Part 3.3 Differentiation Taylor Polynomials

Transcription

1 Part 3.3 Differentiation Taylor Polynomials Definition Taylor 1715 and Maclaurin 1742) If a is a fixed number, and f is a function whose first n derivatives exist at a then the Taylor polynomial of degree n for f at a is T n,a f x) = f a) + f a) x a) + f a) 2! x a) f n) a) n! x a) n. Example Find the Taylor polynomial of degree n for f x) = sin x and a = 0. In this course we have started with the definition of sine as the ratio of sides in a right-angled triangle. We have assumed the addition formula for sine and managed to show that sine is continuous and differentiable of all orders for all x. From such results we have From this we can see that f x) = sin x f 1) x) = cos x f 2) x) = sin x f 3) x) = cos x f 4) x) = sin x. Thus { 1) r f n) sin x if n = 2r is even x) = 1) r cos x if n = 2r + 1 is odd. f n) 0) = { 0 if n = 2r 1) r if n = 2r + 1 Hence the Taylor polynomial for sin x at 0 of degree n is T n,0 sin x) = x x3 3! + x5 5! x7 7! )R x 2R+1 2R + 1)! = R r=0 1) r x 2r+1, 2r + 1)! 1

2 where 2R + 1 n < 2R + 1) + 1. The first few Taylor polynomials can be plotted as 2 T 1,0 =T 2,0 T 5,0 =T 6, T 9,0 =T 10, T 3,0 T 7,0 =T 8,0 =T 4,0 si n x Example Calculate T 8,0 e x sin x). Solution If f x) = e x sin x then f x) = e x sin x f 1) x) = e x sin x + e x cos x f 2) x) = e x sin x + e x cos x + e x cos x e x sin x = 2e x cos x f 3) x) = 2e x cos x 2e x sin x f 4) x) = 2e x cos x 2e x sin x 2e x sin x 2e x cos x = 4e x sin x. The important observation is that f 4) x) = 4f x), for this means that f 5) x) = 4f 1) x), f 6) x) = 4f 2) x), f 7) x) = 4f 3) x) and f 8) x) = 4f 4) x) = 16f x). 2

3 Thus f 0) = 0, f 1) 0) = 1, f 2) 0) = 2, f 3) 0) = 2, f 4) 0) = 0, f 5) 0) = 4, f 6) 0) = 8, f 7) 0) = 8, f 8) 0) = 0. Hence T 8,0 e x sin x) = 0 + x + 2 x2 2! + 2x3 3! + 0x4 4! 4x5 5! 8x6 6! 8x7 7! + 0x8 8! = x + x 2 + x3 3 x5 30 x6 90 x Note When the function f contains trigonometric functions we often find a relationship between f and f 4) as we saw above. Such relations should always be exploited to reduce work. Example With calculate T 5,0 f x). f x) = ex 1 + x Solution Because differentiating quotients leads to complicated expressions we yet again follow the principle of ridding ourselves of fractions by multiplying up as 1 + x) f x) = e x. Then repeated differentiation gives 1 + x) f x) + f x) = e x, thus f 0) + f 0) = x) f x) + 2f x) = e x, thus f 0) + 2f 0) = x) f 3) x) + 3f x) = e x, thus f 3) 0) + 3f 0) = x) f 4) x) + 4f 3) x) = e x, thus f 4) 0) + 4f 3) 0) = x) f 5) x) + 5f 4) x) = e x, thus f 5) 0) + 5f 4) 0) = 1. Starting from f 0) = 1 we can solve to get f 0) = 0, f 0) = 1, 3

