( 1) n (4x + 1) n. n=0

Transcription

1 Problem 1 (10.6, #). Fid the radius of covergece for the series: ( 1) (4x + 1). For what values of x does the series coverge absolutely, ad for what values of x does the series coverge coditioally? Solutio. First, ote that this is as power series, ad it s ceter is 1/4 (divide each term by 4, ad maipulate a little bit, to see why). This series is a geometric series, ad so we ca use the geometric series test to determie where it is absolutely coverget. The geometric series test says that a geometric series coverges absolutely if r < 1, ad diverges for r 1. Here, r = (4x + 1), ad thus we have covergece whe: 1 <4x + 1 < 1, < 4x < 0, 1/ < x < 0. We thus idetify R = 1/4. Because we kow that geometric series do ot coverge whe r = 1, it is ot ecessary to directly test the ed poits. For completeess, however, whe x = 1/, the sum becomes ( 1) ( 1) = 1 which clearly diverges. Whe x = 0, the sum becomes ( 1) (1) = ( 1), which also diverges. Thus the sum coverges absolutely for x i the iterval ( 1/, 0), but does ot coverge coditioally aywhere. It diverges whe x 1/ or whe x 0. Problem (10.6, #19). Fid the radius of covergece for the series: x. For what values of x does the series coverge absolutely, ad for what values of x does the series coverge coditioally? Solutio. It is probably easiest to use the root test for this problem. Recall that this is oly valid for positive fuctios, ad so we assess absolute covergece of the series usig this test. We thus compute: lim ( x ) 1 = lim = x. x 1 Note that I have computed the fial limit usig stadard limit rules ad the fact that 1/ 1. This ca be derived (remember to use a logarithm to simplify), but is well worth committig to memory. I this case, the root test tells us that the series coverges absolutely whe x < 1, or simplified, whe x <. We thus idetify the radius of covergece to be R =. Ulike the geometric series test used i fidig the aswer to problem 10.6 #, the root test is idetermiate i the case that x / = 1. We thus must separately test the edpoits x = ad x =. For x =, the series reduces to ( 1), which diverges by the th term test. The coclusio is the same for x =. Thus the series coverges absolutely whe x is i the iterval (, ), diverges outside of this iterval. Note that if the series were subtly chaged to: x, the iterval of absolute covergece would be the same, but the series would coverge coditioally at x =. 1

2 Problem (10.6, #4). Use Theorem 0 to fid the iterval of covergece of the series: (e x 4). Withi this iterval, fid the sum of the series as a fuctio of x. Solutio. Lookig at Theorem 0, we idetify a = 1 for all, ad f(x) = e x 4. The series x coverges absolutely whe x < 1, ad so by Theorem 0, this sum coverges absolutely whe e x 4 < 1. Simplifyig this iequality: 1 <e x 4 < 1, < e x < 5, l < x < l 5. As explaied i detail i the solutio to problem 10.6 #, it is uecessary to test the edpoits of this iterval for covergece. Thus the series (ex 4) coverges absolutely whe x is i the iterval (l, l 5), ad diverges whe x is ot i this iterval. To fid the sum of this series, we may use the stadard formula for the sum of a geometric series: (e x 4) 1 = 1 (e x 4) = 1 5 e x Problem 4 (10.7 #16). Fid the Maclauri series for the fuctio f(x) = si( x ). Solutio. We begi by fidig a geeral form for the th derivative of f evaluated at x = 0. Note that: f (1) (x) = 1 cos(x/), f () (x) = 1 si(x/), f () (x) = 1 cos(x/), f (4) (x) = 1 si(x/). It s ot particularly easy to write a geeral form for f () (x), although oe is give by: f () (x) = ( 1) + 1 ( x ) +1 cos ( 1) ( 1) ( x ) +1 si ( 1), but fortuately this is uecessary, as we oly eed a geeral form for f () (0), which is quite a bit simpler. Evaluatig the above expressios at x = 0, we fid that: f (1) (0) = 1, f () (0) = 0, f () (0) = 1, f (4) (0) = 0.

