Proofs of identities in FM

Transcription

1 Proofs of dettes FM O The Floor 9 May 6 Bomal expaso (P) ( + x) k + (k )! x for k Q Proof: Notce, at frst, that for oegatve tegers k ad, (k ) For ths s also equvalet to! ( ) k k!!(k )! k(k ) (k + )! We temporarly exted ths otato ( ) k of to all real umbers k ad all oegatve tegers, so what we eed to prove s ( ) k ( + x) k x We splt the problem to few steps, each expadg the set of the umbers for whch the above statemet s true k N Assume k sce k s trval Wrte the expresso as the product of k detcal factors + x, ad we kow that for [, k] we have ( k ) ways to choose x for tmes ad for k tmes Ths yelds the coeffcet of x as that the term (k j) j! ( k ) For x where > k smply ote cotas factor whe j k, so we are doe for ths case k Z Proof: Let s do ducto, each tme reducg k For base case k otce that +x s equal to the sum to fty x+x x + ( ) x ( )( ) ( ) x ()() () ( ) x

2 ( ) k Iductve step: Suppose that ( + x) k x for some k < Dfferetatg both sdes yelds k( + x) k x, or ( + x) k k ( ) k ( ) x k k(k ) ( k + ) x (k ) ( k + ) ( ) k x x k ( ) ( ) k x, as desred k Q ( ( ) ) ( k ( ) ( l ( ) k + l Proof: Frst, we prove that x )x )x Notce ( that ths detty already holds for k, l Z, sce from the frst two subproblems we have ( ) ) ( k ( ) ( l ( ) k + l x )x ( + x) k ( + x) l, whle )x ( + x) k+l To prove ths, we eed to verfy that for each, coeffcet of x s the same for LHS ( ) k + l ( )( ) k l ad RHS For RHS t s For LHS ths s Observe also that ( ) k + l ( )( ) k l s a polyomal varables k ad l f s fxed, so ame t P (k, l) Our am s to prove that P (k, l) s detcally zero The fact P (k, l) for k, l Z already follows from above Now fx k as arbtrary teger ( ) k + l ( )( ) k l ad vary l, we kow that s a polyomal varable l ad has degree at most However, every teger s the root of ths moovarable polyomal, t has ftely may roots ad hace must be detcally zero Also that for a k, l R, P (k, l) P (l, k) (P s a symmetrc polyomal) so we ca assert that P (k, l) wheever oe of k, l s a teger Now tur back to P (k, l) aga, ad fx k as ay real umber ad vary l (so P (k, l) s aga a polyomal wth varable l), ad ths tme P (k, l) for ay teger l (aga!) Therefore ths moovarable polyomal must be detcally zero, aga, ad we have P (k, l) for ay par of k, l R ( Fally, let k p q, ad ( ) ) ( k ( ) ( l ( ) k + l x )x )x meas that ( ( ) ) q ( k ( ) ( kq ( ) p x )x )x ( + x) p (sce p s a teger ad ( ( ) k we already prove ths grad detty true for all tegers) Ths meas )x ( + x) p q ( + x) k, or possbly ( + x) k f q s eve Nevertheless, comparg the costat term (LHSRHS) yelds that ( + x) k s mpossble, so we are doe Commet : It s possbe to establsh P (k, l) by algebrac expaso, but too messy Commet : We ca jump from problem drectly to problem, but the proof of problem shows the beauty of dfferetato

