(x) = 0, \hspace{1in}  x_0 = 4
")

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(func f(x) x ^ 2 + 1)
(newton-root f 1 5)
(newton f 1 5)

(func f(x) ln x)
(newton-root f 4 5)
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(message "
3. The first failure in problem 2 is a quirk, but the second we can fix. Try
solving for the root of ln(x) by considering the even function ln(|x|)
instead. Generate 5 iterates. This trick won't work when x_0 = 10, however.
Can you discover a range of initial values for which Newton will work with this
function?
")

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(func f(x) ln (abs x))
(newton-root f 4 5)
(newton-root f 6 5)

(newton-root f 10 5)

;; When the iterates are exactly opposite each other, i.e. bouncing between x
;; and -x, we're done: That is, we must solve -x = x - f(x)/f'(x):

(func bounce(x) (2 * x  - (ln (abs x)) / (1 / x)))
(domain 3 12)
(newton-root bounce 4)
(newton bounce 4)
(new-plot)

;; (this can also be found analytically, to be e^2!)
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(message "
4. As noted in class, some roots can't be found using Newton's method
when the derivative
of the function of interest is infinite at the root (e.g. f(x)=\sqrt{x}):
one possible solution in such a case is to use bisection. Another
idea is to relax Newton's method: consider

	x_{n+1} = x_{n} - m\frac{f(x_n)}{f^\prime(x_n)}

It turns out that for 0 < m < 1 this method will converge to the
true solution of \sqrt{x}=0.

Write down the equation for Newton's iterates for the case of f(x)=\sqrt{x} ,
and simplify.

Generate 5 iterates with this iteration scheme for m=.25, m=.5, and m=.75.

The iterates' behavior depends on the size of m: what is the difference
for the two cases m < .5 and m > .5?

For what values of m will ``relaxed Newton'' give the roots of
f(x)=x^{\frac{1}{3}}?

Can you explain the geometrical significance of the {\it{m-factor}}?
")

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(func f(x) sqrt x)
(newton-root f 1 10 1e-6 .25)
;;(newton-root f 1 10 1e-6 .5) ; gives the zero then crashes, as f'(0)=infinity.
(newton-root f 1 10 1e-6 .75)

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(message "
5. Yesterday we considered the ``smokestack'' problem, which gave rise to the
function

	S(x) = k_2(\frac{7}{x^2}+\frac{1}{(20-x)^2})

Calculate the derivative, S^\prime, and plot it.

Use Newton's method to approximate the root of S^\prime in [0,20], using a
good initial approximation.

Can you bound your approximation? That is, can you find a small interval such
that the root must be inside? (This is related to the problem of finding a root
to a given accuracy, e.g. 2 decimal places.)
")

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(title "smokestack function")
(func s(x) 7 / x ^ 2 + 1 / (20 - x) ^ 2)
(plot s 3 19)
(diff s x sp)
(newton-root sp 10)

(sp 13.133)
(sp 13.135)
;; These are of opposite sign, so the root is between, i.e. is 13.13something.
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(message "
6. A method related to Newton's method is the secant method, obtained by
approximating the derivative f^\prime(x_n) in \ref{nm} by the discrete
approximation using the two previous guesses: i.e.,

	f^\prime(x_n) \approx \frac{f(x_n)-f(x_{n-1})}{x_n-x_{n-1}}


Rewrite Newton's method using this approximation to get the secant method.
Notice that \u{two} initial guesses must be given to start the secant method:
the two points are joined by a line (the secant line) and the root of that
linear function is used as the next approximation.

Draw a picture illustrating the idea behind the secant method.

(Note: the secant method is particularly useful when a function is continuous
but not differentiable, as is true of many real functions. An example is the
motion of a particle in your favorite beverage: it makes random (but
continuous) movements, called Brownian motion.)
")

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(message "
7. Another method, called Muller's method, relies on parabolas, rather than
secant lines: it thus requires \u{three} initial approximations, which are used
to find the unique parabola passing through them. One root of this quadratic
equation is then used as a next approximation. If you ever take the time to
understand this method, you will say to yourself ``This is a really fantastic 
method!''

Draw the graph of

	f(x)=(x - 1)^3 + 4

and draw a parabola (as closely as you can!) through the points associated with
x values of 1, 0 and -1.

Try to write down the quadratic function that passes through the three points:
if you succeed, what is your next approximation to the root using Muller's
method? (The root of f is approximately -0.5874.)
")

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(func f(x) (x - 1) ^ 3 + 4)
(func g(x) -2 * (x - 1) * x + 2 * (x + 1) * x + -3 * (x - 1) * (x + 1))
(plot f g -1 2)
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