4 f 3) 0) = 2, f 4) 0) = 9 and f 5) 0) = 44. Then ) e x T 5,0 = 1 + 0x + 1 x2 1 + x 2! 2x3 3! + 9x4 4! 44x5 5! = x2 1 3 x x x5. Definition The Remainder, R n,a f x), is defined by R n,a f x) = f x) T n,a f x). 1) TRICK To get estimates for R n,a f x) we first ask how T n,t f x) changes as t varies between a and x. So for the next few results, consider a and x to be fixed. Note that when t = x in the definition of T n,t f x) we get T n,x f x) = f x) + f x) x x) + f x) x x) f n) x) x x) n 2! n! = f x). 2) Thus the remainder can be written as R n,a f x) = T n,x f x) T n,a f x). We start with quite an amazing result, that the derivative w.r.t t of the polynomial T n,t f x) should have such a simply result! Lemma If the first n + 1 derivatives of f exist on an open interval containing a and x then for all t between a and x. Proof in lectures. d dt T n,tf x) = x t)n f n+1) t) n! Note. That the n + 1-th derivative, f n+1) t), exists means that the previous derivatives f i) t) are themselves differentiable w.r.t. t for 1 i n. Yet T n,t f x) is a sum of these f i) t) for 1 i n and thus T n,t f x) is differentiable w.r.t. t. 4

5 An application of the Mean Value Theorem gives Theorem Taylor s Theorem with Cauchy s form of the error. If the first n + 1 derivatives of f exist on an open interval containing a and x then R n,a f x) = f n+1) c) x c) n x a) 3) n! for some c between a and x. Proof Consider R n,a f x) x a = f x) T n,af x) x a = T n,xf x) T n,a f x) x a by definition of R n,a, by 2). Let h t) = T n,t f x) so we can rewrite the last equality as R n,a f x) x a = h x) h a) x a = h c) for some c between a and x by the Mean Value theorem applied to h. Continuing h c) = d dt T n,tf x) x c)n = f n+1) c), t=c n! by Lemma. This result has a weakness in that the unknown c occurs in two terms on the right hand side. An argument, based on Cauchy s Mean Value Theorem, leads to Theorem Taylor s Theorem with Lagrange s form of the error 1797). If the first n + 1 derivatives of f exist on an open interval containing a and x then R n,a f x) = f n+1) c) n + 1)! x a)n+1 4) for some c between a and x. 5

6 So now we have only one occurrence of the unknown c. Proof Let g t) be any function continuous on [a, x], differentiable on a, x) and with g t) 0 for all t a, x). Then R n,a f x) g x) g a) = T n,xf x) T n,a f x) g x) g a) 1 d = g c) dt T n,tf x) t=c as in above proof, by Cauchy s M. V. Theorem, = x c)n n!g c) f n+1) c) by Lemma. If we choose g t) = x t) n the right hand side becomes f n+1) c) /n!, and so we have only one occurrence of the unknown c, as required. Integrate this choice of g to get x t)n+1 g t) =, n + 1 in which case g x) = 0. Multiplying up, R n,a f x) = g x) g a)) f n+1) c) n! = x a)n+1 n + 1) f n+1) c) n! as required. If we set h = x a in Taylor s Theorem with Lagrange s error we get f a + h) = f a) + hf a) + h2 2! f a) hn n! f n) a) + for some 0 < θ < 1. + hn+1 n + 1)! f n+1) a + θh) Taylor s Theorem is often used in Maclaurin s Form which simply has a = 0. 6

7 Example T 4,0 cos 2 x ) = 1 x x4, Solution Let f x) = cos 2 x. Then f 1) x) = 2 cos x sin x = sin 2x. It is important to write it in this way because, continuing, f 2) x) = 2 cos 2x and f 3) x) = 4 sin 2x = 4f 1) x). This relation between third and first derivatives means that f n) x) = 4f n 2) x) for all n 3 which simplifies the calculations of f 0) = 1, f 1) 0) = 0, f 2) 0) = 2, f 3) 0) = 4f 1) 0) = 0 f 4) 0) = 4f 2) 0) = 8. Thus T 4,0 cos 2 x ) = 1 + 0x 2 x2 2! + 0x3 3! + 8x4 4! = 1 x2 + x4 3. Example Use Lagrange s form for the error to show that cos2 x 1 x ) 3 x x 5. Hence show that cos 2 x 1 + x 2 lim = 1 x 0 x 4 3. Solution cos2 x 1 x x4 ) = cos 2 x T 4,0 cos 2 x ) 7 = R4,0 cos 2 x ) = f 5) c) x 5 5!