3 Calculus If a polyomal a x + a x + + a has roots z, z,, z k, each wth multplcty b, b,, b k respectvely (so b + b + + b k ad b ), the the equato d y a dx + a d y dx + + a dy dx + a y has geeral soluto y A j x j e zx for,,, k ad j,, b ad for A j arbtrary costat Proof: Name the polyomal P Frst, otce that ths geeral soluto has exactly parameters (or degree of freedom) because we ca decde arbtrarly o what should the values f(), f (), f (), f ( ) () ad let the rest follow accordg to our dfferetal equato (we ca t go further tha ths sce the dfferetal equato also mples d +k y a dx +k + a d +k y dx +k + + a d k+ y dx k+ + a d k y for ay k ) ad these degrees dxk of freedom gves parameters, accordg to Maclaur s theorem Hece t suffces to show that x j e zx works (we ca omt ay costat sce that wll oly multply LHS by A j Now let s vestgate what happes as we dfferetate x k e ax geeral We clam, by ( ) ducto, that d dx (xk e ax )a x k e ax + a k(k ) (k + )x k e ax (otce that whe > k such terms cota factor k (k + ) + so t ca be dsregarded ( other words, o term has factor x j e ax for j < ) d ( ) Base case s trval Now assume that dx (xk e ax )a x k e ax + a k(k ) (k + )x k e ax, the d+ dx + (xk e ax ) d ( ) dx (a x k e ax + a k(k ) (k ( ) + )x k e ax )a(a x k e ax ) + ka x k e ax + a(a k(k ) (k + )x k e ax ) + ( ) (k )(a k(k ) (k + )x k e ax ) Now the term x k e ax has coeffcet a + ad x k e ax has coeffcet a + k(k ) (k + )+ a + k(k ( ) ( ) ( + ) (k +) )a + k(k ) (k +) so d+ dx + (xk e ax )a + x k e ax + + ( ) + a + k(k ) (k + )x k e ax, cosstet wth our hypothess To complete our proof, we eed to verfy that the coeffcet of x k e ax s s the dfferetal ( ) equato (for k) Now the coeffcet of x k e ax d dx (xk e ax ) s a k(k ) (k + )x ( k ) e ax, so for a z j (a root of the orgal ( polyomal) ) wth k b j we eed a zj k(k ) (k + ) + a zj k(k ) (k + ) + ( ) ( ) + a zj k(k ) (k + ) + a zj k(k ) (k + ) (Of course ( ) k for k < l) Observe that we actually assumed that z j ; ths lmt case l ca be establshed easly wherby the geeral soluto cotas term x ɛ for some ɛ, (so

4 the polyomal s dvsble by x ɛ+ ad a for ɛ But d a dx xɛ for ay ( ) a p z p p j p d dx xɛ for > ɛ so Now wth ths assumpto the above s same as provg that To prove ths, observe that the root x j has multplcty of more tha k tmes, ad k Ths meas that (x z j ) k+ dvdes the polyomal P, ad t s well-kow that d α dx α P (x) s dvsble by (x z j) k+ α for α k so substtutg to alpha yelds that d α dx α P (x) s dvsble by (x z j) k+ (ad k + ) So d P (x) whe dx x z j Also observe that the t-th dervatve of x u s u(u ) (u t + )x u t, so d d P (x) dx dx a p x p p(p ) (p + )a p x p!p(p ) (p + p p p ( ) p )a p x p! a p x p (for x z j ), as clamed Matrces p Row space ad colum space of ay matrx have the same dmeso Proof: Let dmeso of colum space be WLOG let a a m a a a m, a a a m,, be parwse learly depedet The for j >, a j λ kja k for some real umbers λ kj a k Multplyg each elemet row by a a a a a m a m a { a a { a m a m { λ kj a k } λ kj a k a } λ kj a mk a m } we have: where {} meas that j rus from +, +,, g where g s the umber of colums of ths matrx For covece we ame j as aythg betwee + ad g, clusve (We assume that a If t s, the just swap the whole row wth some lower rows wth ozero frst etry) 4

5 Now ref for the frst elemet each row yelds a a a a a m a m a a { a a a { a m a m a { λ kj a k } λ kj ( a k a a k )} λ kj ( a mk a m a k )} If a xy axy a x a y the the orgal matrx becomes a a a k { λ kj } a a { λ kj a k } a m a m { λ kj a mk } Replacg each a xy wth a xy yelds we ca assume that the frst elemet of eaach (except the frst) row s zero Repeatg the ref process yelds: a a ( ) a a k { λ kj } a a ( ) a { λ kj a k } a { λ kj a k } But otce that the frst elemets of j-th row (j > ) are all zero It follows that λ kj a k ( ) must be zero So all other elemets must be zero ad we have colums osstg etrely of zeroes after the -th row Ths yelds that, the dmeso of row space caot exceed (some of the rows to mght be etrely zero, whch case we kow that dmeso of row space < ), e caot exceed the dmeso of colum space Smlarly the dmeso of colum space caot exceed the dmeso of row space Hece they are equal Multplcatve assocatvty of matrces: (AB)C A(BC) Proof: Let s check the dmeso of each matrx for the equato to be vald Let the 5

