Functions of Several Variables PDF

1 FUNCTIONS OF SEVERAL VARIABLES 1.1 Definition and examples Rn = {(x1 , x2 ,..., xn ), x1 , x2 ,..., xn ∈ R}. (x1 , x2 ,..., xn ) is said to be an n-tuple, in geometry it is said to be a point of Rn it is also seen as a vector. A numerical function of n real variables is an applica...

1 FUNCTIONS OF SEVERAL VARIABLES 1.1 Definition and examples Rn = {(x1 , x2 ,..., xn ), x1 , x2 ,..., xn ∈ R}. (x1 , x2 ,..., xn ) is said to be an n-tuple, in geometry it is said to be a point of Rn it is also seen as a vector. A numerical function of n real variables is an application f from a part D of Rn to values in R. We note f : D → R(x1 , x2 ,..., xn ) 7→ f (x1 , x2 ,..., xn ) Or f : D → Rx 7→ f (x) where x = (x1 , x2 ,..., xn ) 1. f (x, y) = x3 + xy − y 2 , D = R2. 1 2. g(x, y, z) = x2 +y 2 +z 2 , D = R3 \ {(0, 0, 0)}. 3. The surface function = xy, volume = xyz. 4. Allometry is the study of scaling relationships between one part of the body and the whole body. An allometric relationship between the mass (M ) and length (L) of fish bodies has the form M = aLb 5. The resistance function of a parallel arrangement of two resistances x and y is given by xy x+y. 1.2 Two-variable function 1.2.1 Domain of definition The domain of definition of a function f (x, y), denoted by Df , is the set {(x, y) ∈ R2 : f (x, y) ∈ R}. In general, to determine Df we go through the following steps: 1. Write the domain. 2. Determine the boundaries. 3. Graphical representation and determination of the regions that constitute Df using par- ticular points located in the regions. p f (x, y) = 4 − x2 − y 2 [-¿] (-4,0) – (4,0) node[right] x; [-¿] (0,-2) – (0,3) node[above] y; [black] (0,0) circle (2pt) node[below left] M1 ; [black] (3,0) circle (2pt) node[below] M2 ; 1. Df = {(x, y) ∈ R2 : 4 − x2 − y 2 ≥ 0}. 2. Determining boundaries: 4 − x2 − y 2 = 0 ⇔ x2 + y 2 = 22 , circle centered at (0, 0) with radius r = 2. 1  3. The circle divides the plane into two regions, take two arbitrary points from these two regions. M1 = (0, 0) and M2 = (3, 0) For M1 one has 4 − 02 − 02 ≥ 0 For M2 one has 4 − 32 − 02 < 0 So Df = the circle and its interior = the closed disk. 1.2.2 Limit and continuity Limit at (0, 0) To calculate lim f (x, y), the first step is to replace x with 0 and y with (x,y)→(0,0) 0, if you find a number or ∞ it’s good. If you find an indeterminate form then you have to make( the change of variable in polar coordinates as follows: x = r cos(θ) y = r sin(θ) θ controls the direction, and so: lim f (x, y) = lim f (r cos(θ), r sin(θ)). (x,y)→(0,0) r→0 Or set y = tx, here t controls the direction, and then lim f (x, y) = lim f (x, tx). (x,y)→(0,0) x→0 If the limit does not depend on θ (or t) and is finite we say it exists. If it depends on θ (or t) or is not finite we say it does not exist. x2 +2y+2 1) lim = 32. (x,y)→(0,0) x+y+3 3 2y 2) lim xx+2x 2 +y 2 = 00 (ID) (x,y)→(0,0) By the change of variable in polar coordinates one finds: lim r(cos3 (θ) + 2 cos2 (θ) sin(θ)) = 0. r→0 x2 +2xy 3) lim 2 2 = 00 (ID) (x,y)→(0,0) x +y This limit depends on θ, so it does not exist. 