13
$\begingroup$

When covering issues related to error estimates in a calculus course, students find the technique of making estimates (definition of limit, Newton's method, numerical integration, remainder formula for power series) quite difficult even in basic situations. I suspect they would all accept the following fake theorem of error analysis: if you apply a recursion and two successive terms have the same first n digits then the sequence converges and those common initial digits are the first n digits of the limit.

This is of course bogus, and a slowly divergent process like the partial sums of the harmonic series provides a counterexample. What I would like to have available to show the class are examples of convergent recursions from a typical (not artificially created) example of the kind met in a first-year calculus course. A parametric family of examples would be even better, but interesting individual examples would be good too. It would be especially nice if the example has terms agreeing to the first 6 or so digits but the limit is completely different than what the data suggest.

$\endgroup$
  • $\begingroup$ A related error is graphing on a calculator to tell differentiability at a point. For this, I like the example $\sqrt{x^2+\frac{1}{1000}}$, which looks like $|x|$ from afar. $\endgroup$ – Steven Gubkin Jun 10 '15 at 15:39
13
$\begingroup$

How about $\frac{1}{3}x^3-\frac{1}{4}x+\frac{1}{12}+\epsilon$? When $\epsilon=0$, this cubic has two distinct roots: a single root at $-1$ and a double root at $\frac{1}{2}$. If we let $\epsilon>0$ be something relatively small, for example $\epsilon=0.0000001$, and start Newton's method at $1$ the procedure will get caught at the local minimum for a little while before finding the true root near $-1$. With this setup, the tenth and eleventh iterations give an approximations of 0.500529571977707 and 0.500076100908512; leading to the incorrect guess.

$\endgroup$
  • 2
    $\begingroup$ More generally the desired behaviour seems available from many dynamical systems. Imagine placing a torus "vertically" and run gradient flow from near the center stable manifolds of one of the two saddle points; this should give something similar to what Adam described. The difficulty for me is trying to figure out how to make it something "naturally occurring" in the context of a first year calculus course. $\endgroup$ – Willie Wong May 28 '15 at 12:00
  • 1
    $\begingroup$ @WillieWong One could vary that example to be a bit more motivated (to Calc students) than Morse theory: Consider the ODE for a non-linear pendulum: $\theta''=-sin(\theta)$. Modulo $2\pi$, there are two constant solutions, $\theta=0$ and $\theta=\pi$, which correspond to the pendulum being motionless and pointed down and up, respectively. If you pick solutions whose energy is either just above or just below that of the $\theta=\pi$ solution, those solutions appear to converge to the constant vertical before moving on. However, this might be more of a Calc 2 or 3 example. $\endgroup$ – Adam May 29 '15 at 14:14
  • $\begingroup$ This idea is nice, but I'll move the double root to $0$. Let $f(x) = 2x^3-2x^2+\varepsilon$, which for $\varepsilon = 0$ has a double root at $0$ and simple root at $1$. Taking $\varepsilon = 1$, $f(x)$ has unique real root around $-.56519$. If $\varepsilon = 1$ and $x_1 = .1$ then $x_2 = 2.98823\ldots$ and $x_3 = 2.11115\ldots$, but the unique real root of $f(x)$ is not near 2. $\endgroup$ – KCd Oct 18 '15 at 19:32
1
$\begingroup$

Take the sequences

  • $a_n=\sqrt{n+1000}-\sqrt{n}$
  • $b_n=\sqrt{n+\sqrt{n}}-\sqrt{n}$
  • $c_n=\sqrt{n+\frac{n}{1000}}-\sqrt{n}$

For $n<10^6$ we have $a_n > b_n > c_n$, but it is

  • $\lim_{n\to\infty} a_n = 0$
  • $\lim_{n\to\infty} b_n = \tfrac 12$
  • $\lim_{n\to\infty} c_n = \infty$

See https://math.stackexchange.com/questions/1151645/how-does-sqrtn1-sqrtn-fracn1-n-sqrtn1-sqrtn for the technique to find the limits. Now you can take sequences like $\tfrac{1}{a_n}$ for a good counterexample to your problem.

$\endgroup$
0
$\begingroup$

This one was the eye opener for me years ago:

1 + 1/2 + 1/4 + 1/8 + ... The digits settle quickly but the sum is equal to 2.

Is it something you look for?

$\endgroup$
  • 3
    $\begingroup$ But they settle down to 1.99999....., which is 2. So this is not really what I am looking for. $\endgroup$ – KCd May 26 '15 at 17:54
  • $\begingroup$ Please excuse me, I have misunderstood the question and focused on the digits. In my example - any fixed number of sum iterations leads to first three digits being 1.99 where in fact the proper digits were 2.00. This fact has ended my trust in comparing digits years ago. $\endgroup$ – Thinkeye May 27 '15 at 12:19
  • 1
    $\begingroup$ The only numbers with multiple decimal representations are fractions with denominator of the form $2^a5^b$, which have both a finite decimal expansion (infinite string of 0s) and a decimal expansion ending in an infinite string of 9s, e.g., $1/8 = 125/1000$ is both $.1250000\ldots$ and $.1249999\ldots$. There is no reason to lose trust in comparing digits. Unless the number has that special form a close enough approximation will not mislead, and if you ever did find an approximation with several 0s or 9s in a row you could guess the possible rational limit to see if it's right. $\endgroup$ – KCd May 27 '15 at 14:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.