M230Lecture 18 fall16 - Lecture notes 18 PDF

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Fall 2016 notes: Quadratic approximation...

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LECTURE 19 In our previous class, we learned that linear approximations are an application of the first partial derivatives. Today, we will learn about quadratic approximations, which are an application of second partial derivatives. Thus, our goal is to “Understand equations both the simple and quadratical” (Major-General’s Song) Quadratic Approximation for Functions of Multiple Variables The material for today’s class is not in your book. There’s a handout you can access on Canvas; go to “Assignments,” and “Taylor polynomials.” The most commonly found type of approximation in math is linear approximation, because linear approximations are so simple. But sometimes, it’s useful to approximate a function by a higher-degree polynomial. Let’s just remind ourselves of a single-variable example. Example Let f (x) = cos x. Find the linear and quadratic approximations to f at x = 0. Solution We will need to find the first and second derivatives. We have f ′ (x) = − sin x, f ′′(x) = − cos x. Thus, f (0) = cos 0 = 1, f ′ (0) = − sin 0 = 0, and f ′′(0) = − cos 0 = −1. The linear approximation is given by L(x) = f (0) + f ′ (0)(x − 0) = 1 + 0(x) = 1. All this says is, when x is close to 0, cos x is close to 1. That’s nice, but not terribly informational. ′′ The quadratic approximation is Q(x) = f (0) + f ′ (0)(x − 0) + f 2(0)(x − 0)2 = 1 + 0x − 2 1 2 x = 1 − x2 . This gives us a much better idea of what cos x is doing near x = 0. Here 2 are cosine and its linear and quadratic approximations (the quadratic approximation is in blue):

You probably remember from Calculus 2 that we don’t have to stop at quadratic approximations. We can take the degree 3 approximation, the degree 4 approximation, and so on. The limit of all these is called a Taylor series. 1

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LECTURE 19

We could go all the way to Taylor series for functions of two variables, but we will stop with the quadratic approximation. Recall that the linear approximation, aka Taylor polynomial of degree 1, to f (x, y) at (a, b) is L(x, y) = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b). The reason why this approximates our function is the following: L(a, b) = f (a, b), Lx (a, b) = fx (a, b), Ly (a, b) = fy (a, b). L(x, y) has the same value and first partial derivatives as f (x, y) at (a, b). To find the quadratic, or degree 2, approximation, we want a function that has the same value, first partials, and second partials as f at (a, b). That function is the following: Q(x, y) = f (a, b)+fx (a, b)(x−a)+fy (a, b)(y−b)+

fyy (a, b) 2 fxx (a, b) (x−a)2 +fxy (a, b)xy+ y . 2 2

Just as a check, let’s see that Qxx(a, b) = fxx(a, b). We have Qx (x, y) = fx (a, b) + fxx (a, b)(x − a) + fxy (a, b)y. If we take the partial with respect to x again, we get Qxx(x, y) = fxx(a, b). Thus, Qxx(a, b) = fxx (a, b). The other second partials of Q also agree with those of f , so Q is the (only) quadratic function that has the same value, first, and second partial derivatives as f at (a, b). 2 +y 2 )

Example Let f (x, y) = e−(x (0, 0).

. Find the linear and quadratic approximations at

Solution We need to calculate the first and second partial derivatives. We have: 2 2 2 2 fx = −2xe−(x +y ) , fy = −2ye−(x +y ) , so 2 2 2 2 2 2 fxx = −2e−(x +y ) − 2x(−2x)e−(x +y ) = (4x2 − 2)e−(x +y ) , 2 2 fxy = 4xye−(x +y ) , and 2 2 fyy = (4y 2 − 2)e−(x +y ) . At (0, 0), we have f (0, 0) = 1, fx (0, 0) = 0 = fy (0, 0), fxx(0, 0) = −2, fxy (0, 0) = 0, fyy (0, 0) = −2. The linear approximation is simply L(x, y) = 1. The quadratic approximation is

LECTURE 19

Q(x, y) = 1 −

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2x2 2y 2 − = 1 − x2 − y 2 . 2 2

Here are the function and its quadratic approximation (the latter in blue):

Example Let f (x, y) = (ex − x) cos y. Find its quadratic approximation at (0, 0). Solution We have fx = (ex − 1) cos y, fy = (ex − x)(− sin y), fxx = ex cos y, fxy = (ex − 1)(− sin y), and fyy = (ex − x)(− cos y ). At (0, 0), these are f (0, 0) = 1, fx (0, 0) = 0 = fy (0, 0), fxx(0, 0) = 1, fxy (0, 0) = 0, and fyy (0, 0) = −1. Note, this is all very similar to what we got in the previous example, except that now fxx and fyy have different signs. Since fxx(0, 0) and fyy (0, 0) are positive and negative respectively, we should anticipate that at (0, 0),f (x, y) curves upward in the x-direction and downward in the y-direction. 2 2 The quadratic approximation at (0, 0) is Q(x, y) = 1 + 21x2 + −1 y 2 = 1 + x2 − y2 . 2 This is a saddle. Here, once again, are the function and its quadratic approximation (the latter in blue):

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LECTURE 19

Example Let f (x, y) = sin(x2 + y 2 ). Find its quadratic approximation at the point p ( π/2, 0). Solution We have fx = 2x cos(x2 + y 2 ), fy = 2y cos(x2 + y 2 ), and fxx = 2 cos(x2 + y 2 ) − 4x2 sin(x2 + y 2 ), fxy = −4xy sin(x2 + y 2 ), fyy = 2 cos(x2 + y 2 ) − 2 4y sin(p x2 + y 2 ). At ( π/2, 0), notice that x2 + y 2 = π/2, and sine of this is 1; cosine of this p is 0. So at this point, f = 1, fx = 0 (since cos = 0), fy = 0 (since y = 0), fxx = 0 − 4( π/2)2 (1) = −2π, fxy = 0, and fyy = 0 − 0 = 0. p Thus, the quadratic approximation is Q(x, y) = 1 − π(x − π/2)2 . Here are the function and its quadratic approximation:

You might ask: what shape can the graph of a quadratic approximation be? Assuming it’s not a flat plane (which would happen if all the second derivatives are 0), it must be a quadric surface, since the degree is 2. But since it’s of the form z = Q(x, y), with z to the power 1, there are only a few possibilities: Elliptic paraboloid (“bowl shape”) – e.g. Q(x, y) = x2 + y 2 Parabolic cylinder – e.g. Q(x, y) = x2 Hyperbolic paraboloid (“saddle”) – e.g. Q(x, y) = x2 − y 2 or Q(x, y) = xy . At some point in the future, we will try to understand local maxima and minima for functions of two variables. Bowl shapes have a local max or min; saddle shapes do not. We’ll develop a “second derivative test” that will tell us, essentially, whether the quadratic approximation at a point is bowl-shaped or saddle-shaped....