Math 17 Fall 013 Homework Solutons Due Thursday Sept. 6, 013 5pm Ths homework conssts of 6 problems of 5 ponts each. The total s 30. You need to fully justfy your answer prove that your functon ndeed has the specfed property for each problem. Please read ths week s lecture notes before workng on the problems. Queston 1. The followng are several possble strateges to prove Cauchy-Schwarz: x y x 1 y 1 + + x N y N (x 1 + +x N ) 1/ (y 1 + + y N ) 1/ x y. (1) Pck any one (or come up wth your own) dea and wrte down a detaled proof. Approach 1. Mathematcal nducton. Approach. Let t R. Then (x ty) (x ty) 0 for all t. Wrte the left hand sde as a quadratc polynomal of t. Approach 3. Use x y ( x ) (y k) 1 k (x k + y k ). Choose approprate k. Soluton. Approach 1. Though the case N 1 s trval. For reasons that wll be clear n a few lnes, we have to prove N. Ths s done n Sept. 16 s lecture and s omtted here. Now we try to prove the case N k + 1 assumng We have x 1 y 1 + +x k y k (x 1 + +x k ) 1/ (y 1 + + y k ) 1/ () x 1 y 1 + +x k y k + x k+1 y k+1 x 1 y 1 + + x k y k + x k+1 y k+1 Note that n the last nequalty we have used the N case. (x 1 + +x k ) 1/ (y 1 + + y k ) 1/ + x k+1 y k+1 ((x 1 + + x k )+ x k+1 ) 1/ ((y 1 + + y k ) + y k+1 ) 1/ (x 1 + +x k+1 ) 1/ (y 1 + + y k+1 ) 1/. (3) Approach. Snce (x ty) (x ty) (y y) t (x y) t+(x x), the fact that t s non-negatve mples whch gves Cauchy-Schwarz. [ (x y)] 4 (y y)(x x) 0 (4) Approach 3. Let k R to be determned later. We have x 1 y 1 + + x N y N 1 [ ] x1 + +x N k +k (y 1 + + y N ). (5) 1
Math 17 Fall 013 Homework Solutons Now take The proof ends. k (x 1 + + x N ) 1/ (y 1 + + y N ) 1/. (6) Queston. Let E R N. Defne ts dstance functon d: R N R as d(x) 6 Prove that x, y R N, d(x) d(y) x y. nf dst(x, y) nf x y. (7) y E y E Proof. Frst we prove d(x) d(y) x y. We have, for any z E, d(x) dst(y, z) nf dst(x, w) dst(y,z) w E dst(x, z) dst(y,z) x z y z x y. (8) Here we appled trangle s nequalty n the last nequalty. Note that x y s ndependent of z. Therefore we can take nfmum and obtan d(x) d(y) d(x) nf dst(y, z) x y. (9) z E Fnally notcng the symmetry between x and y, we have d(y) d(x) y x x y. (10) Summarzang the above, we have d(x) d(y) x y. Queston 3. a) Prove that the followng are both norms on R N : x 6 max { x }; x 1 6 x 1 + x + + x N ; (11) 1,,N b) Let X be a lnear vector space wth norm. Prove the followng: If one can defne an nner product (, ) such that x (x, x) 1/, then for any x, y X, x + y + x y ( x + y ). (1) c) Fnd a norm on R N that cannot be defned through an nner product. Justfy your answer. Soluton. a) We check. x 6 max 1,,N { x } 0; x 0 max x 0 x 0 for all 1,,, N x0; x 1 6 x 1 + x + + x N 0; x 1 0 x 1 + x + + x N 0 x 0 max x 0 x 0 for all 1,,,N x0.. a x max { a x } max { a x } a max { x } a x ; ax 1 ax 1 + ax + + a x N a ( x 1 + x + + x N ) a x 1.