8 for some c between 0 and x, by Lagrange s error. For this we now need f 5) x) = 4f 3) x) = 16f 1) x) = 16 sin 2x. Then f 5) c) 5! x 5 = 16 5! cos 2c x x 5 This gives the first stated result. For the second, divide through by x 4 to get cos 2 x 1 + x 2 1 x x. This can be opened out as x cos2 x 1 + x 2 x x Let x 0 and quote the Sandwich Rule to get result. 8

9 3.3.2 Taylor Series Definition If all the higher derivatives of f exist in some neighbourhood of a R then the Taylor Series of f at a is r=0 f r) a) r! There are two immediate questions. Question 1. Does the series converge? x a) r. 5) This series trivially converges at x = a. Being a power series we can use tests from course MATH10242, such as the Ratio Test or Comparison Tests to find an R 0 for which if x a < R then the series converges at this x, if x a > R then the series diverges at this x, while if x a = R the series may, or may not converge at such x. Here R is called the radius of convergence. It is possible that R =, for example the series for e x. It is possible that R = 0; an example was given by Lerch in 1888 of a function well-defined on R yet whose Taylor series diverges for all x 0. Question 2. If the series converges at x R does it converge to f x)? Even if R > 0 and x 0 a is in the interval of convergence, there is no assurance that the value of the series at x 0 equals f x 0 ). This is the case with Cauchy s example 1823), of f x) = e 1/x2 for x 0 with f 0) = 0. Lemma for all n 1. e 1/x2 lim = 0 x 0 x n Proof Recall that for y > 0, we have from the series defining e y that e y = 1 + y + y2 2 + y3 yn yn ! n! n!, 9

10 throwing away all other terms, allowable since they are positive. Apply this inequality with y = 1/x 2 to get in which case as x 0. e 1/x2 x n e 1/x2 1 n!x 2n, n!x2n x n = n! x n 0 Example The Taylor series for { e 1/x 2 if x 0, f x) = 0 if x = 0. is 0 + 0x + 0 x x3 3! +... which converges for all x R. But it s sum is f x) only when x = 0. Solution We need calculate f n) 0). Starting with n = 1 we find that f 1) 0) = lim x 0 f x) f 0) x 0 e 1/x2 = lim x 0 x = 0, by previous Lemma. For f 2) 0) we need to know f 1) x) for x 0 but by the Rules for differentiation this is f 1) x) = 2 x 3 e 1/x2. Then f 2) 0) = lim x 0 f 1) x) f 1) 0) x 0 2 = lim x 0 x 4 e 1/x2 = 0, by previous Lemma. This can be continued, though the expressions for f n) x) when x 0 get particularly complicated. Nonetheless, it can be shown that f n) 0) = 0 for all n 1. Thus the Taylor Series for f x) is 0 + 0x + 0 x x3 3!

11 which converges for all x R. But it s sum is f x) only when x = 0. Yet Question 2 does have an answer: Recall from the first year course, Sequences and Series, an infinite series is defined to be the limit of the sequence of partial sums, if the limit exists. Thus r=0 f r) a) r! x a) r = lim n n r=0 f r) a) r! = lim n T n,a f x)). x a) r So the Taylor Series of f converges to f for those x for which lim T n,a f x)) = f x) n which can be rearranged to lim n f x) T n,a f x))) = 0, the same as lim R n,af x) = 0. n Lemma For any y R we have y n lim n 0 n! = 0. Proof It is a result from First Year Sequences and Series that {y n /n!} n 1 is a null sequence. It can be noted though that we have been assuming this result implicitly in this course. We have defined e y by the infinite series n r=0 yr /r!. Yet if an infinite series converges its terms tend to zero, which is the statement of this Lemma. Example From an earlier example we have the Taylor Series for sin x is 1) r x 2r+1 2r + 1)! r=0 The ratio test would show that this converges for all x R, but we have to go further and show that, for each x, it converges to sin x. 11

