Solution to HW 3, Ma 1a Fall 2016

Transcription

1 Solution to HW 3, Ma a Fall 206 Section 2. Execise 2: Let C be a subset of the eal numbes consisting of those eal numbes x having the popety that evey digit in the decimal expansion of x is, 3, 5, o 7. Let {c n } be a sequence of elements of C so that c j < fo evey natual numbe j. Show that thee is a subsequence of {c n } which conveges to an element of C. Solution: Take C to be defined as above. We will begin ou poof with a shot lemma. Lemma: Given any x R \ C, thee exists an ɛ > 0 such that fo evey y (x ɛ, x + ɛ), one has that y / C as well. poof We suppose without loss of geneality that x (0, ) \ C, fo if not, we may apply essentially the same poof stating with the highest (nondecimal) place and coecting fo sign. Now, in paticula, we may wite that x =.x x 2 x 3... = x n 0, whee x n n lies between 0 and 9, inclusive, fo any n N. Given that x / C, we have that thee must exist a decimal place such that the value at that decimal place does not lie within {, 3, 5, 7}. Now, by well odeing, thee must be a minimal such decimal place, which we take to be the k-th decimal place. In othe wods, one has that c k / {, 3, 5, 7}, while fo any l such that 0 < l k, one has c l {, 3, 5, 7}. Now, conside ɛ =, 0 k+2 and select y =.y y 2 y 3... = y n 0 n (x 0 k+2, x + 0 k+2 ). We may suppose without loss of geneality that y < x, fo if y = x, we have automatically that y / C, and if y > x, then the poof woks moe o less identically with the subtaction in evese. Then we note that 0 < x y = x y = 0 n < 0 k+2. In paticula, we note that we must have fo all l such that 0 < l k that y l = x l, o y =.x x 2...x k y k y k+..., fo othewise we would have a contadiction, as the distance between them would be geate than. Thus, we in fact 0 k+2 have that x y = n=k 0 n < 0 k+2.

2 Now, futhemoe, if we have that y k = x k, then we have automatically that y / C, so suppose y k x k. Then, in fact, one must have that x k y k =, given that x > y and given that, if the diffeence between the x k and y k wee any lage, we would have x y > 0 k+2 which would be a contadiction. Thus, x y = 0 k + n=k+ 0 n < 0 k+2, which may only be the case if x k+ y k+ = 9. Given that x k+ and y k+ lie between 0 and 9, inclusive, this may only be the case if x k+ = 0 and y k+ = 9. Thus, as y k+ / {, 3, 5, 7}, we have that y / C. Indeed, we have that fo any y (x, x + ), one has y / C. 0 k+2 0 k+2 Now,taking C and {c n } as above, we note that by ou supposition, c n (, ) [, ] fo evey n N, so by the Bolzano-Weiestass theoem, thee exists a convegent subsequence of {c n }. Take {c nk } to be one such convegent subsequence, conveging to a limit c [.]. Now, suppose fo contadiction that c / C. By the lemma we poved above (estated slightly), thee must exist an ɛ > 0 such that fo evey d R such that c d < ɛ, one must have that d / C. Howeve, we note that since lim k c nk = c, thee exists a K N such that fo all k > K, one has that c c nk < ɛ, which immediately implies that fo evey k > K that c nk / C, which is a contadiction. Thus, one must have that c C. In fact, we have poven that, not only must thee exist a convegent subsequence of {c n }, but also that fo EVERY such convegent subsequence of {c n }, its limit must also lie in C. Section 2.2 Execise 2: Pove eithe that the following seies conveges o diveges: n n. Solution: Consideing the seies n n =, we note that = n n. fom hee, we apply the atio test. We see that + = (n + ) n+ /(n + )! n n = (n + ) n+ (n + )206 / (n + )!n 206 = n (n + )n 206. Now, we know that (n + ) 206 = 206 ( 206 ) k=0 k n k is a polynomial in n of degee 206, wheeas (n + )n 206 = n n 206 is a polynomial in n of degee 207. Thus, as the degee of the polynomial in the denominato is geate than that of the polynomial in the numeato, we immediately see that lim + = lim 206 (n + ) 206 = 206 lim n n (n + )n206 2 n (n + ) 206 = 0. (n + )n206