Due Thursday Sept. 6, 013 5pm 3. (Trangle nequalty). x+ y max x + y max ( x + y ) max x +max y x + y. (13) x+ y 1 x 1 + y 1 + x + y + + x N + y N x 1 + y 1 + + x N + y N ( x 1 + x + + x N ) +( y 1 + y + + y N ) x 1 + y 1. (14) b) We have x + y + x y (x+ y,x+ y) +(x y,x y) (x, x) + (x, y) + (y, x) +(y, y) +(x,x)+ (x, y) +( y, x) +( y, y) (x, x) + (x, y) +(y, y) +(x,x) (x, y)+ (y, y) [(x,x) +(y, y)] ( x + y ). (15) c) Take. All we need to show s that t does not satsfy the equalty proved n b). Take xe 1, y e. Then we have x+ y x y x y 1. The equalty s not satsfed. Queston 4. Let O R N N be such that O x x for any x R N. Prove that O s orthogonal. Please prove t drectly and do not use any theorem from lnear algebra. Proof. Frst we show that (O x) (O y) x y for all x, y R N. To see ths we calculate x x+x y + y y (x + y) (x + y) The clam follows. Recallng x y x T y, we have Thus we have shown for all x, y R N. Takng y e 1, [O(x + y)] [O(x+ y)] (O x) (O x) +(O x) (O y) +(O y) (O y) O x + (O x) (O y) + O y x +(O x) (O y) + y x x+(o x) (O y) + y y. (16) (O x) (O y) (O x) T (Oy) x T O T O y [O T O x] T y (O T O x) y. (17), e N, we see that [(O T O x) x] y 0 (18) O T O xx (19)
4 Math 17 Fall 013 Homework Solutons for all x R N. Fnally takng xe 1,, e N we see that O T O I, that s the matrx O s orthogonal. Queston 5. Let D dag(d 1,,d N ) be a dagonal matrx wth all the d s dstnct. Let A R N N be such that AD DA. What can we conclude about A? Justfy your answer. Proof. The (, j) entry for A D s d j a j whle the (, j) entry for D A s d a j. Thus we have (d d j ) a j 0 (0) for all, j 1,, N. As d s are dstnct, ths means a j 0 when j, that s A s dagonal. It s clear that f A s dagonal, then A D D A. Thus we have fully characterzed the matrces that commute wth a dagonal matrx wth dstnct man dagonal entres. Queston 6. (Twn Prme Conjecture) Earler ths year, Prof. Ytang Zhang of Unversty of New Hampshre made hstory through provng the followng result: where p n s the n-th prme number. lmnf (p n+1 p n ) <7 10 7 (1) n a) Prove that the Twn Prme Conjecture There are nfntely many pars of prme numbers wth dfference s equvalent to lmnf n (p n+1 p n ). () b) One step of hs proof s bascally the followng. Assume (θ,d, c) x(log x) A, (3) for some A > 0 and (θ, d, c) x(log x)/d; c C (d) τ 3 (d) ρ (d) d 1 (log x) B (4) Proof. for some B >0. Then we have E 6 τ 3 (d) ρ (d) c C (d) for any A > 0. Prove the above clam usng Cauchy-Schwarz. a) If lmnf n (p n+1 p n ), then there s a subsequence satsfyng Consequently, there s K N such that for all k > K, x(log x)b+1 A. (5) lmnf (p n k +1 p nk ). (6) k p nk +1 p nk <1/. (7) But the left hand sde s an nteger, so t must be 0. That s there are nfntely many pars of prme numbers wth dfference.
Due Thursday Sept. 6, 013 5pm 5 b) We have E ( τ 3 (d) ρ (d) ( ( c C (d) ( ( τ3 (d) ρ (d) (θ, d, c) 1/)( 1/) ( τ3 (d) ρ (d) (θ,d,c) 1/) ) 1/ ( ) ( 1/ (τ 3 (d) ρ (d) (θ,d,c) τ 3 (d) ρ (d) [ c C (d) ) 1/ (θ, d, c) (θ,d, c) ] 1/ (x(log x) A ) 1/ ) 1/ τ 3 (d) ρ (d) d 1 x (log x) (x (log x) A ) 1/ x 1/ (log x) B+1 x 1/ (log x) A/ x (log x) B+1 A. ) 1/ ) 1/ (8)