# Identify the changing parts we see an alternating so

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Identify the changing parts. We see an alternating ± , so we will get ( - 1) k or ( - 1) k +1 . If k = 0, the term is positive, so we take ( - 1) k . In the denominator, we have a power of 2. For the 0-th derivative, the power is 2. In the first derivative, the power is 3, so in each case, the power is one more than the derivative we’re taking. So we get 2 k +2 in the denominator. On top, we see factorials. In the second derivative, we have 3!. In the third derivative, it is 4!, so in general, it will be ( k + 1)!. So, f ( k ) (1) = ( - 1) k ( k +1)! 2 k +2 . Using the general formula for the Taylor formula, we get, T n ( x ) = n X k =0 f ( k ) (1) k ! ( x - 1) k = n X k =0 ( - 1) k ( k + 1)! 2 k +2 · 1 k ! ( x - 1) k = n X k =0 ( - 1) k ( k + 1) 2 k +2 ( x - 1) k . Example 19.4. If f ( x ) = sin(2 x ) , find T 3 ( x ) centered at a = π/ 4 , and then find T n ( x ) for any n . Solution. (a) To find T 3 ( x ), first take three derivatives of f ( x ): f 0 ( x ) = 2 cos(2 x ) , f 00 ( x ) = - 2 2 sin(2 x ) , f 000 ( x ) = - 2 3 cos(2 x ) . So at a = π/ 4, we have f ( π/ 4) = sin( π/ 2) = 1 , f 0 ( π/ 4) = 0 , f 00 ( π/ 4) = - 4 , f 000 ( π/ 4) = 0 , since cos( π/ 2) = 0. So T 3 ( x ) = f ( π/ 4)+ f 0 ( π/ 4)( x - π/ 4)+ f 00 ( π/ 4) 2! ( x - π/ 4) 2 + f 000 ( π/ 4) 3! ( x - π/ 4) 3 = 1 - 4 2! ( x - π/ 4) 2 = 1 - 2( x - π/ 4) 2 . (b) To find T n ( x ), we need a formula for f ( n ) ( π/ 4). We have three derivatives above, and already we notice that every other derivative will be 0. We can calculate a couple more derivatives: f (4) ( x ) = 2 4 sin(2 x ) , f (5) ( x ) = 2 5 cos(2 x ) , f (6) ( x ) = - 2 6 sin(2 x ) , and so on. So at a = π/ 4, the sequence of derivatives we have is: 1 , 0 , - 2 2 , 0 , 2 4 , 0 , - 2 6 , 0 , . . . . Now, you have to deal with the fact that we only have nonzero terms in the even slots (namely, k = 0 , 2 , 4 , 6 , . . . ). Even numbers have the form 2 k , for k an integer, and odd numbers have the form 2 k + 1, where k is some integer. The only nonzero derivatives come from f (2 k ) ( π/ 4), so let’s focus on the even terms: 1 , - 2 2 , 2 4 , - 2 6 , . . . . We see the alternating ± , so we’re going to get a ( - 1) k or ( - 1) k +1 . If k = 0, we have a positive term, so that tells us to take the ( - 1) k . We then see we have 2 to an appropriate power. In the 0-th derivative, the power is 0. In the second derivative, the power is 2, in the fourth derivative the power is 4, so in the 2 k -th derivative, the power will just be 2 k . So f (2 k ) ( π/ 4) = ( - 1) k · 2 2 k . Again, all the odd derivatives are 0 at π/ 4. So instead of giving the polynomial T n ( x ), we can give the polynomials T 2 n ( x ) and T 2 n +1 ( x ). The formula for T 2 n ( x ) is T 2 n ( x ) = f ( π/ 4) + f 0 ( π/ 4)( x - π/ 4) + f 00 ( π/ 4) 2! ( x - π/ 4) 2 + . . . + f (2 n ) ( π/ 4) (2 n )! ( x - π/ 4) 2 n . 51
Math 31B Notes Sudesh Kalyanswamy All the odd terms are 0, and we have a formula for f (2 n ) ( π/ 4), namely f (2 n ) ( π/ 4) = ( - 1) n · 2 2 n . So T 2 n ( x ) = 1 - 2( x - π/ 4) 2 + . . . + ( - 1) n · 2 2 n (2 n )! ( x - π/ 4) 2 n . This only gives the even order Taylor polynomials. We should also give the odd order ones. But the next Taylor expansion, namely T 2 n +1 ( x ), will be the same as T 2 n ( x ) since T 2 n +1 ( x ) = T 2 n ( x ) + f (2 n +1) ( π/ 4) (2 n +1)! ( x - π/ 4) 2 n +1 and f (2 n +1) ( π/ 4) = 0, so T 2 n ( x ) = T 2 n +1 ( x ) . So this describes every Taylor polynomial.
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