3 Finally, as lim n an+ = 0 <, one has by the atio test that n n = conveges (absolutely). Section 2.2 Execise 6: A 30 yea fixed ate motgage is a loan taken out ove a peiod of 360 months. The initial loan amount is M. Each month, the boowe pays a fixed payment p. We define a function f(j) whee j is the numbe of months that have passed. We let f(0) = M, and we let f(j) = ( + )f(j ) p, fo j 360, whee is the fixed monthly inteest ate. Futhe, we equie that f(360) = 0. Deive and pove a fomula fo p in tems of M and in closed fom. Hint: You ll have to use the fomula fo the sum of a finite geometic seies. It helps to eaange things so that you e setting equal the motgage amount M with inteest on it compounded ove 30 yeas and the steam of monthly payments each compounded fom the moment it is made. Solution: Recalling ou definition of f(j), we have that f(0) = M and f(j) = ( + )f(j ) p fo any j 0, whee M is ou initial loan amount, is ou fixed monthly inteest ate, j is the numbe of months that has passed, and p is ou fixed payment amount. Fo any j N, we may continue out the expansion fom befoe as f(j) = ( + )f(j ) p = ( + )(( + )f(j 2) p) p = ( + ) 2 f(j 2) [( + ) + ]p. = ( + ) 2 (( + )f(j 3) p) [( + ) + ]p = ( + ) 3 f(j 3) [( + ) 2 + ( + ) + ]p j = ( + ) j f(0) [( + ) j + ( + ) j ( + ) + ]p = ( + ) j M ( ( + ) k )p Thus, invoking the fomula fo a finite geometic seies, we see that, in fact f(j) = ( + ) j M ( + )j p. Now, one has that f(360) = 0, so we have that ( + ) 360 M ( + )360 p = 0. Reaanging, we get that ( + ) 360 p = ( + ) 360 M, so dividing out fo the tem in font of p, we get ou desied expession: p = ( + )360 M ( + ) 360, 3 k=0

4 o, to simplify it a little, p = M ( + ) 360. Section 2.3 Execise : Show that the seies n n x n diveges fo all x such that x > 0. Solution: Letting S(x) = nn x n, one clealy has that S(0) = 0. Now, let x be such that x > 0. The easiest way to demonstate that nn x n diveges is actually to go back to the pevious section and invoke the ation test. Note that we have, in this case, = n n x n. Thus, we have that + = (n + )n+ x n+ n n x n = (n + )n+ n n x. Now, noting that n n < (n + ) n fo any n N, one has that, in fact + (n + )n+ = n n x = (n + )n n n (n + ) x > (n + ) x. Thus, using the Infinite Squeeze Theoem, one has that since and fo all n N, one has that lim (n + ) x = x lim (n + ) =, n n in fact, one must have that + > (n + ) x, lim + =. n Thus, invoking the atio test, we see that, since the ation diveges, and consequently lim n an+ >, one has that n n x n diveges fo all x such that x > 0. Section 2.3 Execise 3: Let { } be a sequence satisfying = fo n > 2 with a = and a 2 = 2. Following Example 3, find a ational function epesenting the powe seies x n.

5 What is the adius of convegence of this seies? Justify you answe. Hint: The sequence is just the sum of two geometic sequences. Solution: Conside the function x n, defined only fomally at the moment. Using the defining elations among the, we obtain that ( 2x 3x 2 ) ( 2x 3x 2 ) x n = x n 2 x n+ 3 x n+2 n=2 *eindexing* = x n 2 x n 3 2 x n *pulling out leading tems* = a x + a 2 x 2 + x n 2a x 2 2 x n 3 2 x n *plugging in a and a 2 values* = x + 2x 2 2x 2 + *noting the sum is zeo* = x. ( )x n Thus, dividing though by 2x 3x 2, we see that x 2x 3x 2 = x ( 3x)( + x), and, afte applying the patial factions method, that ( 3x) ( + x). We may now apply the fomula fo the sum of a geometic seies to see that ( 3x) ( + x) = ( 3x + x ) = ( 3 n x n ( ) n x n ) 3 n ( ) n = x n 3 n ( ) n = x n, which implies that = 3n ( ) n, so is the sum of two geometic sequences, both of whose esulting seies convege if and only if x = x < and 3x = 3 x <, so one must have x < 3. Thus, the adius of convegence of the powe seies in question is R = 3. 5