Quiz 5 Answers MATH 141

Transcription

1 Quiz 5 Aswers MATH 4 8 AM Questio so the series coverges. + ( + )! ( + )! ( + )! = ( + )( + ) = 0 < We ca rewrite this series as the geometric series (x) which coverges whe x < or whe x <. Thus the radius of covergece is. Questio This is just the sum of a geometric series startig at = with r = /, so the sum is ( ) + =. Applyig the ratio test we have x + ( + )! ( + )! x + x = + = 0 <. Sice this holds for all values of x, the radius of covergece is ifiite, ad thus the iterval of covergece is (, ). 9 AM Questio so the series coverges. + ( + )! ()! ( + )! = ( + ) ( + )( + ) = 0 <, + x + ( + )! x = x + = 0 <. Sice this holds for all values of x, the radius of covergece is ifiite.

2 Questio First ote that by pluggig ito the Taylor series for e x, we have = e. By subtractig the first term of the series ad multiplyig by we get = (e ). We ca rewrite this series as the geometric series (x) which coverges whe x < or whe x <. Whe x =, we have which clearly does ot coverge, ad whe x = we have ( ) which also clearly does ot coverge. Therefore the iterval of covergece is (, ). 0 AM Questio ( + )! ( + ) + ( + )! = ( + )(( + )) ( + )( + ) = ( + ) = e =. Therefore the series diverges. ( + )!x + + x = x = for all values of x except for x = 0. Therefore the radius of covergece is 0. Questio Note that we ca rewrite the series as a two times a geometric series. Evaluatig, we get + = ( ) = = 4.

3 Quiz 5 Aswers MATH 4 ( + )x + x + = x = x which is less tha oe exactly whe x <. Therefore the radius of covergece is, ad all that remais is to check x = ad x =. At x = we get the series which clearly diverges, ad similarly at x = we get the series ( ) which also clearly diverges sice the terms do ot go to 0. Therefore the iterval of covergece is (, ). AM Questio so the series diverges. ( + )! + ( + )! ()! = ( + )( + ) ( + ) =, + x + ( + )! ()! x = x ( + )( + ) = 0 < for all values of x, so the radius of covergece is ifiite. Questio We ca idex the series to start at = 0 istead of =, ad rewritig it we get The last equality holds because ( )! = = = e. is just the Taylor series for e x evaluated at x =. We ca rewrite this series as the geometric series (x) which coverges whe x < or whe x <. Whe x =, we have which clearly does ot coverge, ad whe =

4 x = we have ( ) which also clearly does ot coverge. Therefore the iterval of covergece is (, ). PM Questio Note that l(x) is a strictly icreasig fuctio, so l( + ) =. Sice the terms of the series do ot go to 0, the series diverges. + x + ( + )! x = ( + )! x + = 0 <. Sice this holds for all values of x, the radius of covergece is ifiite. Questio We ca rewrite the series as = = = where the last equality holds because is a geometric series. x + ( + )! ()! x x = ( + )( + ) = 0 <. Sice this holds for all values of x, the radius of covergece is ifiite. Therefore the iterval of covergece is (, ).

5 Quiz 5 Aswers MATH 4 PM Questio This series will diverge because the terms do ot go to 0. For example, for all, ad = so = as well. By factorig out a from the umerator we ca rewrite the series as a geometric series. This gives us + x ( ) x = which coverges oly whe x covergece is. Questio < which happes whe x <. Therefore the radius of If we factor out 5 from the umerator, the we get five times the Taylor series for e x evaluated at. However sice the sum starts at =, we have to subtract the first two terms. Therefore we have 5 = 5 ( + ) x + x = 5(e ) = 5(e ). ( + ) = x = x which is less tha oe exactly whe x <. Therefore the radius of covergece is, ad all that remais is to check x = ad x =. At x = we get the series which clearly diverges, ad similarly at x = we get the series ( ) which also clearly diverges sice the terms do ot go to 0. Therefore the iterval of covergece is (, ).