2.4 Local properties o unctions o several variables In this section we will learn how to address three kinds o problems which are o great importance in the ield o applied mathematics: how to obtain the approximate value o unctions o several variables near a point in their domain, how to obtain and classiy the extreme values maxima, minima and saddle points) o unctions o several variables and inally, how to solve so-called constrained extrema problems. We will start by addressing the irst o these questions: 2.4.1 Taylor series expansions A good way o obtaining the value o a unction near a point at which the value o the unction and its derivatives are known is by means o Taylor s series expansion. Let us recall it or unctions o one variable: Deinition: Let x) be a unction o one variable with continuous derivatives o all orders at a the point x 0, then the series x) = x 0 ) + x 0 )x x 0 ) + = k) x 0 ) x x 0 ) k, 2.141) is called the Taylor series expansion o about x = x 0. I x 0 = 0 the term Maclaurin series is usually employed in place o Taylor series. Note: In practice when employing Taylor s series we will only consider the irst contributions to the sum above and suppose that they provide already a good enough approximation. In act, i we cut the series 2.141) at k = n we will obtain the best order-n polynomial approximation o the unction x) near the point x 0. The Taylor series expansion can be easily generalized to unctions o more than one variable. Let us as usual consider the case o unctions o two variables: Deinition: Let x, y) be a unction o two real variables which is continuous at a certain point x 0, y 0 ) and such that all its partial derivatives are also continuous at that point. Then the Taylor series expansion o x, y) about the point x 0, y 0 ) can be obtained exactly in the same way as or unctions o one variable. We can irst apply 2.141) to expand the unction on the variable x about x 0 keeping y ixed: 1 k x k x0,y) x x 0 ) k. 2.142) We can now take the expansion 2.142) and treat it as a unction o y. I we do so, we can use again Taylor s expansion or unctions o one variable on 2.142) and expand it about the point y = y 0, p=0 1 p! p k ) y p x x0 k x x 0 ) k y y 0 ) p. 2.143) 29
Note: Notice that i we would have irst expanded about y 0 and then about x 0 the derivatives with respect to x and y in 2.143) would appear in the reverse order. However, since we assume that is continuous and has continuous partial derivatives o all orders at the point x 0, y 0 ) this implies that p k ) y p x x0 k = k p ) x k y x0 p, 2.144) and thereore both ormulae are equivalent. Let us consider the irst terms on the expan- An alternative version o Taylor s ormula: sion 2.143). They are x 0, y 0 ) + x x 0, y 0 )x x 0 ) + y x 0, y 0 )y y 0 ) + yx x 0, y 0 )x x 0 )y y 0 ) + 1 2 yyx 0, y 0 )y y 0 ) 2 + 1 2 xxy 0, x 0 )x x 0 ) 2 + = ϕ 0) x 0, y 0 ) + ϕ 1) x 0, y 0 ) + 1 2 ϕ2) x 0, y 0 ) + 2.145) with ϕ n) x 0, y 0 ) = x x 0 ) x + y y 0) ) n x, y) y x0. 2.146) It is easy to prove that the appearance o the operator ϕ n) x 0, y 0 ) extends to all other terms in the Taylor expansion in act it can be proven by induction, in the same way described ater 2.112)). This means that we can write 2.143) as n=0 ϕ n) x 0, y 0 ). 2.147) n! Note: Notice that the operation 2.146) means that ater expanding the n-power we act irst with the partial derivatives on and then take those derivatives at the point x 0, y 0 ). For example ϕ 2) x 0, y 0 ) = x x 0 ) x + y y 0) ) 2 x, y) 2.148) y x0 ) = x x 0 ) 2 2 x 2 + y y 0) 2 2 y 2 + 2y y 0)x x 0 ) 2 x, y) y x x0 = x x 0 ) 2 xx x 0, y 0 ) + y y 0 ) 2 yy x 0, y 0 ) + 2y y 0 )x x 0 ) xy x 0, y 0 ). Let us see the working o these ormulae with one example: Example: Let x 2 y 3. 2.149) Obtain the Taylor expansion o this unction about the point 1, 1) including up to second order terms. 30
