Anonymous Math 361: Homework 5. x i = 1 (1 u i )


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1 Anonymous Mth 36: Homewor 5 Rudin. Let I be the set of ll u (u,..., u ) R with u i for ll i; let Q be the set of ll x (x,..., x ) R with x i, x i. (I is the unit cube; Q is the stndrd simplex in R ). Define x T (u) by Show tht x u x ( u )u... x ( u ) ( u )u x i ( u i ) i Show tht T mps I onto Q, tht T is  in the interior of I, nd tht its inverse S is defined in the interior of Q by u x nd x i u i x x i for i,...,. Show tht i J T (u) ( u ) ( u ) ( u ) nd J S (x) [( x )( x x ) ( x x )] We wish to show the first equlity by induction on. In the bse cse we hve x u ( u ) which is true. Now we ssume tht for n rbitrry we hve i x i i ( u i), nd show tht this lso holds for +. We note tht + i x i x + + i x i, nd we my plug in the definition of x + to get ( u ) ( u )u + + ( u ) ( u ). We my fctor out most of the terms from the products to get + ( u ) ( u )(u + ) ( u ) ( u )( u + ) + i ( u i) s desired. Tht T mps I to Q follows esily from the definition. If we te ny x T (u), then note tht for ech u i we hve u i which implies u i. Then ech x i is product of nonnegtive numbers, which mens x i. Since ech fctor is, we lso hve x i. From our first equlity we hve i x i i ( u i), nd we now from the exctly sme rgument tht i ( u i), which implies i x i. Thus T mps I to Q. We show the surjectivity of T by showing tht the bove formuls for S, when plugged into those for T, give us bc x. This proves tht x hs the preimge u. We gin proceed by induction. First let x Q with i x i <. For the bse cse, x u x wors. For some g, ssume x g ( u ) ( u ). If we show tht S is the inverse mp of T, then tht will prove tht T is injective nd surjective. We show the vlidity of S by proving tht S(T (u)) u. We my do this elementwise. For u x u it is evident. For other elements we recll our definitions nd plug in.
2 x i u i x x i x i i j x j ( u ) ( u i )u i i j ( u j) Now we see tht ll the fctors cncel except for u i itself s long s i j ( u j) ( u j ) j. This mounts to stying wy from ny of the upper (u j ) borders of the unit cube. Since we only were only concerned with the interior of I, this is fine. Thus, on the interior of I, S(T (u)) u so S is the inverse of T nd T is bijective. The Jcobin mtrices follow esily from the mp definitions s the mtrices of prtil derivtives. We note tht if j > i, then xi u j nd ui x j from the definitions of the function. x u ( u ) x 3 x 3 J T (u) u u ( u )( u ) x u x u x u 3 ( u ) ( u ) u x x u J S (x) 3 u 3 x x x x u u u x x x 3 x x Since the mtrices re tringulr, the determinnts re the product of the digonl elements, nd we do not cre bout the complicted, nonzero derivtives offdigonl. The determinnts cn then be computed s J T (u) ( u ) ( u ) ( u ) nd J S (x) [( x )( x x ) ( x x )] Rudin.3 Let r,..., r K be nonnegtive integers, nd prove tht x r r xr! r! dx Q ( + r + + r )! Hint: Use Exercise, Theorems.9 nd 8.. Note tht the specil cse r r shows tht the volume of Q is /!. We use the chnge of vrible formul nd the mp T from the prior problem to rewrite the integrl nd simplify it. x r xr dx u r [( u ) ( u )] r ( u ) ( u ) ( u ) du Q I u r u ( u ) r+ +r ( u ) r ( u ) ( u ) du du I u r u ( u i ) i+ ji+ rj du du I i i u ri i ( u i) i+ ji+ rj du i