12 Solution Let f x) = sin x, so Lagrange s form of the error term is, for any x R, R n,0 sin x) = f n+1) c) n + 1)! xn+1 for some c between 0 and x. Yet as seen earlier, f n+1) c) is either sin c or cos c and both are 1, so R n,0 sin x) x n+1 n + 1)! 0 as n for any x R, by the Lemma above. Hence, for each fixed x R, we have lim n R n,0 f x) = 0 and so the Taylor series converges to sin x, i.e. sin x = x x3 3! + x5 5!... = r=0 1) r x 2r+1. 2r + 1)! This series be taken as the definition of sine but this would have made the proofs of this course more difficult. For example, to prove d sin x/dx = cos x, we would need to be able to differentiate an infinite series term by term.) Example Find the Taylor Series for cos 2 x around a = 0 and show that the series converges to cos 2 x for all x R. Solution If f x) = cos 2 x then we have seen before that f n) x) = 4f n 2) x) for all n 3. We can repeatedly apply this to get f n) x) = 4f n 2) x) = 4) 2 f n 4) x) = 4) 3 f n 6) x) =... = 4) l f n 2l) x), for any l with n 2l 1, i.e. 2l n 1. Given n it is either even or odd, i.e. of the form 2r or 2r + 1. If n = 2r can choose any l : 2l 2r 1, i.e. l r 1. With l = r 1, when n 2l = 2, we find f n) x) = 4) r 1 f 2) x) = 2 4) r 1 cos 2x = 1) r 2 n 1 cos 2x. If n = 2l is odd we can choose l = r to find f n) x) = 4) r f 2r+1 2r) x) = 4) r f 1) x) = 4) r sin 2x = 1) r+1 2 n sin x. 12

13 Thus { 1) r 2 n 1 cos 2x if n = 2r even f n) x) = 1) r+1 2 n sin x if n = 2r odd. 6) Choosing x = 0 gives f n) 0) = Thus the Taylor Series for cos 2 x is 1 + n=1 n=2r even { 1) r 2 n 1 if n = 2r is even 0 if n 3 is odd 1) r n 1 xn 2 n! = 1 + r=1 1) r 2r 1 x2r 2 2r)!. For what x does lim n R n,0 cos 2 x) = 0? By Lagrange s form of the error R n,0 cos 2 x ) = f n+1) c) x n+1 n + 1)! for some c between 0 and x. In the present case, we look at the modulus so we don t worry about the sign, when f n+1) c) { 2 n cos 2c = if n + 1 is even, 2 n+1 sin c if n + 1 is odd, from 6) above. Then f n+1) c) 2 n+1 in both cases. Thus, by the Lemma above, Rn,0 cos 2 x ) 2 x ) n+1 0 n + 1)! as n for any x R. Hence the Taylor Series for cos 2 x converges to cos 2 x for all x R. Example In the Appendix we show that the Taylor Series for 1 + x) t for t R, is t t 1)... t r + 1) x r, r! r=0 and this converges to 1 + x) t when 1 < x < 1. This is a generalisation of the Binomial Theorem. 13

14 Example In the Appendix we show that the Taylor Series for ln 1 + x) is 1) r 1 x r, r r=1 and this converges to ln 1 + x) when 1 < x 1. If we put x = 1 we get ln 2 = ) 6 14

