Multivariate Calculus - Bruce E. Shapiro
Multivariate Calculus - Bruce E. Shapiro Multivariate Calculus - Bruce E. Shapiro
98 LECTURE 13. TANGENT PLANES Therefore f (4) (x) = − 15 16 (x + 1)−7/2 ⇒ f (4) (0) = − 15 16 P 3 (x) = f(0) + xf ′ (0) + x2 2 f ′′ (0) + x3 3! f (3) (0) = 1 + 1 2 x + 1 ( − 1 ) x 2 + 1 ( 3 x 2 4 6 8) 3 = 1 + 1 2 x − 1 8 x2 + 1 16 x3 and R 3 (x) = f (4) (c) x 4 = −15(c + 1)−7/2 x 4 15x 4 = − 4! 384 384(c + 1) 7/2 for some c between 0 and x. Since smaller denominators make larger numbers, the remainder is maximized for the smallest possible value of c, which occurs when c = 0. So the error at x is bounded by |R 3 (x)| ≤ 15x4 384 For example, the error at x = 1 is no more than 15/384. Looking more closely at the Taylor series f(x) = f(a) + (x − a)f ′ (a) + 1 2! (x − a)2 f ′′ (x 0 ) + · · · we observe that the first two terms f(x) = f(a) + (x − a)f ′ (a) + · · · give the equation of a line tangent to f(x) at the point (a, f(a)). The next term gives a quadratic correction, followed by a cubic correction, and so forth, so we might write f(x near a) = (equation of a tangent line through a) + (quadratic correction at a) + (cubic correction at a) + (quartic correction at a) + · · · For a function of two variables z = f(x, y), the tangent line because a tangent plane; the quadratic correction becomes a parabolid correction; and so forth. The explicit result is the following. Theorem 13.3 Let f(x, y) be infinitely differentiable in some open set J that contains the point (a, b). Then the Taylor Series of f(x,y) about the point (a,b) is f(x, y) = f(a, b) + f x (a, b)(x − a) + f y (a, b)(y − b) + 1 [ fxx (a, b)(x − a) 2 + 2f xy (a, b)(x − a)(y − b) + f yy (a, b)(y − b) 2] 2 + · · · for all points (x, y) ∈ J. Revised December 6, 2006. Math 250, Fall 2006
LECTURE 13. TANGENT PLANES 99 According to the text, “The details are best left to higher-level books.” [8th ed., page 669]. Nevertheless we will at least do one example. Example 13.6 Find a Taylor Approximation to f(x, y) = √ x + y + 1 near the origin Solution. Our point (a, b) is the origin, so that (a, b) = (0, 0). Differentiating, Hence f x (x, y) = f y (x, y) = 1 2 √ x + y + 1 f x (a, b) = f x (0, 0) = 1 2 , f y(a, b) = f y (0, 0) = 1 2 Similarly so that f xx (x, y) = f yy (x, y) = f xy (x, y) = −1 4(x + y + 1) 3/2 Therefore the Taylor expansion is f xx (0, 0) = f yy (0, 0) = f zz (0, 0) = − 1 4 f(x, y) ≈ f(0, 0) + [xf x (0, 0) + yf y (0, 0)] + 1 [ fxx (0, 0)x 2 + 2f xy (0, 0)xy + f yy (0, 0)y 2] + · · · 2 = 1 + 1 2 x + 1 2 y + 1 [− 1 2 4 x2 − 2 4 xy − 1 ] 4 y2 + · · · = 1 + 1 2 (x + y) − 1 8 (x2 + 2xy + y 2 ) + · · · = 1 + 1 2 (x + y) − 1 8 (x + y)2 + · · · Math 250, Fall 2006 Revised December 6, 2006.
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LECTURE 13. TANGENT PLANES 99<br />
According to the text,<br />
“The details are best left to higher-level books.” [8th ed., page 669].<br />
Nevertheless we will at least do one example.<br />
Example 13.6 Find a Taylor Approximation to f(x, y) = √ x + y + 1 near the<br />
origin<br />
Solution. Our point (a, b) is the origin, so that (a, b) = (0, 0). Differentiating,<br />
Hence<br />
f x (x, y) = f y (x, y) =<br />
1<br />
2 √ x + y + 1<br />
f x (a, b) = f x (0, 0) = 1 2 , f y(a, b) = f y (0, 0) = 1 2<br />
Similarly<br />
so that<br />
f xx (x, y) = f yy (x, y) = f xy (x, y) =<br />
−1<br />
4(x + y + 1) 3/2<br />
Therefore the Taylor expansion is<br />
f xx (0, 0) = f yy (0, 0) = f zz (0, 0) = − 1 4<br />
f(x, y) ≈ f(0, 0) + [xf x (0, 0) + yf y (0, 0)]<br />
+ 1 [<br />
fxx (0, 0)x 2 + 2f xy (0, 0)xy + f yy (0, 0)y 2] + · · ·<br />
2<br />
= 1 + 1 2 x + 1 2 y + 1 [− 1 2 4 x2 − 2 4 xy − 1 ]<br />
4 y2 + · · ·<br />
= 1 + 1 2 (x + y) − 1 8 (x2 + 2xy + y 2 ) + · · ·<br />
= 1 + 1 2 (x + y) − 1 8 (x + y)2 + · · · <br />
Math 250, Fall 2006 Revised December 6, 2006.
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