Recitation 3: More Applications of the Derivative


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1 Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech edge consists of distinct unordered pir of distinct vertices For exmple the pentgon 4 3 cn be thought of s the grph with V = { } E = { {1 2} {2 3} {3 4} {4 5} {5 1} }. Given grph G nd vertex v V we cn define the degree of v to be the number of edges tht hve v s one of their endpoints. For exmple every vertex in the pentgon bove hs degree 2 s ech vertex is the endpoint for precisely two edges. A tringle decomposition of grph G is wy to brek its edge set E prt into disjoint tringles. It s not too hrd to show tht if tringle decomposition exists then every vertex needs to hve even degree: this is becuse ech of tringle s vertices hs precisely two edges from the tringle hitting it so if we ve broken ll of our edges into tringles every vertex hs degree equl to 2 (the number of tringles tht use tht vertex) which is even. The number of edges in E hs to be multiple of 3 becuse ech tringle uses three edges nd they re ll disjoint. Suppose tht G is grph on n vertices 1. (Known not too hrd:) Show tht if the degree of every vertex is n 1 (i.e. every vertex is connected to every other vertex) these conditions re lso sufficient: i.e. if ll of the degrees re even nd the number of edges is multiple of 3 tringle decomposition exists. 2. (Known firly hrd:) Show tht if the degree of every vertex is ( ) n i.e. the vertices re connected to lmost every other vertex these conditions re still sufficient. 1
2 3. (Current reserch results I m working on) Show tht if the degree of every vertex is ( ) n then these conditions re still sufficient. 4. (Conjecture probbly wildly difficult) Show tht if the degree of every vertex is 3 4 n then these conditions re still sufficient. 2 Directionl Derivtives In our lst clss we spent lot of time working on the ide of derivtive in R n. Specificlly we introduced the ide of prtil derivtives used these to construct the totl derivtive ( liner pproximtion to our function t point which ws similr to the liner pproximtion ide tht we hd for the derivtive in R 1 ) nd discussed some of the geometric interprettions nd pplictions of these ides. However these techniques were not the only wys to tlk bout the derivtive! Another different wy is to generlize the ide of slope from R 1 vi the concept of the directionl derivtive: we discuss how to do this below. In R 1 the derivtive of function f : R R t some point is the slope of the grph f(x) = y t the point ( f()): essentilly we re mesuring the chnge in f(x) s we move long the xxis. However for functions R n R we re no longer restricted to just the xxis; insted we cn move long ny vector in R n! This leds us to define the directionl derivtive of function f : R n R t some point long some direction v s the slope of f t the point s mesured in the direction v. More formlly: Definition. The directionl derivtive of function f : R n R t some point long some direction v is the derivtive f (; v) := d dt (f( + t v). To illustrte wht s going on here consider the following exmple: Question 2 Consider the function f(x y) = x 2 + y 2. Wht is the directionl derivtive of this function t the point (0 1) in the direction (0 1)? Solution. First to get good ide of wht s going on in this problem we grph our function: z y x f (;v) v 2
3 Visully if we look t the point (0 1) nd its slope in the direction (0 1) we cn see tht it should be 1 just by exmintion. So let s clculte nd see if our visul intuition mtches our mthemticl definition: f ((0 1); (0 1)) = d dt (f((0 1) + t (0 1))) = d dt (f(0 t 1)) = d dt ( (t 1) 2 ) = d dt ( t 1 ) = d dt ( ( (t 1))) [ becuse ner 0 (t 1) is negtive] = d dt (t 1) = 1 = 1. t=0 This mtches our visul picture nd our intuition. The following result cn mke clculting the directionl derivtive esier in the cse tht we lredy know the grdient of function: Theorem 3 Suppose tht f is differentible t some point : one notble cse where this hppens is when ll of f s prtil derivtives re continuous t s we mentioned lst clss. Then the directionl derivtive of function f : R n R t some point long some direction v is given by the dot product of the grdient of f t with v/ v. In other words f v (; v) := f v. To illustrte the use of this theorem return to our cone problem from erlier. There we hd f(x y) = x 2 + y 2 ; thus if we hold y constnt we cn see tht Similrly by holding y constnt we hve f x = 2x 2 x 2 + y = x 2 x 2 + y. 2 f y = y x 2 + y 2. 3