15 Appendix Contents Taylor Polynomial for e x sin x + cos x), T 8.0 cos 2 x) Taylor Polynomial and error terms for sin 2 x, Taylor Series for sin x around π/2, Taylor s Theorem without error term, Taylor s Theorem with error implies the Mean Value Theorem, Integral form of the error, lim x 0 cos x 1) /x 2, Taylor Series for e x sin x, Binomial Theorem with real exponent, Taylor Series for e x sin x + cos x), Taylor Series for ln 1 + x). 1. Example With f x) = e x sin x + cos x) calculate T 3,0 f x). Solution f 1) x) = e x sin x + cos x) + e x cos x sin x) = 2e x cos x, f 2) x) = 2e x cos x 2e x sin x = 2e x cos x sin x), f 3) x) = 2e x cos x sin x) + 2e x sin x cos x) = 4e x sin x. f 4) x) = 4e x sin x 4e x cos x = 4f x). 15

16 The fact that a derivative is connected to the function simplifies matters greatly. For now f 5) x) = 4f 1) x), f 6) x) = 4f 2) x), f 7) x) = 4f 3) x) and f 8) x) = 4f 4) x) = 16f x). Thus f 0) = 1, f 1) 0) = 2, f 2) 0) = 2, f 3) 0) = 0, f 4) 0) = 4 f 5) 0) = 8 f 6) 0) = 8 f 7) 0) = 0 and f 8) 0) = 16. Hence T 3,0 f x) = 1 + 2x + 2 x2 2! + 0x3 3! 4x4 4! 8x5 5! 8x6 6! + 0x7 7! + 16x8 8! = 1 + 2x + x 2 x4 6 x5 15 x x Example Calculate T 8.0 cos 2 x ) Solution If f x) = cos 2 x then f x) = 2 cos x sin x = sin 2x. This last equality will save a lot of effort when differentiating. Leave it to the student to check that T 8.0 cos 2 x ) = 1 x x x x8. 3. Example Let f x) = sin 2 x. Calculate T 5,0 f x) and use Lagrange s form of the error to bound sin 2 0.1) T 5,0 f 0.1). Solution f x) = 2 sin x cos x = sin 2x f 0) = 0, f x) = 2 cos 2x, f 0) = 2, f x) = 4 sin 2x = 4f x), f 0) = 0, f 4) x) = 4f x), f 4) 0) = 8, f 5) x) = 4f x) = 16f x) f 5) 0) = 0. f 6) = 16f x). 16

17 Thus T 5,0 f x) = 0 + 0x + 2 x x3 3! 8x4 4! + 0x5 5! = x x4. And Lagrange s form of the error states that, for some c between x and 0, R 5,0 f x) = x6 d 6 6! dx 6 sin2 x = x6 32 cos 2c). x=c 6! First note that with x = 0.1 > 0 we have R 5,0 f x) > 0, in which case sin 2 0.1) = T 5,0 f 0.1) + R 5,0 f 0.1) > T 5,0 f 0.1). Yet so T 5,0 f 0.1) = 0.1) )4 = , sin 2 0.1) > For any upper bound we have Thus R 5,0 f 0.1) 32 6! 0.1) sin 2 0.1) = T 5,0 f 0.1) + R 5,0 f 0.1) 0.1) ) ! 0.1)6 = Hence In fact < sin 2 0.1) < sin 2 0.1) =

18 4. Example Calculate the Taylor Series for sin x around π/2. Solution Consider d n dx sin x n x= π 2 π ) = sin 2 + nπ = 2 = { 0 if n = 2r + 1 1) r if n = 2r. Hence the Taylor Series around π/2 is 1) r x π ) 2r. 2r)! 2 r=0 0 if n odd 1 if n = 0, 4, 8,... 1 if n = 2, 6, 10,... The same proof as for sin x around 0 will show that this converges to sin x for all real x. 5. Taylor s Theorem without an error term would have stated that if the first n derivatives of f exist and are continuous at a then R n,a f x) lim x a x a) n = 0. 8) In assuming a little more, namely that the first n + 1 derivatives of f exist which implies continuity of f i), 1 i n) we can deduce a little more, namely 3) or 4). 6. Putting n = 0 in 1), the definition of the remainder gives f x) = T 0,a f x) + R 0,a f x) = f a) + R 0,a f x). Using Lagrange s Theorem gives f x) = f a) + f c) x a) for some c between a and x, by 4). Rearranging, f x) f a) = f c) x a which is the Mean Value Theorem seen earlier. But this is not a proof of the Mean Value Theorem since we used the ideas of the Mean Value Theorem to prove Lagrange s form of the error, 4). 18

