Summary - Linearization and differentials - review of logarithms PDF

Preview

Summary

LINEARIZATION AND DIFFERENTIALS - Review of Logarithms...

Description

LINEARIZATION AND DIFFERENTIALS KEITH CONRAD

For a function y(x) that is differentiable at a number a, the function L(x) = y(a) + y′ (a)(x − a)

is called the linearization of y(x) at a. This is the linear function whose graph is the tangent line to the graph of y(x) at x = a. Here are several examples of linearizations, with the graph of y(x) in blue and the graph of L(x) in red. Example 1. Linearize x3 − x2 + x at 1. Here y(x) = x3 − x2 + x and a = 1. Since y′ (x) = 3x2 − 2x + 1, the linearization of y(x) at 1 is L(x) = y(1) + y′ (1)(x − 1) = 1 + 2(x − 1) = 2x − 1. y

x 1 1 + 2x at 0. 3 + 4x Here y(x) = (1 + 2x)/(3 + 4x) and a = 0. Using y′ (x) = 2/(3 + 4x)2 , the linearization of y(x) at 0 is 1 2 L(x) = y(0) + y′ (0)(x − 0) = + x. 3 9 Example 2. Linearize

y

x

1

2

KEITH CONRAD

Example 3. Linearize 1/x at 2. Here y(x) = 1/x, so y′ (x) = −1/x2 and the linearization of y(x) at 2 is 1 1 1 L(x) = y(2) + y′ (2)(x − 2) = − (x − 2) = − x + 1. 2 4 4 y

x 2 √ Example 4. Linearize 3 x at 1. √ Here y(x) = 3 x, so y′ (x) = 13 x−2/3 . The linearization of y(x) at 1 is L(x) = y(1) + y′ (1)(x − 1) = 1 +

1 1 2 (x − 1) = x + . 3 3 3

y

x 1

To illustrate the use of linearizations in making approximations, suppose we want to √ estimate 3 1729.03. The number 1729.03 is close to 1728 = 123 , a perfect cube, so write r r r r √ 1729.03 √ 1.03 3 3 3 3 1729.03 3 3 1729.03 = 1728 . 1729.03 = 1728 = 12 1 + = 12 1728 1728 1728 1728 √ 1 + x when x = 1.03/1728, a rather small number. The linearization We want to estimate 3 √ of the function y(x) = 3 1 + x at x = 0 is 1 1 y(0) + y′ (0)(x − 0) = 1 + (x − 0) = 1 + x. 3 3

LINEARIZATION AND DIFFERENTIALS

3

Therefore r

  1.03 1 1.03 1.03 = 12.002384 . . . 12 1 + = 12 + 4 ≈ 12 1 + · 1728 3 1728 1728 √ and by comparison the actual value of 3 1729.03 is 12.002383 . . ., so the linearization of √ 3 1 + x at 0 gave us an estimate of the cube root of 1729.03 that is correct to 5 digits after the decimal point. This cube root calculation is based on a story the physicist Richard Feynman told about being challenged to make calculations with pencil and paper against an abacus salesman to see who worked faster. See http://www.ee.ryerson.ca/∼elf/abacus/feynman.html Feynman wrote in the middle of his story “I had learned in calculus that for small fractions, the cube root’s√excess is one-third of the number’s excess,” which is precisely the linear approximation 3 1 + x ≈ 1 + 13 x when x is small. 3

Besides their role in making approximations, linearizations are useful in error estimates: they help us estimate the error in y(x) when x undergoes a small change. For historical reasons a small change in x is written in calculus as dx instead of ∆x, and it is just any small number. The corresponding change in the linearization of y at x is called the differential of y and is denoted dy. If x changes by dx and the corresponding change in the linearization of y at x is written as dy then dy = L(x + dx) − L(x) = (y(x) + y′ (x)(x + dx − x)) − (y(x) + y′ (x)(x − x)) = (y(x) + y′ (x)dx) − y(x) = y′ (x)dx.

dy dx, which looks like a rule of dx fractions. But watch out: in this equation, dy on the left and dy in dy/dx are not the same thing, and likewise the dx in dy/dx and the second factor dx are not the same thing. The whole expression dy/dx is a symbol for the derivative y′ (x), while the separate symbol dx is a small change in x and the separate symbol dy is the change in the linearization of y at x dy dx is suggestive, but writing correponding to the change by dx in x. The equation dy = dx ′ it as dy = y (x)dx may help in working with this formula. Let’s go back to the previous four examples and write down dy. We just multiply the derivative y′ (x) by dx each time. Writing dy = y′ (x)dx in Leibniz notation makes it dy =

Example 1. If y = x3 − x2 + x then dy = (3x2 − 2x + 1)dx. 1 + 2x 2 dx. Example 2. If y = then dy = (3 + 4x)2 3 + 4x 1 1 Example 3. If y = then dy = − 2 dx. x x √ 1 Example 4. If y = 3 x then dy = x−2/3 dx. 3 What do these equations with differentials really mean? They tell us a good estimate for the change in y when x changes by a small amount dx. We will look at the four examples using dx = .01 each time. Example 1. If y = x3 − x2 + x, then when x = 1 and dx = .01 we have dy = (3x2 − 2x + 1)dx = (3 − 2 + 1)(.01) = .02. For comparison, the exact change in y when x

