Math F15 Rahman

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1 Math F5 Rahma Week 0.9 Covergece of Taylor Series Sice we have so may examples for these sectios ad it s usually a simple matter of recallig the formula ad pluggig i for it, I ll simply provide the aswers. Also, may of these problems are i the book so I would suggest lookig up the problems i the book as well. Ex x si 2 x x 3 x 5 /3 + 2x 7 /45 + coverges everywhere by the theorem o products of power series i i a previous sectio. Theorem. Taylor: If f has derivatives up to order at a eighborhood of x a, the f(x) f(a)+f (a)(x a)+ f (a) (x a) f () (a) (x a) +R (x). () 2 a! Where, R (x) f (+) (ξ) ( + )! (x a)+ for some ξ betwee x ad a. (2) Theorem 2. Remaider: If f (+) (x) M for x a d, the e x ( + )! x a +. (3) Ex We may use remaiders to fid a alterate proof of the covergece of e x x. Note, this is ot a very efficiet proof, so the proof by ratio test is preferred. But evertheless, we fid the remaider ad take it s limit, lim R e ξ lim ( + )! x+ 0 for ay x. Sice this is true for all x, it must coverge everywhere. Ex Fid the Taylor series of e x about x 2, ad boud it s remaider. Solutio: For this problem lets fid the boud i the geeral case, however o the exam you will be asked to fid it for a certai domai. e 2 (x 2) ; R (x) e ξ ( + )! (x 2)+ (x 2)+ ( + )! { e 2 x < 2, e x x > 2; Ex Fid the boud for the remaider of si x. Solutio: We foud the taylor series of si x i a previous sectio. The boud o the remaider will be, ( + )! x + x + ( + )!. We choose M because si x for all x.

2 Ex Fid the Taylor series of x cos x. Solutio: We foud the Taylor series of cos x already, which is x2 ( ) (2)!. Therefore, x cos x x2+ ( ) (2)!. It is useful to kow a few commo Taylor series. Of course, you should be able to derive these. Commo Taylor Series x x + x + x 2 + (4) e x si x cos x x + x + x2 2 + x3 3! + (5) ( ) x2+ (2 + )! x x3 3! + x5 5! + (6) ( ) x2 (2)! x2 2 + x4 4! +. (7) Ex Evaluate e x2 dx as a series ad calculate 0 e x2 dx correct to withi.00. Solutio: We kow e x x, the e x2 ( x 2 ) ( ) x2 We ca itegrate this term by term, x 2 + x4 2 x6 3! +. e x2 dx C + x x3 3 + x5 0 x7 7 3! + x9 9 4! x 5! + x 2+ (2 + ) + Now, lets fid what will give us our desired accuracy. I order to do these we employ our formula for remaider, ( + )! x + < M ( + )! (2 + 3) ( + )! <.00. We see that 4 does the trick. This meas that, 0 e x2 dx

3 Ex lim e x x x 2. Solutio: We employ the Taylor series of e x, but we oly eed the first few terms. e x x ( + x + 2 lim x 2 lim x2 + 6 x3 + ) x 2 x 2 lim x2 + 6 x3 + x 2 lim x ) Fid the Taylor series of e x si x up to the first three ozero terms. Solutio: Note, this is ot a recommeded techique to solve the problem. Uless you are extremely cofidet about this you should ot use it! The best way would be to just take three derivatives ad compute it maually. Regardless, this is a way to solve the problem, e x si x ) ) ( + x + (x x2 2 + x3 6 + x3 6 + x + x x3 +. Ex Fid the Taylor series of ta x up to the first three ozero terms. Solutio: We kow the Taylor series of si x ad cos x, ad sice ta x si x/ cos x we ca use log divisio, ta x si x cos x x x3 /6 + x 5 /5! + x 2 /2 + x 4 /4! + x + 3 x x5 +. Ex Fid the Taylor series of cosh x. Solutio: This is pretty easy if we remember the defiitios. If we do t remember the defiitios, just take a buch of derivatives as per usual. cosh x 2 (ex + e x ) 2 [ ( x ) ] x + ( ) x 2 (2)!.

4 0.0 Biomial Series ad More Taylor Series Cosider the series f(x) ( + x) m, lets calculate the Taylor series. We otice that f () (x) m(m ) (m ( ))( + x) m, ad whe this is evaluated at x 0 we get f () (x) m(m ) (m ( )). The, f(x) m(m ) (m ( )) (m )! x. Techically, the factorials should be replaced with Gamma fuctios, i.e. Γ(m + ), however we re i Calc II, so lets ot ad say we did. I statistics we call (m )! ( m ), m choose. Remids me of Ralph Wiggum haha - I choo choo choose you. This series is called a biomial series, ( + x) m ( ) m x, x < ; ( ) m (m )!. (8) Agai, to be perfectly accurate those factorials should be Gamma fuctios. Lets do a example that we attempted out of the book, but could t solve fully because the book put it i the wrog sectio... ) If we wat to fid the geeral formula for + x (+x) /2. Now, if they ask us to fid, say the first four terms always calculate the derivatives! ( ) /2 + x x. Somethig that is metioed i the book, but somethig we wot be seeig too much of is Euler s Idetity: e iθ cos θ + i si θ. Lets look at some examples from the book, 27b) We kow what the power series for /(+x 2 ), ow we ca get ta x by itegratig term by term, ad agai we get the itegral of that i a similar fashio. Sice we are evaluatig these at x 0 ad ta (0) 0 ad x 0 ta x x0 0, so we eed ot be cocered with costats of itegratio. + x 2 ( ) x 2 ta x ( ) 2 + x2+ x 0 ta xdx ( ) x 2+2 (2 + )(2 + 2). From this we ca fid a formula for the Remaider. Notice that sice this is a alteratig series we use the alteratig series remaider because it s more accurate, istead of the Taylor series remaider. Also, otice that our domai is [0, ]. R x 2+4 ( ) (2 + 3)(2 + 4) (2 + 3)(2 + 4) So, we eed 5. <

5 27a) This oe is fairly easy if you use trial ad error but a lot more difficult if you try what we did for the 27b. Basically try it for 0, it wot work, so try it for, that will work. 47) For this we otice that we ca pull out a x 3, x 3 + x 4 + x 5 + x 6 + x +3 x 3 x x3 x. 49) Agai we ll pull out a x 3, but we ll have to massage it a bit before we ca put it ito a form that we ca use, x 3 x 5 +x 7 x 9 +x + ( ) x 2+3 x 3 ( ) x 2 x 3 ( x 2 ) 37) For this, of course you ca use L Hôpital, but the problem asks us to use Taylor series. So, first we compute the Taylor series for l(+x 2 ) about x 0. Why do we calculate it about x 0. Thik about that. L lim l( + x 2 ) cos x lim l( + x 2 ) ( x 2 /2 + x 4 /4! + ) lim l( + x 2 ) x 2 /2 x 4 /4! +. Notice that the derivative of l( + x 2 ) is 2x/( + x 2 ). So we ca fid the power series of 2x/( + x 2 ) ad itegrate term by term, 2x + x 2 2x ( ) x 2 2 ( ) x 2+ l(+x 2 ( ) ) + x2+2. Agai we do t worry about the costat of itegratio because l() 0. The, x 2 x 4 /2 + L lim x 2 /2 x 4 /4! + lim x 2 /2 + /2 x 2 /4! + 2. x3 + x 2.