19 7. An alternative form of the remainder which is sometimes useful is: Integral Form: Cauchy 1821) If the first n + 1 derivatives of f exist and are continuous on an open interval containing a and x then R n,a f x) = x a f n+1) t) n! x t) n dt. We are jumping the gun here since we have to wait until the next chapter before we define integration!) Note that we have to assume that f n+1) t) not only exists but is continuous on a, x). This is more than is required for either Lagrange s or Cauchy s forms of the error. 8. Taylor s Theorem in the form cos x = 1 x2 2! + x3 3! sin c for some c between 0 and x leads to cos x 1 lim = 1 x 0 x 2 2, which we have seen in Part 1 of this course. But now we see that 1/2 arises as a coefficient in the Taylor series. 9. Example Calculate the Taylor series for f x) = e x cos x around 0. Solution f 1) x) = e x cos x e x sin x, so f 1) 0) = 1, f 2) x) = e x cos x e x sin x e x sin x e x cos x = 2e x sin x, so f 2) 0) = 0, f 3) x) = 2e x sin x 2e x cos x, so f 3) 0) = 2, f 4) x) = 2e x sin x 2e x cos x 2e x cos x + 2e x sin x = 4e x cos x = 4f x) so f 4) 0) = 4, 19

20 The fact that f 4) x) = 4f x) makes life easy, we start repeating ourselves. f 5) x) = 4f 1) x) so f 5) 0) = 4, f 6) x) = 4f 2) x) so f 6) 0) = 0, f 7) x) = 4f 3) x) so f 7) 0) = 8, f 8) x) = 4f 4) x) = 16f x) so f 8) 0) = 16. So the Taylor series starts as 1 + x + 0 2! x2 2 3! x3 4 4! x4 4 5! x ! x7... 9) = 1 + x 1 3 x3 1 6 x x x7... The question must then be whether this is the same as we would obtain from multiplying together the series for e x and cos x? Try it and see... Question Does the series 9) converge to e x cos x? Solution We first need a bound on the size of f n) x). From the first list above we see that f n) x) 4e x for all x and 0 n 4. But then f 4) x) = 4f x) which means that f n) x) = 4) k f n 4k) x) as long as n 4k 0. We can choose k 1 such that 0 n 4k 1 < 4 which means that f n) x) = 4 k 1 f n 4k 1 ) x) 4 k 1 +1 e x. Finally 0 n 4k 1 implies k < n when n 2. Thus, for such n we have the bound f n) x) 4 n e x for all x. Hence, for each fixed x R, there exists c between 0 and x for which f n+1) c) R n,0 f x) = x n+1 4n+1 e c n + 1)! n + 1)! x n+1 x 4 x )n+1 e n + 1)! 0 20

21 as n by Lemma above. Hence, for each fixed x R, we have lim n R n,0 f x) = 0 and so the series 9) converges to e x cos x for all x R ) The Binomial Expansion for 1 + x) t = e t ln1+x), for any exponent t R, not just t N. Note though, that for general t, the function 1 + x) t is only defined for x > 1 for only then is ln 1 + x) well-defined). Since d n 1 + x) t dx n = t t 1)... t n + 1) 1 + x) t n, the Taylor Series for 1 + x) t is 1+tx+ t t 1) x 2 t t 1) t 2) + x = 2! 3! r=0 t t 1)... t r + 1) x r. r! To prove that lim n R n,0 1 + x) t ) = 0 it transpires that it is easier to use Cauchy s form of the error. I leave it to the interested student to check this, and thus find that the Taylor Series converges to 1 + x) t for 1 < x < Example Show that the Taylor Series for f x) = e x cos x + sin x) converges to f x) for all x R. Solution We have already calculated that the Taylor series of f x) starts as 1 + 2x + x x x x x8 +..., 10) and f 4) x) = 4f x). From Taylor s Theorem with Lagrange s form of the error we have R n,0 f x) = f n+1) c) n + 1)! xn+1 for some c between 0 and x. As for the e x sin x example above we can show that f n) c) 4 n e c 4 n e x, 21