4

KEITH CONRAD

changes from 1 to x + dx = 1.01 is ∆y = y(1.01) − y(1) = 1.020201 − 1 = .020201,

and dy = .02 is a good approximation to this. If instead x = 2 and dx = .01 then dy = (3x2 − 2x + 1)dx = (3 · 22 − 2 · 2 + 1)(.01) = .09 while the exact change in y when we pass from x = 2 to x + dx = 2.01 is not far from this: ∆y = y(2.01) − y(2) = .0905. Example 2. If y =

2 1 + 2x , then when x = 0 and dx = .01 we have dy = dx = 3 + 4x (3 + 4x)2

2 (.01) = .0022 . . .. The exact change in y when we move from x = 0 to x + dx = .01 is 32 1.02 1 − = .00219 . . . , ∆y = y(.01) − y(0) = 3.04 3 and dy = .0022 . . . is a good approximation to ∆y. 1 1 1 Example 3. If y = , then when x = 2 and dx = .01 we have dy = − 2 dx = − (.01) = x x 4 −.0025. The exact change in y when x changes from 2 to x + dx = 2.01 is 1 1 − = −.00248 . . . , ∆y = y(2.01) − y(2) = 2.01 2 and the differential dy = −.0025 approximates this well. √ 1 Example 4. If y = 3 x, then when x = 1 and dx = .01 we have dy = x−2/3 dx = 3 1 (1)(.01) = .00333 . . .. The exact change in y when x changes from 1 to x + dx = 1.01 is 3 √ √ 3 3 ∆y = y(1.01) − y(1) = 1.01 − 1 = .003322 . . . , which is approximated well by dy = .00333 . . .

REVIEW OF LOGARITHMS KEITH CONRAD

For a number b > 0 with b 6= 1, the function bx has a graph that looks like one of those below, depending on whether b > 1 (left) or 0 < b < 1 (right). y

y

y = bx

y = bx x

x

b>1

0 1) or decreasing (if 0 < b < 1). In both cases the function bx is one-to-one: x1 6= x2 =⇒ bx1 6= bx2 .

Therefore the function bx has an inverse with domain (0, ∞) and range (−∞, ∞): the inverse function at a positive number x is the number y fitting by = x. We call y the base-b logarithm of x, written as log b x, so logb x is the one number for which b raised to that power is x. That is, blogb x = x, and also log b (bx ) = x. Example. Since 4 = 22 and 8 = 23 , log2 4 = 2 and log2 8 = 3. Since 21 < 3 < 22 , log2 3 lies between the exponents 1 and 2. More precisely, log 2 3 = 1.5849 . . .: this solves 2y = 3. A graph of y = logb x is formed by flipping the graph of y = bx across the line y = x, and is illustrated below. It has the y-axis as a vertical asymptote and no other asymptotes. y

y

x

x b>1

0 0, (2) logb y (3) logb (xy ) = y log b x for x > 0. To derive each of the formulas in (1)–(3) we rely on the characteristic property of a logarithm value: logb x is the only number y satisfying the equation by = x. Proof of (1): Let u = log b x and v = logb y, so bu = x and bv = y. Thus xy = bu bv = bu+v , so logb (xy) = u + v = log b x + logb y. Proof of (2): Let u = log b x and v = logb y, so bu = x and bv = y. Thus x/y = bu /bv = u−v b , so logb (x/y) = u − v = log b x − logb y.

Proof of (3): Let u = log b (xy ), so bu = xy . Also x = blogb x , so xy = (blog b x )y = b(logb x)y = b . Since bu = by logb x we get u = y logb x, so logb (xy ) = y log b x. √ Example. Using formula (3), log2 ( 3) = log2 (31/2 ) = 12 log2 3 and log 5 (1/32 ) = log5 (3−2 ) = −2 log 5 3. Example. While x2 > 0 for x 6= 0, the formula logb (x2 ) = 2 log b x is only true for x > 0. The correct formula when x 6= 0 is log b (x2 ) = 2 logb |x| since x2 = |x|2 and |x| > 0. Warning. Avoid bogus algebraic identities. While there are formulas for logarithms of multiplicative expressions like xy, x/y, and xy , there is no formula for logarithms of additive expressions: log b (x +y) or logb (x − y) can’t be written in terms of log b x and logb y. Unless you can write the number inside a logarithm as a product, ratio, or power, there is no identity for it. The logarithm formulas above all involve a single base. There is an additional formula for logarithms involving two bases b and c, called the change of base formula: log c a . logb a = logc b To derive this formula, rewrite it as a product: (logb a)(logc b) = log c a. We will show this equation is true by raising c to the left side: y logb x

c(logb a)(logc b) = (clog c b )logb a = blog b a = a. Therefore the exponent in the first expression must be logc a, so (log b a)(logc b) = log c a. The change of base formula lets us write a logarithm function in any base b in terms of a logarithm function in any other base c: 1 log c x = log c x. logb x = logc b logc b This means that up to a scaling factor there is basically only one logarithm function. For example, base 2 logarithms can be written in terms of base 10 logarithms and (base e) natural logarithms: ln x log 10 x = . log2 x = log10 2 ln 2 This kind of formula is important to be aware of if you want to calculate a logarithm to base 2 on a calculator and you only have buttons for log10 and ln. While base 10 logarithms are the main kind of logarithm seen in high school, in math and physics the preferred base for logarithms is e on account of special properties of natural logarithms in calculus. In computer science, due to the use of binary representations, the preferred base for logarithms is often 2....