Practice Problems: Taylor and Maclaurin Series

Transcription

1 Practice Problems: Taylor ad Maclauri Series Aswers. a) Start by takig derivatives util a patter develops that lets you to write a geeral formula for the -th derivative. Do t simplify as you go, because it might hide the patter: f 0) x) = x ) f ) x) = )x ) ) f ) x) = ) )x ) 3 ) f 3) x) = ) ) 3)x ) 4 3 ) Pull out the -) factors from all the leadig terms, ad fill i some missig expoets of or 0 to complete the patter. Writig with similar terms aliged to make it clearer, f 0) x) = ) 0 x ) 0 ) f ) x) = ) ) x ) ) f ) x) = ) )) x ) 3 ) f 3) x) = ) 3 ))3)x ) 4 3 ) The patter is fairly clear at this poit remember that 0! = ): f ) x) = )!)x ) ) ) = )! x ) We ca ow geerate the Taylor series aroud x =, usig the formula f ) ) x )! Usig our derived formula for the -th derivative, with substituted for x, )! )! x ) We ca simplify a little by cacelig a! ad doig a little arithmetic: ) x ) This is our Taylor series. You ca check it by graphig the first few partial sums ad seeig whether it looks like it s becomig a better approximatio of the origial fuctio. Now to fid the iterval of covergece. I ll use the ratio test: ) ) x ) ) x lim ) ) x ) ) x ) ) x

2 This ratio is less tha o the iterval, 3 ), so our Taylor series coverges there. At the left ad right edpoits, our series becomes: ) ) = ad ) ) = ) Neither of these coverge -th term test), so the Taylor series coverges oly o the iterval, 3 ). b) Start by takig derivatives util a patter develops: f 0) x) = si x f ) x) = cos x f ) x) = si x f 3) x) = cos x This group of four fuctios the repeats forever, but how ca we write it as a series, if the fuctio chages betwee sie ad cosie each time? Look at the sie terms by themselves: f 0) x) = si x f ) x) = si x f 4) x) = si x f 6) x) = si x This gives us a patter for terms with eve values, say where = k. Lookig ow at the cosie terms: f k) x) = ) k si x f ) x) = cos x f 3) x) = cos x f 5) x) = cos x f 7) x) = cos x This gives us a patter for terms with odd values, say where = k. f k) x) = ) k cos x So ow if we look at the Taylor series formula for a series cetered at x = π, f ) ) π x π )! we could break it up ito a sum of two series, oe with the eve umbered terms, ad oe with the odd umbered terms f k) ) π x π ) k f k) ) π x π ) k k)! k )!

3 Now, I ca substitute the formula for the eve terms ito the first oe, ad the odd terms ito the secod oe: ) k si π k)! x π ) k ) k cos π k )! x π ) k Now, we re helped by the fact that we kow si π = ad cos π = 0, so that makes the secod series completely vaish every term is zero), ad we re left with just the first: ) k k)! x π ) k There s o more simplifyig we ca do, so that s our Taylor series. To check the iterval of covergece, I ll use the ratio test, ) k ) lim k))! x π k) ) ) ) k k)! x π k k ) k))! x π ) k ) k k )! k)! ) x π ) k )k ) x π ) k )k ) x π ) k)! ) k Now, we ca brig the x π ) outside the limit, ad what s left will have limit 0, o matter what x is. That meas our series coverges o, ). c) Start by takig derivatives product rule) util a patter develops: The patter is pretty easy to see: f 0) x) = xe x f ) x) = xe x e x f ) x) = xe x e x e x f 3) x) = xe x e x e x e x f ) x) = xe x e x = x )e x So ow if we look at the Taylor series formula for a series cetered at x =, the substitutig our formula for f ) x), f ) )! )e x ) =! x ) e! x )

4 That s our Taylor series. Now use the ratio test to fid its iterval of covergece. e)! x ) lim e! x ) e)! x ) e! x ) e )! e! x ) ) ) The fractio ivolvig is positive for all >, so we ca brig it outside the absolute values, ad what s left does ot deped o, so we ca brig it outside the limit: x lim ) ) The limit goes to zero there s a term i the deomiator if you expad the product), o matter what x is, so this series coverges o, ).. For these, we just eed to take three derivatives, there s o eed to try to establish a patter ad fid a geeral -th derivative. a) Fidig the first three derivatives: f 0) x) = ta x f ) x) = x ) f ) x) = )x ) x) = xx ) f 3) x) = x )x ) 3 x) )x ) = 8x x ) 3 x ) you ca see where fidig a geeral patter would be hard for this fuctio) Evaluatig each of these at the ceter poit x =, f 0) ) = ta = π 4 f ) ) = f ) ) = f 3) ) = The pluggig those ito the first four terms i the Taylor series formula we eed four, because the first term has power 0, ad the fourth term has power 3 - the degree we re lookig for), P 3 x) = f ) ) x )! π 4 = 0! x )0! x )! x ) x )3 3! = π 4 x ) 4 x ) x )3 We could simplify this further by expadig all powers the combiig like powers of x, but this form is good eough...

5 b) Fidig the first three derivatives: f 0) x) = x 5 f ) x) = 5x 4 f ) x) = 0x 3 f 3) x) = 60x Evaluatig each of these at the ceter poit x = 0, f 0) 0) = 0 f ) 0) = 0 f ) 0) = 0 f 3) 0) = 0 The pluggig those ito the first four terms i the Taylor series formula, P 3 x) = f ) 0) x = 0! 0! x0 0! x 0! x 0 3! x3 = 0 The third degree Taylor polyomial is just zero! That meas the lie y = 0 is the best third-degree polyomial approximatio of x 5 ear x = 0. c) Fidig the first three derivatives: f 0) x) = l x f ) x) = x f ) x) = x f 3) x) = x 3 Evaluatig each of these at the ceter poit x = e, f 0) e) = f ) e) = e f ) e) = e f 3) e) = e 3 The pluggig those ito the first four terms i the Taylor series formula, P 3 x) = f ) e) x e)! = 0! x e)0 = x e) e e! x e) e! x e) e x e) x e)3 3e3 e 3 x e)3 3!