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Real Analysis

by: Briana Cassin

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Real Analysis MATH 30750

Briana Cassin
ND
GPA 3.65

Nancy Stanton

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Nancy Stanton
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This 0 page Class Notes was uploaded by Briana Cassin on Sunday November 1, 2015. The Class Notes belongs to MATH 30750 at University of Notre Dame taught by Nancy Stanton in Fall. Since its upload, it has received 17 views. For similar materials see /class/232699/math-30750-university-of-notre-dame in Mathematics (M) at University of Notre Dame.

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Date Created: 11/01/15
Math 30750 Spring 2010 Solutions Assignment 1 11 2 By the distributive law P9 yfyx yWw f wy 0 By the commutative property for multiplication P5 wmzamymzmyam 5 Applying the distributive law P9 to each term gives zzyyxyx2xyyxyz 3 By the commutative property for multiplication P5 x2wyy222wy2 4 By 1H4 x y2 2 2mg y2 In addition to being a reminder of how to use some of the properties of real numbers this gives you something you need for 8 11 4 By P5 P3 and P9 0zz0z00z0200202 5 Combining 5 with P1 and P3 gives 0920w0920 0 Q Hence by Proposition 111a 00 m In addition to giving you practice with the order properties this gives you something else you need for 8 11 8 By 04 with z 72y the inequality is equivalent to 0 S x272xyyz 8 By P4 and P9 0 2951 010 2w 294 9 1 Hence7 by Proposition 111b7 72y 27y 10 By 107 Proposition 111c and 2 zl7 wy27z22zbw7w27yf OD The result will follow from showing that for any real number 2 0 S 22 Either 0 g z or 2 S 0 by 01 Case i If 0 S 2 by 4 and 057 0ozlt on Case ii If 2 g 07 72 2 0 by Proposition 111 Hence by Case i7 0 7z2 By Proposition 111 c7 7z2 22 So7 in either case7 0 S 22 The inequality in this problem is one of the most useful inequalities in real analysis 12 8 The inde nite integral of a polynomial is a polynomial7 so I 73 7 73 If p and q are polynomials with p 7E q7 then Ip 7 q 7E 0 Since Ip 7 q Ip 7 Iq7 Ip 51 Iq Hence I is one to one I is not onto because nonzero constant functions are not in its range7 or7 equivalently7 Ip0 0 for all p so any polynomial in its range vanishes at 0 Here7s a linear algebra proof that I is one to one7 or7 what7s hiding in the rst proof I is a linear transformation on the vector space 73 so it is one to one if and only if its null space is If p an l alz a0 is a polynomial with an 7E 07 Ip has leading term zn so Ip 7E 0 and the null space is 13 8 Since An is countable7 there is a one to one map fn N 7 An which is onto Let ank denote Here are two methods of de ning f N 7 UnAn Only one is needed Method 1 Let f N 7 UnAn be given by f10117 f2 0217 3 0127 4 0317f5 0227 Q 1137m and in general7 nn 7 1 fj an7j1j7 7 17 771 80 using the following array 111701270137 12170227 1237 0317 13270337 the values of f are obtained by going up the diagonals which start at the left7 beginning with the one starting at all then the one starting at 121 then 131 etc So7 eg7 the 2 diagonal starting at 151 is 1517 14241337 1247 115 Students dont need to give a formula for Method 2 Let f N gtlt N 7 A be given by fm7n Then f is onto It might not be one to one For method 17 de ne a subset B of N by k E B if fj 7E fk for j lt k For method 2 de ne a subset B of N gtlt N by mm 6 B if fkj 31 fnm for k lt n and any j So7 you might describe this by saying B picks out the rst time each value is taken Then f is a one to one correspondence from B onto UnAn Each An is in nite7 so UnAn is7 hence B is By Proposition 1327 B is countable Hence UnAn is 14 12 a capzi j7k1 j l 13 IfjkZ forsomeZZ27 j k1k2kz l71timeS gt k1k1k1 k1k1quot1 14 and l gt7 1 1 H w mltmgt 39 15gt M Using this in 13 gives 675 lt 1Lltgt2 16 k k1 k1 k1 b The expression in brackets in 16 is a geometric series7 with sum 1 k 1 7 17 17 1 k Using 17 in 16 gives 6 7 s lt 18 k kk c If e where mm are integers7 then 7116 mn 7 1 is an integer and nlsn is an integer since nn 71n 7 j 1 is an integer for j S n d If e g by c nle 7 5 is an integer It is positive because 0 lt e 7 5 By b 7116 7 5 lt g So 7116 7 5 is a positive integer less than 1 This is impossible7 so 6 is irrational 21 2d To do this7 rst you need to gure out what the limit a should be The limit is 1 To prove this7 note that each an lt 1 Given 67 we need to nd Ne so that lan71l17 n 6ifn2Ne n1 176 n n1 Then Squaring both sides gives 1 i e g 19 Solve the inequality 71 117 62 S n 20 17 E n1 7 1 7 62 7126 7 627 21 216 g n 22 So7 we can take Ne to be any integer greater than or equal to 2156 and an converges to 1 This can also be done the way Example 2 is 21 6 If an 7gt a and b lt 17 let 6 Any 6 less than a 7 b will do There is N such that lan 7a S e if n 2 N Then a7b aiange 2 7 7b 17 S anifnZN 23 The left side in the last line is ab 7gtb 2 since a gt b Hence7 an gt b if n 2 N 21 8 There are many examples Here is one of each aan 7 bnnsocn b an bnn2socnn c an72 bnn2 so cn17 all n d an 71 bn n so on 71 n

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