6 dmesos of A, B, C be a a, b b ad c c, respectvely For AB to be vald we must have a b ad the resultg matrx has dmeso a b For (AB)C to be vald we eed b c ad the matrx at LHS has dmeso a c A smlar check at RHS yelds a b, b c ad resultg matrx wth dmeso a c, hece the dmeso check s doe Now let s vestgate every sgle etry x at -th row ad j-th colum Deote [ab] xy as the elemet of -th row ad j-th colum of matrx AB; defe [bc] xy smlarly Now b c b c a b for (AB)C ths etry s [ab] k c kj ( a d b dk )c kj a d b dk c kj d d a, k b a b a b b c For A(BC) ths etry s a d [bc] dj a d ( b dk c kj ) a d b dk c kj d d d a, k b Hece the every correspodg etres of (AB)C ad A(BC) are equal QED Gve a matrx P wth o-zero determat, ad r, r, r are vectors such that P r T r T (where T stads for traspose) The P ( ) r r r r r r r T det(p ) Proof: Let P ( ) s s s Observe that r s j for j ad otherwse Ths meas that s s a scalar multple of r + r + (dces take modulo ) (Why? s s perpedcular to both r + ad r + ) Now let costats k be that s k (r + r + ), so k r (r + r + ) However, t s well-kow that r (r r ) r (r r ) r (r r ) det(p ) so k k k det(p ) ad we are doe To verfy ths for k, f r a a, r a a a, r a a a a a a a a a + a a a a a a det the r (r r ) a a a a + a a a a a a a a det(p ) 4 If a matrx Q has learly depedet egevectors r, r, r ad correspodg egevalues a, a,, a, the Q P DP, where P (r r r ) ad D s the dagoal matrx wth a at -th row ad -th colum Proof: Deote, aga, T as the traspose of vector If P s T s T s T the r s j f j ad otherwse Now let matrx x be (r ), the P x has for all elemets except the -th etry, whch s Ths meas DP x D has for all elemets 6

7 except the -th etry, whch s a Therefore (r r r ) a a (r ) a x Ths meas that P DP has exactly the same egevectors (wth correspodg egevalues) as Q, ad we show that such matrx s uque To ths ed, let s show that P DP Q Now we have Q(r ) P DP (r ) a (r ), so (P DP Q)(r ) for,,, But sce r, r,, r are learly depedet, we kow that the dmeso of ull space of P DP Q s, or the rak of ths matrx s Hece P DP Q s a zero matrx QED 5 Let A be a m matrx If there exsts a matrx B such that AB ad BA are detty matrces, the m Proof: If AB ad BA are both defed, the the dmeso of B s m, ad AB ad BA have dmesos m m ad, respectvely If m, wthout the loss of geeralty we ca assume that m < We prove that the matrx BA has rak of at most m, whch forces ts determat to be zero (ad therefore caot be detty matrx) Now let C BA, ad deote ts etry vector maer (v v v ) The v b a + b a + + b m a m b a + b a + + b m a m (a w + a w + + a m w m ) where w j represets the vector b a + b a + + b m a m b j b j, or the j-th colum of matrx B Notce that every vector v b j ca be represeted as x w + x w + + x m w m, where x, x,, x R Ths meas the set {v, v,, v } ca oly spa at most m dmesos, e the dmeso of colum space of ths matrx BA s at most m 6 det(a) det(b) det(ab), where A, B are matrces Proof: Let C AB, ad deote a j as the etry of -th row ad j-th colum of A Defe b j ad c j smlarly The term a pq b rs wll appear as a term C ff q r, whch case t wll be cotaed the etry c ps Moreover, to be cosdered to the determat of C, each term must be the form of c σ() a j b jσ(), where {σ() } s the permutato of {,,, } Idvdually speakg, the product of sums above comprses terms the form a α() b α()σ(), where α(), α(), α() s a sequece of umbers satsfyg α(j), j [, ] Notce also that det(c) 7 j c c c