4 Therefore we know tht the directionl derivtive of f t (0 1) in the direction (0 1) is given by ( ) ( ) f f (0) (0 1) (0 1) (0 1) = x y ( 1) ( 1) (0 1) ( 1) 2 = (0 1) (0 1) = 1 which mtches our erlier nswer. 3 HigherOrder Derivtives nd their Applictions Another thing we could wnt to do with the derivtive motivted by wht we were ble to do in R 1 is the concept of higherorder derivtives. These re reltively esy to define for prtil derivtives: Definition. Given function f : R n R we cn define its secondorder prtil derivtives s the following: = ( ) f. x i x j x i x j In other words the secondorder prtil derivtives re simply ll of the functions you cn get by tking two consecutive prtil derivtives of your function f. A useful theorem for clculting these prtil derivtives is the following: Theorem 4 A function f : R n R is clled C 2 t some point if ll of its secondorder prtil derivtives re continuous t tht point. If function is C 2 then the order in which secondorder prtil derivtives re clculted doesn t mtter: i.e. for ny i j. x i x j = x j x i It bers noting tht if the conditions of this theorem re not met then the order for computing secondorder prtil derivtives my ctully mtter! One such exmple is the function { x 3 y xy 3 (x y) (0 0) f(x y) = x 2 +y 2 0 (x y) = (0 0) At (0 0) you cn clculte tht xy = 1 1 = yx : result tht occurs becuse the secondorder prtils of this function re not continuous. 4
5 However the interesting spects of higherorder prtil derivtives re not relly in their clcultion; rther the pplictions of higherorder prtil derivtives re the things worth studying. In R for exmple we could turn the second derivtive of function into lot of informtion bout tht function: in prticulr we could use this second derivtive to determine whether given criticl point ws locl minim or mxim whether function is concve up or down t given point nd wht the secondorder Tylor pproximtion to tht function ws t point. Cn we do the sme for functions from R n to R? As it turns out the nswer is yes! The tool with which we do this is clled the Hessin which we define here: Definition. The Hessin of function f : R n R t some point H(f) (h) is the following function from R n to R: H(f) (h) = 1 x 2 (h 1 x 1 ()... h x 1 x n () 1 1 h 2... h n ).... h 2. x nx 1 ().... x nx n () h n The min useful property of the Hessin is the following: Theorem 5 Let f : R n R be function with welldefined secondorder prtils t some point nd H = H(f) be its Hessin. Pick ny two co ordintes x i x j in R n : then x i x j () = 2 H h i h j. In other words H s secondorder prtil derivtive re precisely the secondorder prtil derivtives of f t! So H is bsiclly function designed to hve the sme secondorder prtils s f t. One quick thing this theorem suggests is tht we could use H(f) to crete secondorder pproximtion to f t in similr fshion to how we used the derivtive to crete liner (i.e. firstorder) pproximtion to f. We define this below: Theorem 6 If f : R n R is function with continuous secondorder prtils we define the secondorder Tylor pproximtion to f t s the function T 2 (f) ( + h) = f() + ( f)() h + H(f) (h). You cn think of f() s the constnt or zeroth order prt ( f)() h s the liner prt nd H(f) (h) s the secondorder prt of this pproximtion. To illustrte how this process ctully cretes pretty decent pproximtion to f we clculte n exmple: 5
6 Exmple. Clculte the secondorder Tylor pproximtion to the function f(x y) = e xy t the origin. Answer. Clculting the second derivtives of f is pretty strightforwrd: f x = yexy f y = xexy x 2 = y2 e xy y 2 = x2 e xy xy = xyexy + e xy = yx. If we evlute these prtils t 0 nd plug them into the definition bove for T 2 (f) (00) we get T 2 (f) (00) ((0 0) + (h 1 h 2 )) = f(0 0) + ( f)(0 0) (h 1 h 2 ) + H(f) (00) (h 1 h 2 ) = 1 + (0 0) (h 1 h 2 ) + 1 ( ) ( ) (h h1 1 h 2 ) 1 0 h 2 = ( ) 2 (h h2 1 h 2 ) h 1 = (2h 1h 2 ) = 1 + h 1 h 2. So t the origin the secondorder Tylor pproximtion for f is just T 2 (f)(x y) = 1 + xy. The following grph with e xy in solid red nd T 2 in dshed blue shows tht it s ctully somewht decent pproximtion t (0 0): z y x 6
