Taylor Series 6B. lim s x. 1 a We can evaluate the limit directly since there are no singularities: b Again, there are no singularities, so:
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1 Taylor Series 6B a We can evaluate the it directly since there are no singularities: b Again, there are no singularities, so: + + c Here we should divide through by in the numerator and denominator and then use the fact that 4 : d Again, dividing through by, we find: a We make use of the Taylor epansion about : 4 sin4 4 sin !! sin4 4 4 b Here we use cos + + to see that:! 4! cos cos + + co s c We use e to see:! 9 e + + 9! e + + e! d We use arctan to see: arctan arctan4 4 Pearson Education Ltd 8. Copying permitted for purchasing institution only. This material is not copyright free.
2 a We use the Taylor series for sin about π: sin cosπ( π cosπ( π + 6 sin ( π + ( π + 6 Then we find: π π sin π ( π + 6 b Here we look for a Taylor epansion about for sin( 4: d sin( 4 cos( 4 d sin( 4 sin( 4 sin( + ( + ( + cos( 4 4 4( + ( So we find that: sin( 4 ( 4+ ( a We make use of the Taylor series for ln( + about to deduce that: 4 ln( + + Then we see: ln(+ + b Now we want the Taylor series for ln about given by: Then we note that: ( ( ( + ln ( ( ln ( ( + And observe that in the it as, the second and remaining terms do go to zero, so: ln + Pearson Education Ltd 8. Copying permitted for purchasing institution only. This material is not copyright free.
3 4 c We calculate the leading order terms in the denominator and numerator: ( 6 ( e e ( ln( Then we can take the it after dividing through by e e ln( to obtain: d We perform a similar procedure: e sin ( + + ( +! ! ln(+ ( + + Then in the it we have that: 5 e sin ln(+ 6 5 a The first few leading order terms in the Taylor series for sin and 5 sin + +! 5! e + +!! which can be found easily by differentiating the two functions using: d d d sin cos, cos sin, e e d d d e are: Pearson Education Ltd 8. Copying permitted for purchasing institution only. This material is not copyright free.
4 5 b Now we compute: e sin sin e sin ( e Plugging in the Taylor epansions: e sin + + +! + sin ( e ( ( ( ( ( Then we can take the it: e sin sin sin ( e e + sin e + 6 a Using: d d,, d d 4 d d d ln,,, d d d we find that the first few leading order terms in the Taylor series for ln and about are given by: ln ( ( + ( +!! ln ( ( + ( + + ( ( + ( +! 4! 8 + ( ( + ( b We have that: + ln ln (ln ( ( + ( ( + + ( 8( + ( + ( ( + + ( ( 8 + ( +, as + ( + Pearson Education Ltd 8. Copying permitted for purchasing institution only. This material is not copyright free. 4
5 7 a We use d (sinh coshas well as d d d (cosh sinhto write down: 5 sinh where we use sinh,cosh (cosech sinh b Note that cosech sinh, so ( ( (cosech d d d 8 a We compute the necessary derivatives: ( + 4 d! + 4 (+ 4 4( ( ( + 7 b To find the it we can find a finite Taylor series for ( + ( + 8( + ( Then writing + 4 ( ( such that The it is given by: ( ( + 6 Challenge a We do the required differentiation: d dy y (+ 5 y d 5 5 (+ 5 y (+ 5 y dy 4 d ( + 5 y (+ 5 y dy y + 5y 5y 5 y Pearson Education Ltd 8. Copying permitted for purchasing institution only. This material is not copyright free. 5
6 Challenge b Now we make the substitution y in the epression it as is the same as taking the it as y y y y Then: (( 5y 5y y ( y+ y y and note that taking the Pearson Education Ltd 8. Copying permitted for purchasing institution only. This material is not copyright free. 6
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