Worksheet 9. Topics: Taylor series; using Taylor polynomials for approximate computations. Polar coordinates.

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1 ATH 57H Spring 0 Worksheet 9 Topics: Taylor series; using Taylor polynomials for approximate computations. Polar coordinates.. Let f(x) = +x. Find f (00) (0) - the 00th derivative of f at point x = 0. Solution. Since we know that + t = t + t t +..., t < then, taking t = x, we obtain + x = x + x 4 x () On the other hand, we can find the power series expansion of the same function f using Taylor s formula: + x = f (n) (0) x n. () n! n=0 Series () and () have to be equal, because they represent the same function. Therefore, in particular, the coefficients by x 00 have to be equal. In series () this coefficient is equal to, while formula () tells us that the coefficient by x 00 is equal to f (00) (0) 00!. Therefore, f (00) (0) 00! =, and we conclude that f (00) (0) = 00!. Fact. Let f = f(x) be a function smooth in some interval containing point a. If there exists a constant such that f (n+) (c) for any c in that interval, then f(x) T n (x) }{{} R n(x) where T n is the n-th degree Taylor polynomial.. Find ln. with two-decimal-place accuracy. (n + )! x a n+, Solution. Let f(x) = ln( + x). We know that ln( + x) = x x + x x4 +..., x <. 4 Thus, we can find (approximately) ln. by taking a partial sum of this series and then letting x = 0.. We just need to figure out how long this partial sum has to be, in order to guarantee the desired precision. Here s how we can do it: That means, having derivatives of all orders.

2 Step. We want to find the formula for the (n+)-th derivative of our function f(x) = ln(+x): f (x) = + x f (x) = ( + x) f (x) = ( )( ) ( + x) f (4) (x) = ( )( )( ) ( + x) 4... f (n+) (x) = ( )n n! ( + x) n+ On the interval 0 x 0., we have f (n+) (x) = n! n! ( + x) n+ Hence, we can take = n!. That s an upper bound for the (n + )-th derivative of our function f on the interval 0 x 0.. Step. We can guarantee that the Taylor polynomial of degree n will produce an approximate value of ln. precise up to m decimal places, provided that the following inequality holds: (n + )! x a n m. In our case, Then the above inequality takes the form = n!, a = 0, x = 0., m = n + 0.n () Step. We would find the smallest natural number n that satisfies inequality (). Let us start with n =. Then the left-hand side of () becomes 0. = It s greater than = Thus n = does not work. Let us take n =. The left-hand side of () becomes It s still greater than = Let us take n =. Then the left-hand side of becomes < It works! Thus the Taylor polynomial that gives us the desired precision has order three. ore precisely, it has the form T (x) = x x + x.

3 Now, we are ready to finish our computations. ln. (x x + x Hence, ln Find e 0. ) x=0. Remark. A calculator returns ln. = with three-decimal-place accuracy = = 0.64 Solution. Let f(x) = e x. Then e 0. = f( 0.). We know the power series expansion of f(x) = e x : e x = + x + x! + x! + x4 4! +... Thus, we can approximate e 0. by taking a partial sum of this series and plugging x = 0. into it. We just need to figure out how long should this partial sum be, in order to guarantee that our approximation is good enough. Step. We should start by finding the formula for the (n + )-th derivative of f. In our case it s super-easy: f (x) = f (x) = = f (n+) (x) = e x. Since f (n+) (x) = e x is monotonically increasing, then on the interval 0. x 0 the greatest value that f (n+) (x) takes is e 0 =. Thus, we take = - it s an upper bound for the (n + )-th derivative of f. Step. We can guarantee that the Taylor polynomial of degree n will produce an approximate value of e 0. precise up to m decimal places, provided that the following inequality holds: (n + )! x a n m, where, in our case, Then the above inequality takes the form =, a = 0, x = 0., m =. (n + )! 0.n (4) Step. Now, we would like to find the smallest natural number n such that inequality (4) will hold. Let s start with n =. The left-hand side of (4) becomes 0.0 = This is larger than = Let s try n =. The left-hand side of (4) becomes 0.00 = < It works! Thus the Taylor polynomial that will provide the desired precision has order. ore precisely, it has the following form: T (x) = + x + x!.

4 Now we can finish our computations e 0. ( + x + x! ) x= 0. = = Find sin 4 with three-decimal-place accuracy. A sketch of solution. Since 4 is close to 45, find the Taylor series expansion (at least, the first three or four terms) of f(x) = sin x at point a = π 4. Then find a formula for the (n + )-th derivative of f(x) = sin x and estimate how large it can be on the interval 4 x 45. Let s say an upper bound that you found is. Next, solve (by trial and error) for n the inequality (n + )! x a n+ < m, where, in our case, a = π 4 and m =. Once you found n, take Taylor s polynomial of order n and plug x = 4 (convert it to radians!) into it. The number you that you will get is the desired approximation for sin 4. 4

5 5. Plot the points with the given polar coordinates: ( ) ( ) ( ) ( ) ( ) ( ), π 6,, 5π,, 7π 6, π,, π 5, 0, 7π.. Points A and B have polar coordinates (, π ) ( ) and, π. Find the polar coordinates of the midpoint of the line segment AB.. Points A and B have polar coordinates (, π ) and (, 5π ). Find the distance between the points A and B. 5