3 Where the lst step is justified becuse the integrls re ll independent. Recll from 8. we hve t x ( t) y dt (x)(y) (x + y) We my use tht here with x r i + nd y + i + ji+ r j to get Q x r xr dx i (r i + )( + i + ji+ r j) ( + i + ji r j) Now we note tht the fctor in the numertor for the l th term is ( + l + jl+ r j), nd the fctor in the denomintor for the l + th term is ( + l + jl+ r j), which re the exct sme. Thus ll of these fctor out except for the first denomintor term nd the lst numertor term. So we hve x r () xr dx Q ( + + j r j) (r i + ) We recll tht for positive integers n, (n) (n )!. Remembering tht!, we hve As we wnted to show. Q x r xr dx i r i! i ( + j r j))! 3. Evlute x sin y dx + y cos x dy where is the prmetrized curve : [, ] R defined by where m is some fixed nonzero rel number. (t) (t, mt) t [, ] We use the formul for evluting the line integrl of vector field, F(r) dr b F(r(t)) r (t)dt. Here we hve F (x sin y, y cos x), r(t) (t, mt), dr (dx, dy), r (t) (, m) nd, b. Now we my plug in nd evlute using dimensionl clculus. x sin y dx + y cos x dy (t sin mt, mt cos t) (, m)dt t sin mt + m t cos t dt m (sin mt mt cos mt) + t sin t + cos t (sin m m cos m) + sin + cos m 3
4 4. Evlute x x + y dx + y x + y dy where is the prmetrized pth : [, π] R defined by (t) (e t cos t, e t sin t) t [, π] b We gin use F(r) dr F(r(t)) r (t)dt. x Here we hve F ( x + y, y x + y ), r(t) (et cos t, e t sin t), dr (dx, dy), r (t) (e t cos t e t sin t, e t sin t + e t cos t) nd, b π. We plug in, simplify, nd evlute: x x + y dx + y x + y dy π π π π π ( e t cos t e t cos t + e t sin t, e t ) sin t e t cos t + e t sin (e t cos t e t sin t, e t sin t + e t cos t ) dt t e ( t cos t sin t cos t ) + e ( t sin t + sin t cos t ) e t cos t sin t cos t + sin t + sin t cos t dt dt dt 5. Difference between line integrls of functions nd line integrls of forms: Let : [, b] R n be smooth prmetrized pth in R n. Define : [, b] R n by (t) ( + b t) for t b, thus is lso smooth prmetrized pth in R n. () For continuous function f : R n R show tht f ds f ds Here we simply follow the definitions of prmetriztion of line integrls. For the left integrl, we hve b b f ds f((t)) (t) dt nd for the right one we hve f ds f((t)) (t) dt. We note tht by b the chin rule, (t) ( + b t) ( ). We plug in nd get f ds f(( + b t))) ( + b t) dt. Now we let perform chnge of vribles nd let τ + b t, which mens dτ dt, nd we must chnge the integrl bounds t τ b, t b τ. Then we hve: 4
5 f ds b b b b f((t)) (t) dt f(( + b t))) ( + b t) dt f((τ))) (τ) dτ f((τ))) (τ) dτ Since the nmes of our vribles don t mtter, we now hve f ds f ds. (b) For continuous function f : R n R nd i n, show tht f dx i f dx i This cse is nerly identicl but tht we re no longer ting the modulus of the pth length. We initilly b b get f dx i f((t)) i (t) i dt nd f dx i f((t)) i (t) i dt where we remember tht we re only integrting over one component. Then we my chse the definitions: f dx i b b b b f((t)) i (t) i dt f(( + b t)) i ( ( + b t) i )dt f((τ)) i (τ) i dτ f((τ)) i (τ) i dτ 6. () Suppose is smooth closed simple curve which surrounds (simply connected) region D in R. Show tht the line integrl (x dy y dx) computes the re of D. (Hint: pply Green s Theorem) In terms of differentil forms, Green s Theorem sys tht if w P dx + Q dy is form, then δd D dw. If we let w (x dy y dx), then dw d(x dy y dx) [d(x dy) d(y dx)]. We my then use the product rule nd recll tht the derivtive of differentil form is zero, which gives us dw f f [d(x) dy) d(y) dx)]. Then we recll tht for function f, d(f) x dx + y dy, i.e. the grdient of f. We use tht definition here, long with the rule dx dx nd get dw [dx dy) dy dx)]. Now we remember dx dy) dy dx, nd hve dw dx dy dx dy. Now we hve (x dy y dx) dx dy, which is the re of D. D 5
6 (b) Use the integrl bove to compute the re of the ellipse D { (x, y) R x / + y /b } where, b >. We prmetrize the border or this ellipse with the smooth, closed, simple curve (t) ( cos t, b sin t) t b [, π]. We use the sme formul s before, F(r) dr F(r(t)) r (t)dt, now with F ( y, x), r(t) ( cos t, b sin t), r (t) ( sin t, b cos t), dr (dx, dy), nd, b π. We get: (x dy y dx) Which is the correct re of n ellipse. π π π ( b sin t, cos t) ( sin t, b cos t) dt b(sin t + cos t) dt b dt b π πb 6
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