8 ( ( ) N(γ) c γ() ), where he sum s take across all permutatos γ ad N(γ) s the umber of versos γ Ths meas that the coeffcet of a α() b α()σ() det(c) s well-defed, e the sums of coeffcet of ths term all the terms ( ) N(γ) c γ() We show that f {α() } s ot a permutato of {,,, } (e some of them are equal), the the product a α() b α()σ() has total coeffcet zero det(c) Ad f t s a permutato (meag each term s dfferet), the the coeffcet s + f {σ() } s a eve permutato, ad - otherwse Now for each partcular permutato σ, the terms the (mult-)set {a j b jσ(), j } s parwse dstct, whch meas the term a α() b α()σ() has coeffcet of exactly the expaso of c σ() a j b jσ() (Ie every term the form j )x a xy b yσ(x) (σ permutato of x) ca oly appear at most oce ths expaso) Defe the occureces of the term a α() b α()σ() as the total umber of the permutato σ such that ths term s a term the expaso of c σ () The for each k we ame f(k) as the umber of [, ] st α() k We show that the occureces of a α() b α()σ() s f()! Ideed, ths s equvalet to sayg that a α ()b α ()σ () a α() b α()σ(), or smply b α()σ () b α()σ() as we ca take α α for the purpose of establshg ths proof of our clam Now defe S k as {, α() k} (for k,,, ) ad what we eed s {σ () S(k)} {σ() S(k)}, k [, ] Now σ σ s deftely a soluto, ad for each such k we kow that {σ() S(k)} has f(k)! solutos For each k, there are f(k)! possbltes to very σ, ad multplyg ths for all such k yelds the aswer f()!f()! f()! To fsh ths lemma, we eed to prove that f f(k) for some k, the amog all f()! permutatos σ, exactly half of them s odd ad half eve Ths meas that there must be f()! occureces of a α() b α()σ() that cotbutes postvely to the determat of C (e have coeffcet + ths determat calculato) ad f()! that cotrbutes egatvely, whch gves a overall coeffcet of zero We eed the followg well-kow lemma (ot hard to prove, though) : If teger x, swappg ay two elemets a permutato of x dstct objects wll chage the permutato from odd to eve, ad vce versa Ths allows us to par up all feasble permutatos σ to f()! pars such that ech par has a odd permutato ad eve permutato Let j be a dex [, ] such that f(j), ad choose j ad j such that α(j ) α(j ) j The for each permutato σ, par t up wth aother permutato σ such that: 8

9 σ ( ) σ ( ) σ ( ) σ ( ) σ () σ (), [, ]\{, } Clearly, f σ fulflls the clam that {σ () S(k)} {σ() S(k)}, k [, ] the σ also fulflls the clam that {σ () S(k)} {σ() S(k)}, k [, ] Moreover, each par s dsjot (meag o elemet ca exst two groups) sce the other elemet (e permutato) the same par wth a permutato s defed uquely before Ths completes our bjecto (ad hece our clam) If f(k) for all k, the there s oly oe such occurece of a α() b α()σ(), or σ σ s the oly possble permutato The two prevous terms have coeffcet det(c) ff c σ() has a coeffcet det(c), e σ s a eve permutato or N(σ) s eve Fally, let s prove that for f α s a permutato, the the coeffcet of det(c) s the product of the coeffcet of a α() b α()σ() a α() det(a) ad the coeffcet of b α()σ() det(b) Let I be the detty permutato, e I(x) x, x [, ] De- ote also the permutato that brgs α to σ as ω (meas f α() j ad σ() k, the ω(j) k Try to thk ω as σα ) Now we kow that f a permutato σ s eve, the I ca be mapped to σ by swappg two eghbourg elemets oly by a eve umber of tmes (ad vce versa) I other words, f N deotes the mmum umber of swappgs eeded, the we ca map I to σ by N(α) + N(ω) of eghbourg swappgs (frst N(α) swappg from I to α, the N(ω) swappgs from α to σ) Ths etals N(σ) N(α) + N(ω) (mod ), or ( ) N(σ) ( ) N(α) ( ) N(ω) Sce b α()σ() b ω(), we have: coeffcet of a α() b α()σ() s ( ) N(σ), coeffcet of a α() s ( ) N(α), ad b α()σ() s ( ) N(ω) Ths proves our clam, ad det(c) ( ) N(σ) c σ() ( ) N(σ) a α() b α()σ() (σ ay permutato, α ay - tuple) ( ) N(σ) a α() b α()σ() (same defto for σ but restrct α to permutatos, sce the other case has coeffcet zero ad ca be dsregarded) ( ) N(α) a α() ( ) N(ω) b ω() ( ) N(α) a α() ( ) N(ω) b α()σ() det(a) det(b), as desred 4 Cetrod of bodes Sold hemsphere wth radus r Aswer: 8r from the cetre, ad lyg o the le through cetre ad perpedcular to the plae of the great crcle Proof: Let the radus of the hemsphere be r, the the equato of ths hemsphere s just y r x from to r havg rotated 6 We kow that ȳ To fd x smply 9