7 As well we cn use the second derivtives to serch for nd find locl minim nd mxim! We define these terms here: Definition. A point R n is clled locl mxim of function f : R n R iff there is some smll vlue r such tht for ny point x in B (r) not equl to we hve f(x) f(). A similr definition holds for locl minim. So: how cn we use the derivtive to find such locl mxim? Well it s cler tht (if our function is differentible in neighborhood round this point) tht no mtter how we move to leve this point our function must not increse in other words for ny direction v R n the directionl derivtive f ( v) must be 0. But this mens tht in fct ll of the directionl derivtives must be equl to 0! becuse if f ( v) ws < 0 then f ( v) would be > 0. This motivtes the following definitions nd bsiclly proves the following theorem: Definition. A point is clled sttionry point of some function f : R n R iff (f) = (0... 0). A point is clled criticl point if it is sttionry point or f is not differentible in ny neighborhood of. Theorem 7 A function f : R n R ttins its locl mxim nd minim only t criticl points. However it bers noting tht not every criticl or sttionry point is locl mxim or minim! A trivil exmple would be the function f(x y) = x 2 y 2 : the origin is sttionry point yet neither locl minim or mxim (s f(0 ɛ) < 0 < f(ɛ 0) nd thus there re positive nd negtive vlues of f ttined in ny bll round the origin where it is 0.) How cn we tell which sttionry points do wht? Well in onevrible clculus we used the ide of the second derivtive to determine wht ws going on! In specific we knew tht if the second derivtive of function f t some point ws negtive then tiny increses in our vrible t tht point would cuse the first derivtive to decrese nd tiny decreses in our vrible t tht point would cuse the negtive of the first derivtive to increse i.e. cuse the first derivtive to decrese nd therefore mke the function itself decrese! Therefore the second derivtive being negtive t sttionry point implied tht tht point ws locl mxim. In higher dimensions things re tricker t given point we no longer hve this ide of single second derivtive but insted hve mny different second derivtives like xy () nd (). Yet we cn still use the sme ides s before to figure out wht s going z 2 on! In prticulr in one dimension we sid tht we wnted tiny positive chnges of our vribles to mke the first functions decrese. In other words given ny of the prtils f x i we wnt ny positive chnges in the direction of this prtil to mke our function decrese i.e. we wnt the directionl derivtive of f x i to be negtive in ny direction v where ll of the coördintes of v re positive. (Positivity here stems from the sme reson tht in one dimension we hve tht the first derivtive is incresing for ll of the points to the left of mxim nd decresing for ll of the points to the right of mxim.) 7
8 So: this condition if we write it out is just sking tht for every i nd nonzero v tht ( 2 ) f () ()... () (v x 1 x i x 2 x i x n x 1 2 v vn) 2 i is negtive. If you choose to write this out s mtrix this ctully becomes the clim tht for ny v 0 we hve x v T 1 x 1 ()... x 1 x n ()..... v < 0. x nx 1 ()... x nx n () liner lgebr you my hopefully remember tht ny mtrix stisfying this condition is clled being negtivedefinite nd is equivlent to hving ll n of your eigenvlues existing nd being negtive. Similrly if we were looking for locl minim we would be sking tht the bove mtrix product is lwys positive: i.e. tht the mtrix is positivedefinite which is equivlent to ll of its eigenvlues being positive. But we ve seen this construction before! In prticulr this mtrixproduct thing bove is just the Hessin! Bsed on these observtions we mke the following definitions nd observtions: Definition. The Hessin H(f) of function f t some point is clled positivedefinite if for ll h 0 H(f) (h) > 0 nd similrly tht the Hessin is negtivedefinite if for ll h 0 H(f) (h) < 0. Theorem 8 The Hessin H(f) is positivedefinite if nd only if the mtrix x 1 x 1 ()... x 1 x n ()..... x nx 1 ()... x nx n () is positivedefinite. (The sme reltion holds for being negtivedefinite.) Theorem 9 A function f : R n R hs locl mxim t sttionry point if ll of its secondorder prtils exist nd re continuous in neighborhood of nd the Hesssin of f is negtivedefinite t. Similrly it hs locl minim if the Hessin is positivedefinite t. If the Hessin tkes on both positive nd negtive vlues there it s sddle point: there re directions you cn trvel where your function increse nd others where it will decrese. Finlly if the Hessin is identiclly 0 you hve no informtion s to wht your function my be up to: you could be in ny of the three bove cses. A quick exmple to illustrte how this gets used: 8
9 Exmple. For f(x y) = x 2 + y 2 g(x y) = x 2 y 2 nd h(x y) = x 2 y 2 find locl minim nd mxim. Solution. First by tking prtils it is cler tht the only point t which the grdient of these functions is (0 0) is the origin. There we hve tht H(f) = (00) [ ] H(g) = (00) [ ] H(h) = (00) [ nd thus tht f is positivedefinite t (00) g is negtivedefinite t (00) nd h is neither t (00) by exmining the eigenvlues. Thus f hs locl minim t (0 0) g hs locl mxim t (00) nd h hs sddle point t (00). ] 9
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