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 pre-imge 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 element-wise. 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, non-zero derivtives off-digonl. 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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