10 take the fracto π xy dx π y dx x(r x ) dx r x dx [xr x4 4 ]r [xr x ]r r 4 4 r 8 r Hemsphercal shell wth radus r Aswer: r from the cetre, ad lyg o the le through cetre ad perpedcular to the plae of the great crcle π ( ) r dy xy + dx dx Proof: same equato as above for x ad y, but ow we eed x π ( ) r dy y + dx dx Observe that d ( ) dx ((r x ) 5 x dy ) (r x so y + (r x ) 5 r ) 5 dx (r x ) r The orgal tegral the becomes x xr dx [r( )]r r dx [r(x)] r r r r Crcular sector wth radus r ad subteded agle α r s α Aswer: from the cetre of the crcle (e sector) ad lyg o the agle bsector α of the sector Proof: Same equato as above, for x from r cos α to r, ad y ta α for x from r cos α xy dx to r cos α Now x y dx x ta α dx + r cos α x r x dx cos α x ta α dx + r cos α r x dx Now to deal wth the secod tegral umerator we eed the substtuto x r cos θ, ad r we have dx r s θdθ r cos α x r x dx α r cos θ r r cos θr s θdθ [ ] α α r cos θ s θdθ r s θ r s α (We kow from the area of sector that the deomator s r α) cos α x ta α dx r s α cos α Ths yelds x r s α [ x ta α Addg the two terms up yeld r s α ] r cos α (r cos α) s α cos α (cos α + s α) r s α r s α r α α It s obvous that the cetre of mass of ths sector les o the agle bsector of the sector due to symmetry, hece have gradet cos α rom cetre O (the org) The legth from the org to the cetre of mass has therefore legth r s α r s α α cos α α 4 Crcular arc wth radus r ad subteded agle α Aswer: r s α from the cetre of the crcle (e sector) ad lyg o the agle bsector α of the sector Proof: The equato s gve by y r x, x from rcos α to r As show above, ( ) r + dy xr r dx Here, x r cos α r x dx r x r r The deomator s smply the arc r cos α r dx x legth rα, whle the umerator s equal to [ r r x ] r r cos α r r r cos α r s α Therefore x r s α Aga, dvde t by cos α to fd he dstace from the rα cetre ad we have r s α rα cos α r s α α

