Question

Quadratic approximation... I am trying to solve 2A-7: sec x / (srt (1-x^2)) for x ~0. I have been trying to calculate derivatives and second derivatives (which has been impossible given all the trig fxns.) Is there a faster way of doing this? The solution given for the exercises is much simpler, but I can't figure out how they got it. Thanks.

What does this have to do with the fact that the quad approx of cos x = 1 - x^2/2? (The problem above can be re-written as 1/cosx (srt (1-x^2)), but I don't know how this connects)

Answer 1

Perhaps the main point of this exercise is to show that the shortcut formulas can be used to quickly solve some approximation problems that would otherwise be almost insanely difficult. Let's begin by putting the problem into a form that works well with the shortcut formulas:\[\frac{ \sec x }{ \sqrt{1-x^2 } }=\frac{ 1 }{ \cos x(1-x^2 )^\frac{ 1 }{ 2 } } \]Now we have a denominator that contains a product of two functions for which we have shortcut formulas. If you viewed the recitation videos, you may recall that there was one showing that you can find the linear approximation of a product by finding the linear approximations of the elements separately and then multiplying those approximations together, discarding anything of higher order than 2. For quadratic approximations near zero we have\[\cos x \approx 1-\frac{ 1 }{ 2 }x^2 \]so that's going to be one part of the denominator. The quadratic approximation of the other part is a little tricky. We're using the quadratic approximation for (1+x)^r, which is\[(1+x)^r \approx 1+rx+r(r-1)\frac{ x^2 }{ 2 }\]In this context, r = 1/2 (no problem there), but we also have to plug in -x^2 in place of x, because we're dealing with a power of (1 - x^2) rather than a power of (1 + x). Now, notice that the last part of the shortcut formula includes x^2. If we replace x with -x^2 in that formula we'll get something with x^4, and we're doing a quadratic approximation so we discard anything of order higher than 2. That means we discard that portion of the shortcut formula and we get this:\[(1-x^2 )^\frac{ 1 }{ 2 }\approx 1-\frac{ 1 }{ 2 } x^2 \]The shortcut formula is 1 + rx (plus a quadratic element that gets discarded in this case), where r is the exponent, 1/2, and x is replaced by -x^2.

At this point you may want to note that, by coincidence (or perhaps not), the quadratic approximations of each of the two elements in the denominator are the same.\[\cos x \approx 1-\frac{ 1 }{ 2 }x^2\]\[(1-x^2)^\frac{ 1 }{ 2 }\approx 1-\frac{ 1 }{ 2 }x^2 \]These are good quadratic approximations of two different functions that happen to come out the same. We plug them into the denominator of our formula in place of the elements they represent and we get\[\frac{ 1 }{ \cos x(1-x^2 )^\frac{ 1 }{ 2 } }\approx \frac{ 1 }{ (1-\frac{ 1 }{ 2 }x^2)(1-\frac{ 1 }{ 2 }x^2) }\]Hang in there, because we're almost home. The next step is to multiply the two elements in the denominator together, but once again discard the x^4 term:\[(1-\frac{ 1 }{ 2 }x^2)(1-\frac{ 1 }{ 2 }x^2)\approx 1-x^2 +(discarded term)\]Just one more step! The expression above is in the denominator, so we apply the quadratic approximation formula for (1+x)^r one more time, where r = -1. That gives us the final answer, 1 + x^2.

This may seem like an awful lot, but if you get good at working with these approximation formulas you can solve something like this almost as fast as you can write the answer -- and one heck of a lot faster than you can find the second derivative of a messy function like this.

Answer 2

By the way, there's a fuller discussion of approximations in materials accompanying one of the other calculus courses: http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/readings/a_approximations.pdf

Answer 3

Thanks!!