Found this exercise in this PDF pamphlet on some basic problems in numerical analysis.
The exercise, as stated:
Even innocent-looking calculus problems can lead to num-
bers beyond the calculator range. Show that the maximum
value of the function
f(x) = [x^2] / (1.0001)^x
is greater than 10^124
[Hint: Use logarithms.]
Sure it grows large, but I don’t think it’s that big. Is this just a typo, or am I missing something about how calculators compute (ln 1.0001)?
(While I’m at it for anyone who clicks the link, is there a simple way to figure the limit in the first exercise there? So far I can only think of a tedious application of L’Hopital’s rule, but maybe there’s a trig identity that simplifies it.)
Supposing they mean to restrict x to positive values (for there is no maximum if x is allowed to go arbitrarily negative), then, yeah, I believe the book is mistaken; the maximum value is just under 5.5 * 10^7.
As for the limit of 1/sin(x)^2 - 1/x^2 as x goes to 0, recall that the Taylor series for sin(x) is x - x^3/3! + … = x * (1 - x^2/6 + …). Thus sin(x)^2 is x^2 * (1 - x^2/3 + …), and thus 1/sin(x)^2 is 1/x^2 * (1 + x^2/3 + …) = 1/x^2 + 1/3 + … [it is convenient to remember that multiplicative operations on series of the form “1 + linear term + ignored higher order terms” simply amount to corresponding additive operations on the linear terms]. Subtracting 1/x^2 from this and dropping the elliptical terms as x goes to 0, we are left with 1/3. [Note: This is essentially the same reasoning as in the L’Hopital’s rule derivation, but presented in a manner perhaps easier to manipulate for this purpose]
Calculation: The quantity is maximized where its logarithmic derivative is 0. The logarithmic derivative is 2/x - ln(1.0001), so we’re looking for x = 2/ln(1.0001). For small d, we have that ln(1 + d) ~= d, so this is approximately 2/0.0001 = 2 * 10^4. Plugging this into x^2/1.0001^x, we get 4 * 10^8 / (1 + 1/10^4)^(2 * 10^4) ~= 4 * 10^8 / e^2. As e is approximately 2.7, we have that 4/e is approximately 1.5, and thus 4/e^2 is approximately 1.5/2.7 = 5/9 ~= 0.55. Thus we can conclude the maximum value is approximately 5.5 * 10^7. [And paying more careful attention to the error inequalities, we can conclude that the maximum value is actually slightly under this.]
Actually, I’ve phrased that circuitously, so let me rephrase it properly (as you no doubt already have):
Calculation: The quantity is maximized where its logarithmic derivative is 0. The logarithmic derivative is 2/x - ln(1.0001), so we’re looking for x = 2/ln(1.0001). Plugging this into x^2/1.0001^x, we get (4/ln(1.0001)^2)/e^2. As ln(1 + d) is slightly less than d for small d, the numerator is slightly less than 4 * 10^8. Furthermore, as e is slightly more than 2.7, and 4/2.7 is slightly below 1.5, and 1.5/2.7 is 5/9, we see that the answer is slightly below 5/9 * 10^8.
[If forced to manually demonstrate, as I noted before, that this is in fact furthermore less than 0.55, I would at this point calculate out the inequality 4 < 2.7^2 * 0.55 = 7.29 * 0.55 = 4.0095]
I don’t know, I think it’s interesting to watch your sequences of posts asymptotically converge on correctness and concision. It’s like there’s a squeeze theorem involved, balancing your desire for more information between your desire for conveying information as elegantly as possible.
Just wanted to say thanks for the reply (and the explanation on the first exercise too).
I’d been examining it from the point of trying to determine why 10^124 was chosen, and figuring the number of digits, so here’s a bit more on that:
Call the maximum value M, and the value of x at that point x’.
Then M = (x’)^2 / 1.0001^x’
Take the logarithm base 10 (this is the number of digits in the answer, to compare to 124), and rewriting a bit:
log[sub]10[/sub]M = 2*log[sub]10[/sub]x’ - x’*log[sub]10[/sub]1.0001
I didn’t plug in the exact value of x’ (as derived above), but was calculating it as ~= 20001. Further calculated values gives the number of digits as roughly 7.7, much less than 124.
The assumption of considering positive values makes sense in context, although it probably ought to have been explicit. The thrust of that section is that if you are ‘guessing’ at what the value of a limit is by using numerical methods, you can be deceived if you see something blow up really big and think that’s what happens as x goes to infinity. The limit as x -> +infinity in this case is an example (it’s 0, as the exponential growth of the denominator eventually ‘overcomes’ the growth of x^2).
Some day, this board will grant me the unlimited edit window I need. In the meanwhile, all my posts are doomed to linger in various states of imperfection…
No problem. (I also felt like I should have looked for those solutions earlier!)
The third section heading on that PDF says “WHERE CALCULUS IS MORE POWERFUL THAN CALCULATORS AND COOMPUTERS”. At first I thought it was a joke but now I think that’s another typo.
I wonder how many more typos there might be in that document.