By second-order terms it is meant that we take the terms on the Taylor expansion until second-order partial derivatives. That means that we need to calculate the ollowing sum 1, 1) + x 1, 1)x 1) + y 1, 1)y 1) + 1 2 xx1, 1)x 1) 2 + 1 2 yy1, 1)y 1) 2 + xy 1, 1)x 1)y 1) +. 2.150) Thereore the irst thing we need to compute are the 1st- and 2nd-order partial derivatives o, Thereore we have x = 2xy 3 y = 3x 2 y 2 xx = 2y 3, 2.151) yy = 6x 2 y xy = yx = 6xy 2. 2.152) 1, 1) = 1 x 1, 1) = 2 y 1, 1) = 3, 2.153) xx 1, 1) = 2 yy 1, 1) = 6 xy 1, 1) = 6. 2.154) Theore the expansion 2.150) is given by x, y) 1 + 2x 1) + 3y 1) + x 1) 2 + 3y 1) 2 + 6x 1)y 1) = 6 6x 9y + x 2 + 3y 2 + 6xy. 2.155) We can actually check how good this approximation is near the point 1, 1) by plotting the exact unction and the approximate unction 2.155): Figure 8: The unction x 2 y 3 a) and its Taylor approximation b). In the picture above you can see that near 1, 1) both unctions are very similar. Thereore the approximation 2.155) is quite good there. However i we go a bit ar rom 1, 1), or example the point 0, 0) both unctions are already very dierent. In act unction a) takes the value 0 at 0,0), whereas unction b) is 6 at the same point!. As we see rom the example, it is common to take only a ew terms o the Taylor expansion o a unction around a certain point. It is thereore convenient to have a ormula which tells us precisely the order o magnitude o the error we make when we take only n terms in the expansion 2.147). This ormula is the so-called Taylor expansion ormula with Lagrange s remainder and has the ollowing orm: 31
Deinition: Given a unction x, y) with Taylor expansion 2.147) we can write n ϕ k) x 0, y 0 ) where R n x, ỹ) is Lagrange s remainder or error term and is given by + R n x, ỹ), 2.156) R n x, ỹ) = ϕn+1) x, ỹ), 2.157) n + 1)! where x, ỹ) is a point such that x is a number between x and x 0 and ỹ is a number between y and y 0. Example: Estimate the value o the remainder o the Taylor expansion given in the previous example at the point 1.1, 0.9). In the previous example we considered the unction x 2 y 3 and carried out its Taylor expansion around the point 1, 1) up to 2nd-order terms. We obtained the result which means that x, y) 6 6x 9y + x 2 + 3y 2 + 6xy, 2.158) 1.1, 0.9) 0.88. 2.159) The problem is asking us what error we make when we approximate 1.1, 0.9) by the value 2.159). The answer to this question is given by computing the remainder o the Taylor expansion at the point 1.1, 0.9). According to our deinition o the remainder, we need to compute R 2 x, ỹ) = ϕ3) x, ỹ), 2.160) 3! where x, ỹ) is by deinition a point in between 1, 1) and 1.1, 0.9). We have also seen that ϕ 3) is given by ϕ 3) x, ỹ) = x x 0 ) x + y y 0) ) 3 x, y) y x,ỹ) = x x 0 ) 3 3 x 3 + y y 0) 3 3 y 3 + 3x x 0) 2 3 y y 0 ) x 2 y +3x x 0 )y y 0 ) 2 3 ) x y 2 x, y) x,ỹ) = x x 0 ) 3 xxx x, ỹ) + y y 0 ) 3 yyy x, ỹ) + 3x x 0 ) 2 y y 0 ) xxy x, ỹ) +3x x 0 )y y 0 ) 2 xyy x, ỹ). 2.161) where we have used the act that the order o the derivatives does not matter i and its derivatives are continuous. This allows us to assume xxy = xyx = yxx and yyx = yxy = xyy. Notice that the coordinates o all three points x, y), x 0, y 0 ) and x, ỹ) are involved in the ormula! In order to evaluate the remainder we need to obtain all 3th order partial derivatives o the unction x 2 y 3. They are given by xxx x, ỹ) = 0, yxx x, ỹ) = 6ỹ 2, yyy x, ỹ) = 6 x 2, xyy x, ỹ) = 12 xỹ, 2.162) 32
Thus we obtain ϕ 3) x, ỹ) = 6y 1) 3 x 2 + 18ỹ 2 x 1) 2 y 1) + 36 xỹx 1)y 1) 2. 2.163) For x, y) = 1.1, 0.9) the remainder becomes, R 2 x, ỹ) = ϕ3) x, ỹ) 6 = 0.1) 3 x 2 + 3ỹ 2 0.1) 2 0.1) + 6 xỹ0.1) 0.1) 2. 2.164) As we said beore x, ỹ) is some point lying between 1, 1) and 1.1, 0.9), which means that 1 < x < 1.1 and 0.9 < ỹ < 1. 2.165) Let us or example take the middle point x = 1.05 and ỹ = 0.95 and substitute into the remainder, R 2 1.05, 0.95) = 0.1) 3 1.05) 2 +30.95) 2 0.1) 2 0.1)+61.05)0.95)0.1) 0.1) 2 = 0.002175. 2.166) we ind that the maximum error we make should be smaller than 0.002175. We can now check i this is true by computing the exact value o 1.1, 0.9) = 1.1) 2 0.9) 3 = 0.88209, 2.167) and comparing it to 2.159). We ind that the dierence between the two values is 0.00209 which is indeed smaller than 0.002175. 33