Math Is Fun Forum

Phew, ok, I finally worked this out.  I'll hide it behind a button so the page doesn't explode.

I don't think you quite understood what I meant by 'optimal' answer.  What I meant was that my answer allows the widest range of values for x that still satisfy the requirements of a limit.  However, in order to provide an epsilon-delta limit proof you only need to provide any delta that satisfies the epsilon requirements.  In the case of problem #1, the book's answer of δ = min{1/6, ∈/18} is more restrictive than my answer.  So it is not 'optimized' in the sense that there are values of δ that satisfy the proof that are not covered by their answer, but it is correct in that it satisfies the requirements.

As for how they reached their answer, I'm not sure.  I haven't done an epsilon-delta proof in a while and don't remember how to simplify them.

For question 1, assume we have a variable

This is equivalent to

Now we just need an expression for δ such that

We go with the 2nd inequality since it is smaller for all positive epsilons.

Edit: My answer is the loosest possible expression for δ.  Obviously the text sacrificed some optimization for the sake of simplicity.  Since you only need to prove that there is some δ > 0 such that |f(x) - L| < ∈ this is fine, you don't need the most optimized expression.

Judging by the thread title I think the number of revolutions per minute is a geometric progression, starting at 1500 and decreasing by 20% each minute.

Multiply the number of revolutions in the 4th minute by 0.8 for your answer.

You can brute force this question, but the easier way is to come up with a function to determine the number of revolutions for the x-th minute.  f(1) = 1500, f(2) = 1500 * 0.8, f(3) = 1500 * 0.8 * 0.8, etc.

Use the function you determined in part ii and solve for f(x) < 100.

Again, find a function s(x) to calculate this value for the x-th minute.

Use the function from part iv to solve for s(x) > 7400.

Use the function from part ii to solve for f(x) <= 0.

I feel sorry for people who had to get through life before calculus.

Right?

Probably true, but I'm something of a 1-trick pony

Since there is a limit at x=3, you can safely cancel as long as you keep in mind that the function is undefined at that point.  More formally, you can rewrite your function as

It doesn't matter that your function is undefined at x = 1/3, since that's the case both before and after the cancellation.  You didn't change that fact by simplifying the function.

First of all, I think you need to be more specific when you say "shortest solution".  I would guess, although I'm not sure, that the algorithm with the best worst-case performance would be different from the algorithm with the best average performance.

Reviving this problem, you can carry bobby's work a little further.  For the sake of redundancy I want to point out that both A and B must be greater than 1 for logs to be meaningful, otherwise you get into divide by 0 problems.  Not a huge issue since you can test A = B = 1 by hand, just wanted to cover all of our bases.

Now, we got to

The left side is clearly an integer.  For the right side to also be an integer we must have B = A^n, where n is a rational number greater than 1 (I won't show the work here, but you can pretty easily see that B > A and thus n > 1).  Rewriting that equation gives us

Now take the log again:

There are 3 possibilities: n < A, n = A, and n > A.  If n = A then the LHS is 0 but the RHS is 1, so n cannot be equal to A.  If n > A the LHS is negative but the RHS is positive, so n cannot be greater than A.  If n < A the RHS is less than 1, so we can narrow our search to A - 1 < n < A.

Off the top of my head I can't think of a way to prove that the solution is unsolvable, but I'm pretty sure it is.

No worries ganesh, just takes switching around some numbers.