4) lim √ xy 2 2 = 00 (ID) (x,y)→(0,0) x +y By the √ change of variable y = tx one has: x2 t lim 1+t2 = 0. √ x→0 xy 5) lim 2 2 = 00 (ID) (x,y)→(0,0) x +y This limit depends on t, so it does not exist. Limit at (x0 , y0 ) Set X = x − x0 and Y = y − y0 lim f (x, y) = lim f (X + x0 , Y + y0 ). (x,y)→(x0 ,y0 ) (X,Y )→(0,0) x+y−3 (X+1)+(Y +2)−3 X+Y L= lim 2 = lim 2 = lim 2. (x,y)→(1,2) x +y−3 (X,Y )→(0,0) ((X+1) +(Y +2)−3) (X,Y )→(0,0) X +2X+Y Now by the change Y = tX one obtains L = lim X 2X+tX +2X+tX 1+t = 2+t. X→0 2 Limit at (x0 , ∞) Set X = x − x0 , Y = y1. lim f (x, y) = lim f (X + x0 , Y1 ). (x,y)→(x0 ,∞) (X,Y )→(0,0) ln(X+1+Y ) L= lim y ln(x + y1 ) = lim Y. (x,y)→(1,+∞) (X,Y )→(0,0) By posing Y = tX one obtains L = lim ln(1+(1+t)X) tX = 1+t t. X→0 Limit at (∞, ∞) Set: X = x1 and Y = y1. lim f (x, y) = lim f ( X1 , Y1 ). (x,y)→(∞,∞) (X,Y )→(0,0) sin(X+Y ) L= lim x sin( x1 + y1 ) = lim X. (x,y)→(+∞,+∞) (X,Y )→(0,0) By setting Y = tX one obtains L = lim sin((1+t)X) X = 1 + t. X→0 Continuity: f is continuous at (x0 , y0 ) if: lim f (x, y) = f (x0 , y0 ). (x,y)→(x0 ,y0 ) Study the(continuity at (0, 0) of the function x2 y x2 +y 2 if (x, y) ̸= (0, 0) f (x, y) = 0 if (x, y) = (0, 0) One has by the change y = tx: 2 lim x2x+yy 2 = lim 1+t tx 2 = 0 = f (0, 0). (x,y)→(0,0) x→0 So f is continuous at (0, 0). 1.2.3 Partial derivatives We first give the definition for the general case. The partial derivative of the n-variable function f (x1 , x2 ,..., xn ) with respect to the variable xk (k = 1,..., n), is the derivative of the function xk 7→ f (x1 , x2 ,..., xk ,..., xn ) of the variable xk , considering all the other variables xj as constants (or parameters). This partial derivative of f with respect to xk remains an n-variable function and it is ∂f denoted ∂x k. The partial derivatives of the function: f (x1 , x2 , x3 , x4 ) = 3x21 + 5x23 + ln(x3 x4 ) + x1 x2 are given by ∂f ∂x1 = 6x1 + x2 , ∂f ∂x2 = 15x22 + x1 , ∂f ∂x3 = x13 , ∂f ∂x4 = x14 The partial derivatives of a three variable function f (x, y, z) are denoted by: ∂f , ∂f , ∂f. ∂x ∂y ∂z For f (x, y, z) = xe2z + ln(xyz) one has ∂f ∂x = e2z + x1 , ∂f ∂y = y1 , ∂f ∂z = 2xe2z + z1. For the two-variable function g(x, y) = x2 + xy 2 + 3y 3 + exy one has: ∂f ∂x = 2x + y 2 + yexy , ∂f ∂y = 2xy + 9y 2 + xexy. Now we give the definition of the second order partial derivatives for a two-variable function. 3  The second order partial derivatives of the two-variable function f (x, y) are the partial derivatives of the functions ∂f ∂x and ∂f ∂y. We enumerate four: ∂2f 1) the second order partial derivative with respect to x denoted ∂x2 ∂2f 2) the second order partial derivative with respect to y denoted ∂y 2 ∂2f 3) the second order partial derivative with respect to x and then y denoted ∂y∂x ∂2f 4) the second order partial derivative with respect to y and then x denoted ∂x∂y For the function g(x, y) = x2 + xy 2 + 3y 3 + exy from example 10 one has: ∂2f ∂x2 = 