3 If you would like, it may be useful to write out a few more terms i order to see the patter. The geeral form is: f () (0) = ( 1) +1 1, for a odd umber, ad f () = 0 for a eve umber. The Maclauri series is give by: f(x) = f () (0) x!. The easiest way to deal with the fact that all of the -eve terms are zero is to represet the sum as: f(x) = f ( 1) x 1 (0) ( 1)!. =1 Thik about what is happeig here - for = 1, 1 = 1. for =, 1 =. for =, 1 = 5. So this is a way of automatically skippig the -eve terms whe writig dow the sum. All that remais is to compute: f ( 1) (0) = Pluggig ito what we already kow, we have: f(x) = =1 ( 1) ( 1)+1 1 1, = ( 1) 1 1. ( 1) 1 1 x 1 ( 1)!. Remark 1. The above problem is a good demostratio as to why it is a very good idea to commit certai thigs to memory. O a exam, you might eed to make use of the MacLauri series for si(x/). It would be a pai to go through all of this. Istead, if you just remembered that: si(y) = ( 1) 1 y 1 ( 1)! = y y! + y5..., =1 you could plug i y = x/ ad immediately fid the formula derived i the above problem. The formula give i the back or your book for the MacLauri series for si x differs slightly from the oe I gave; you should be able to covice yourself they are equivalet. I persoally fid it far easier to just remember that si y = y y! + y5..., ad recostruct the idexed sum from that, if ecessary. Problem 5 (10.7 #7). Use the Taylor series geerated by e x at x = a to show that: [ ] e x = e a (x a) 1 + (x a) + +.! Solutio. Let f(x) = e x. Luckily, ulike i problem 10.7 #16, fidig a geeral form for f () (a) is trivial - f () (a) = e a. Thus the Taylor series for e x cetered at x = a is give by: e x = f () (x a) (a),! a (x a) = e,! [ ] = e a (x a) 1 + (x a) + +.! Note that this series coverges for all x.

4 Problem 6 (10.9 #19). Use power series operatios to fid the Taylor series at x = 0 for the fuctio: f(x) = x 1 x. Solutio. We first recall that y=0 y = 1 1 y, at least for y < 1. Sice we are fidig the power series cetered at x = 0, this is fie. We substitute y = x to fid that: (x) = 1 1 x. This series coverges so log as x < 1, or more simply, x < 1/. Now we ca simply multiply both sides of the equatio by x : x (x) = x 1 x. Sice x does ot deped o, we ca pull it iside of the sum. There are a umber of ways to simplify the result, ad you ca choose your favorite. I ll simplify it to: x 1 x = x +. Although it is ot asked for i the questio, you may choose to use a test (for example, the root test), to verify that the radius of covergece is uaffected by multiplyig each term of the series (x) by x. Problem 7 (10.9 #6). Estimate the error if P 4 (x) = 1 + x + (x /) + (x /6) + (x 4 /4) is used to estimate the value of e x at x = 1/. Solutio. Beig able to approximate somethig is t very useful if you have o idea how far you might be off by, hece the importace of problems such as these. The formula for the remaider R 4 (x) is: R 4 (x) = f (5) (c) x 5, for some c betwee x ad 0. This is give as equatio () o pg 607 of your text, i the box etitled Taylor s Formula. I have merely replaced with 4 ad a with 0 (the ceter or the Taylor polyomial P 4 (x) give above). Notig that: f () (x) = e x, for all values of, we see that: R 4 (x) = ec x5, Now we do t kow which value of c makes the above equality true, so we istead ask the simpler questio: what is the biggest that e c ca be for 0 c x? This will the give us a iequality istead of a equality, but such is life. Sice e x is a mootoe fuctio, we kow that e c e x for all 0 x c. Thus: R 4 (x) ex x5. (This is basically the cotet of Theorem 4 from your book). Now pluggig i x = 1/, we have that: R 4 (1/) e1/ (1/)

5 Remark. What I did i the above problem is, I thik, what the book wats you to do, but if you thik about it for a momet, it s kid of ridiculous! We re supposedly estimatig e 1/, ad the fidig the error i our estimatio. But the, to evaluate our error boud, we eeded to kow what e 1/ was! Of course, if we kew e 1/ already, we would t eed to estimate it. Istead, let s come up with a cruder, but easier to evaluate boud for e 1/. Well, we kow that e < 4, ad so e <. Thus a cruder, but easy to evaluate (by had, if you had to) boud for R 4 (1/) is: R 4 (1/) (1/) This is of course a bit larger, due to our cruder estimate, but it ca legitimately be doe without preordaied kowledge of what we re tryig to fid, ad is still quite small. Fially, we ca compare this to the true error: R 4 (1/) = P 4 (1/) e 1/ 1 = + 1/ + ((1/) /) + ((1/) /6) + ((1/) 4 /4) e 1/ , which is quite a bit less tha either estimate made above. 5