11 5 Tragular lama ABC Aswer: Let the mdpot of BC to be M, the the cetre of mass s the pot G o segmet AM, satsfyg AG : GM : Proof: We prove that ths cetrod must le o all medas of a tragle by establshg that the magtude of momet wth respect to AM at both sdes of AM must be the same To do ths, we cut the tragle to very ty strps, each oe parallel to AM, ad prove that for each strp, we ca fd aother strp o the other sde of the meda that has the same legth (or same mass) ad wth the same dstace from the le AM We eed to prove frst that the umber of strps o both sdes must the same, gve the each has wdth d (fsmal) Now, the product of d ad the total umber of strps o sde BM s the perpedcular dstace from B to AM, whch s BM s AMB; the product of d ad the total umber of strps o sde CM s the perpedcular dstace from C to AM, whch s CM s AMC However, BM CM (M s the mdpot of BC) ad s AMB s AMC (these agles are supplemetary) so ths clam s prove Next, cosder a pot P o sde BM, ad Q a pot o sde AM such that P Q s parallel to AM Next, reflect P across M to get P, ad let Q be a pot o AC such that P Q AM Now P Q AM BP BM CP CM P Q AM so P QP Q (The frst equvalece s because tragles BQP ad BAM are smlar; the secod equvalece follows whe BP CP by the defto of reflecto ad BM CM; the thrd equvalece s the cosequece of the fact that tragles CAM ad CQ P are smlar) Fally, the dstace betwee strp P Q ad le AM s P M s BAM ad dstace betwee strp P Q ad le AM s P M s CAM It s easy to verfy that the are the same The smlar proof above ca be used to verfy that the other medas cota the cetre of mass of tragle ABC as well 5 Momet of erta Th rod of mass m legth r about the perpedcular axs through the mdpot Aswer: mr r Proof: Momet of ertam x x dx [ m ]r r r dx [x] r m r r mr r Rectagular lama of mass m ad dmesos a b about the perpedcular axs through the cetre of mass Aswer: m(a + b ) Proof: Cosder al the partcles o the rectagular lama (wth mass dm), so the mass of the partcles s actually dm ad the momet of erta s (x + a x a, b y b a x a, b y b y )dm Turg ftely may partcles to double tegrato (because there are two dmesos!) we have a a ( b b dy) dx for mass, ad a a ( b b x + y dy) dx for momet of erta Now ( b b dy)[y]b b b so a a ( b b dy) dx a a b dx [bx]a a 4ab ( b b x + y dy)[x y + y ]b b x b + b so a a x b + b dx b + xb [x ] a a 4a b + 4ab ( a + b ) 4ab Dvdg t by 4ab ad multplyg by m yelds the momet

12 ( a + b ) of erta as m Dsc/sold cylder wth mass m ad radus r about the axs pepedcular to the crcular plae ad pasg through ts cetre mr Aswer: Proof: Let the heght of sold cylder be h ( the evet of a crcular lama we ca treat ths h as fsmal (whch does t affect our aswer ayway) Now cosder the sub-rg (or sub-cylderc shell) wth cetre the axs ad radus x, ad the surface area s deftely πxh The desred rato ow becomes whch leads to our aswer πx h dx πxh dx x dx [x4/4]r x dx [x /] r r, 4 Sold sphere of mass m ad radus r about a axs passg through ts cetre mr Aswer: 5 Proof: Treat ths sphere as x + y + z r about the z-axs, ad cosder the locus of pots of dstace d from ths z-axs Now d x + y, ad d + z r Therefore z r d or r d z r d The locus s therefore the curved face of cyldrcal shell wth radus d ad heght r d, ad the surace area s πd( r d ) 4πd r d The desred fracto s ow (x )4πx r x r dx 4πx r x dx x r x dx x Now for both de- r x dx omator ad umerator, we use the substtuto x r s θ to get dx r cos dθ, so that x r x dx [ π (r s θ)(r cos θ)(r cos θ dθ) π cos r s θ cos θ dθr ] π θ r Smlarly, x r x dx π (r s θ) (r s θ)(r cos θ)(r cos θ dθ) π r5 s θ cos θ dθ π r5 s θ( cos θ) cos θ dθ π r5 s θ cos θ dθ - π [ cos - r 5 5 θ 5 r 5 5 r 5 ] π [ cos r5 s θ cos 4 θ dθ r 5 θ r5 5 r5 r5 Summarzg above yelds the desred fracto as 5 r, whch yelds the aswer 5 Sphercal shell of mass m ad radus r about a axs passg through ts cetre mr Aswer: Proof: Treat ths shell as the equato y r x, rotated through 6 the x-axs (x rgg from r to r) Assume, too, that the x-axs s our axs of referece Now the dstace of each pot from ths axs s ts y- coordate, so for each y, x ± r x ( ) dy For each of these two x, a crcle wth crcumferece πy + s geerated upo dx reoluto (recall how we foud surface area of revoluto of a graph!) As always, + ( ) dy ( ) r dy dx r Ths meas πy + π r x dx x r πr Ths meas r x ] π