2 + y 2 exy , ∂2f For the function g(x, y) = x2 + xy 2 + 3y 3 + exy from example 10 one has: ∂2f ∂x2 = 2 + y 2 exy , ∂2f ∂y 2 = 2x + 18y + x2 exy , ∂2f ∂y∂x = 2y + (xy + 1)exy , ∂2f ∂x∂y = 2y + (xy + 1)exy ∂2f ∂2f If at a point (x, y) the second order derivatives ∂x∂y and ∂y∂x are continuous, then ∂2f ∂2f ∂x∂y = ∂y∂x. 1.2.4 Critical points and extrema A critical point for a two-variable function f is a couple (x, y) satisfying ∂f ∂x = ∂f∂y =0 A point (x0 , y0 ) is a local maximum of f, if there exists an interval ]a, b[ such that, f (x, y) ≤ f (x0 , y0 ) ∀x, y ∈]a, b[. A point (x0 , y0 ) is a local minimum of f, if there exists an interval ]a, b[ such that, f (x, y) ≥ f (x0 , y0 ) ∀x, y ∈]a, b[. If a function f has a local minimum or maximum at a point (x, y), then that point is a critical point. Let (x0 , y0 ) be a critical point of a two-variable function f, denote: 2 ∂2f 2 R = ∂∂xf2 , S = ∂x∂y , T = ∂∂yf2 and W = RT − S 2. Then 1) If at (x0 , y0 ) one has W > 0, f has at (x0 , y0 ) a maximum if R < 0 and a minimum if R > 0. 2) If at (x0 , y0 ) one has W < 0, f does not have an extremum at (x0 , y0 ). We speak of a saddle point. 3) If at (x0 , y0 ) one has W = 0, one cannot conclude. Study the existence of extrema of the function f (x, y) = x3 + y 3 − 3x − 3y. One has: ∂f 2 ∂2f 2 ∂x = 3x2 − 3, ∂f ∂y = 3y 2 − 3, R = ∂∂xf2 = 6x, S = ∂x∂y = 0, T = ∂∂yf2 = 6y. The ( critical points are solutions of the system 3x2 − 3 = 0 3y 2 − 3 = 0 ⇔ ( x = ±1 y = ±1 4  So there are 4 critical points which are M1 = (1, 1), M2 = (1, −1) M3 = (−1, 1) M4 = (−1, −1). Apply Theorem 3 at these points: 1) At M1 = (1, 1) one has: W = RT − S 2 = 36 > 0 and R = 6 > 0. So, the function f has a minimum at M1. 2) At M2 = (1, −1) and M3 = (−1, 1) one has: W = RT − S 2 = −36 < 0. So, f does not have an extremum at either of these two points. 3) At M4 = (−1, −1) one has: W = RT − S 2 = 36 > 0 and R = −6 > 0. So, f has a maximum at M4. 1.2.5 The differential The differential at the point (x, y) of a two-variable function f is the expression df = ∂f ∂x dx + ∂f∂y dy. More generally the differential at the point (x1 , x2 ,..., xn ) of an n-variable function f is given by: n ∂f ∂f ∂f P ∂f df = ∂x 1 dx 1 + ∂x2 dx 2 + · · · + ∂xn dx n = ∂xk dxk. k=1 The differential is used to calculate errors. For a function z = f (x, y), the question is: What is the error on z knowing the errors on x and y? Let ∆x, ∆y and ∆z be the errors on x, y and z. These errors are positive and one has: x ± ∆x, y ± ∆y and z ± ∆z. From the differential of f one has: (∆z is considered to be the maximum error on z) ∆z = | ∂f ∂x |∆x + | ∂f ∂y |∆y. The area of a rectangle with sides x and y is S = xy. The error ∆S is given by: ∆S = | ∂S ∂x |∆x + | ∂S ∂y |∆y = |y|∆x + |x|∆y. If x = 10 ± 0.1 and y = 20 ± 0.2 then ∆S = (20)(0.1) + (10)(0.2) = 4. If the unit is meters, then S = 200 ± 4 m2. This tells us that in case of sale of this plot of land (by committing these errors) with a price (for example) of $50,000 per m2 there would be a loss of $200,000. 5