Question

I am not quite satisfied with how quadratic approx formula is derived. Referring to lecture 10, 10:35.
It uses f(x) = a + bx + cx^2 as a model, then recover a, b and c. I am fine with that. But we made special case for x=0, and that's why we end up getting a = f(0), b = f'(0), c = f''(0)/2. But since we made special case for x=0, how can we generalize this formula into any point a, f(x)=f(a) + f'(a)(x-a) + 1/2 f''(a) (x-a)^2. If x is not zero, we won't recover this formula in the first place. Could someone please explain?? Thanks in advance.

Answer 1

I had trouble with this also. Here is how I finally made sense of it. The following argument is similar to the one Prof. Jerison used, but to me at least seems a bit clearer.

The quadratic function Q(x) that best approximates some other arbitrary function f(x) near the point (a, f(a)) has to have three characteristics. First, Q(a) = f(a). In other words, both curves contain the point (a, f(a)). Second, Q’(a) = f’(a). In other words, at point (a, f(a)) both curves have the same slope. And third, Q’’(a) = f’’(a). In other words, the rate at which the slope is changing at point (a, f(a)) is the same for both curves. The claim is that the following function satisfies these criteria:\[Q(x)=f(a)+f'(a)(x-a)+f''(a)\frac{ (x-a)^2 }{ 2 }\]Let’s check to see if it satisfies these criteria. First, it is easy to see that Q(a) = f(a), because when x = a, the second and third terms of the formula become zero. To check the second requirement we need to find the derivative of Q(x). Note that for purposes of finding this derivative, f(a) is a constant, as are f’(a) and f’’(a). This is because Q(x) is constructed by finding the actual values of these expressions and then using them as coefficients in this formula. Here is what we get:\[Q'(x)=f'(a)+f''(a)(x-a)\]We’re hoping to find that Q’(a) = f’(a), and we can now confirm that this is true because when x = a, the second term in this derivative equals zero. To check the third requirement we need to find the second derivative:\[Q''(x)=f''(a)\]And that’s it. Clearly we wouldn’t get this result if we failed to divide by two in the last term of the quadratic formula. We need the factor of one-half to cancel out the 2 that comes from differentiating f’’(a)(x-a)^2.

Just for fun I prepared three images. The first one shows the linear approximation of a function at (a, f(a)). The second one shows a “bad” quadratic approximation, omitting the factor of one-half. You’ll see that it isn’t much improvement, if any, over the linear approximation. But the third one, with the correct quadratic approximation, is much better.

Answer 2

Hmm.. I see. I was also suspecting that Prof Jerison used some kind of "simplified" explanation to the formula. The way you described it make much more sense. And, thanks also for the diagrams. Which f(x) you are using in the diagram by the way?

Answer 3

This is a graph of f(x)=x^3-4x^2+10, but it isn't to scale (the y-axis is compressed by about a factor of 10). Just an arbitrary choice to get something that would look good. L(x) (the linear approximation at x=4) is 10+16(x-4), and Q(x) is 10+16(x-4)+8(x-4)^2. BadQ has 16 in place of 8, i.e., the coefficient that would result if we didn't divide by 2.