13 ( ) dy r (y )πy + dx dx that our desred fracto s ow ( ) dy r πy + dx dx [πr x rx ]r r [πrx] r r 4πr4 4 r4 4πr r mr so the aswer s r πr(r x ) dx r πr dx 6 Perpedcular axs theorem (apples oly to a lama) Proof: We assume that the lama s o the x y plae, ad Let f(x, y) be the desty at pot x, y Now the mass of ths lama s (gorg lmts) ( f(x, y) dx) dy For each partcle at pot (x, y), the dstace from x-axs s y ad the dstace from y-axs s x Its dstace from z-axs s x + y (e dstace from the org) Therefore I z ( (x +y )f(x, y) dx) dy ( x f(x, y) dx+ y f(x, y) dx) dy ( x f(x, y) dx)dy + ( y f(x, y) dx) dyi y + I x, as desred 7 Parallel axs theorem (apples to a sold ad a lama) Proof: Let the desty at pot (x, y, z) be f(x, y, z), ad ts mass s m ( ( f(x, y, z) dx) dy) dz (aga, droppg lmts) Assume, WLOG, that the cetre of mass les at pot (,,), meag that ( ( xf(x, y, z) dx) dy) dz ( ( yf(x, y, z) dx) dy) dz ( ( zf(x, y, z) dx) dy) dz Now let the axs through the cetre of mass to be the z-axs, ad the other axs to be x a, y b The dstace of sold from the frst axs s x + y ad dstace from the secod axs s (x a) + (y b) Now the momet of erta about the secod axs s ( ( ((x a) + (y b) )f(x, y, z) dx) dy) dz ( ( (x + y + a + b ax by)f(x, y, z) dx) dy) dz ( ( (x + y )f(x, y, z) dx) dy) dz+ ( ( (a + b )f(x, y, z) dx) dy) dz - ( ( axf(x, y, z) dx) dy) dz - ( ( byf(x, y, z) dx) dy) dz I z axs + (a + b ) ( ( f(x, y, z) dx) dy) dz - a ( ( xf(x, y, z) dx) dy) dz - b ( ( yf(x, y, z) dx) dy) dz I z axs + (a + b )m -, e the momet of erta about the frst axs + mr where r s the dstace of the secod axs from the org 6 Regresso les Gve pots (x, y ),,,, The the equato of le of regresso of y o x (x x)(y ȳ) s gve by y ȳ (x x) (x x) Proof: Frst, we prove that for a set of les wth commo gradet, the sum of square of vertcal devato of each pot from the le s mmum whe the le passes through the cetre of mass of the pots Suppose that the gradet s k, ad let θ ta k be the agle (atclockwse) of the le wth the x-axs Now perform a rotato of everythg (the pots ad the le) of agle θ clockwse about ay pot, say, the org The the le ow becomes horzotal, the tal vertcal devato of each pot to the le becomes the θ -degree to the vertcal devato to the le (whch s, cos θ tmes the ew vertcal devato (or perpedcular dstace) to the le: multplato by a costat across all pots) Moreover, the cetre of mass of the pots also follows the same ( rotato (It makes ) ( prefect ) sese tutvely, but to cos α s α x rgorze the argumet, otce that rotates the pot (x, y) as α cos α y

14 gle α atclockwse If α θ the the rotato of pots (x, y ),,,, have cetre of mass ( ) ( ) cos α s α x ( ) ( ) cos α s α x, whch s α cos α s α cos α s the orgal cetre of mass havg rotated agle α) y Now let (x, y ) be our ew pots, ad sce the le s ow horzotal we ca assume that ts equato s y w, w costat Now the vertcal dstace of each pot to the le s y w so the sum of square of dstace s (y w) w w y + (y ) w y ( y ) + (y ) Now the θ degree to the ormal dstace ca be obtaed by dvdg each term by cos θ, whch s a costat across each term! Therefore, the mmum dstace ca be acheved whe w the cetre of mass y y, e passg through To fd the gradet of ths le, otce that the equato of the le must be the form y ȳ b(x x) The vertcal devato from each pot to the le s thus y b(x x) ȳ Therefore the sum of square of vertcal devato s [(y ȳ) b(x x)] b (x x) b (x x)(y ȳ) + (y ȳ) Aga, by completg the square we kow that the least square sum s acheved whe b umerator s equvalet to to (x ) ( ) x x y (x x)(y ȳ) x y Notce, also, that the (x x) ad that the deomator s equvalet 4