22 since we need a bound not containing the unknown c. Thus x 4x)n+1 R n,0 f x) e n + 1)! 0 as n, by the Lemma above. Hence 10) converges to e x cos x + sin x) for all x R. 12. In this course we have defined the natural logarithm as the inverse of e x. Thus we can calculate the Taylor series of ln 1 + x). First published by Mercator in 1668, the series is x x2 2 + x3 3 x4 4 + x The ratio test shows the series converges for x < 1, while the Alternating Series Test shows that it converges when x = 1. But again we have to show that it converges to ln 1 + x). Writing f x) = ln 1 + x) then f j) x) = 1)j+1 j 1)! 1 + x) j for all j 1. The integral form of the error states R n,0 f x) = x 0 f n+1) t) n! x t) n dt = 1) n+2 x 0 x t) n 1 + t) n+1 dt. The most interesting case because it is the most difficult) is x = 1 when we get the integral 1 ) n 1 t dt I n = 1 + t 1 + t). Substitute w = 1 t) / 1 + t) to transform into I n = w n 1 + w dw 1 0 w n dw = 1 n as n. Thus lim n R n,0 ln 1 + x)) x=1 = 0. This justifies 7). 22

23 13. Example Find the Taylor Series for sin 2 x around a = 0 and show that the series converges to sin 2 x for all x R. Solution Let f x) = sin 2 x. Then Hence f 1) x) = 2 sin x cos x = sin 2x, f 2) x) = 2 cos 2x, f 3) x) = 4 sin 2x f 4) x) = 8 cos 2x f 5) x) = 16 sin 2x f 6) x) = 32 cos 2x. f 0) = 0, f 1) 0) = 0, f 2) 0) = 2, f 3) 0) = 4f 1) 0) = 0, f 4) 0) = 4f 2) 0) = 8, f 5) 0) = 16f 1) 0) = 0, f 6) 0) = 16f 2) 0) = 32, From the first list above we see that. { 1) t 1 f r) 2 x) = 2t 2 sin 2x if r = 2t 1 is odd 1) t 1 2 2t 1 cos 2x if r = 2t is even. Thus f r) 0) = { 0 if r is odd 1) t 1 2 r 1 if r = 2t 2. Therefore the Taylor series is 0 + 0x + r=2 r even r=2t 1) t 1 2 r 1 x r = r! 1) t 1 2 2t 1 x 2t. 2t)! t=1 23

24 The first few terms are x x x x To show that the series converges to sin 2 x for all x R we need show that lim R n,0 sin 2 x ) = 0 n for all x R. To do this we use Lagrange s form of the error so, for any x R we have R n,0 sin 2 x ) = f n+1) c) n + 1)! xn 11) for some c between 0 and x. From the first list above we see that f r) x) { 2 2t 2 = 2 r 1 if r = 2t 1 is odd 2 2t 1 = 2 r 1 if r = 2t is even. Thus f r) x) 2 r 1 for r 1. Hence 11) becomes Rn,0 sin 2 x ) f n+1) c) = x n n + 1)! 2n n + 1)! x n = 2 x )n n + 1)! 0 as n by Lemma above. Thus R n,0 sin 2 x ) 0 as n and the series converges to sin 2 x for all x R. 14. Example f x) = sin x and a = 0 In this course we have started with the definition of sine as the ratio of sides in a right-angled triangle. We have assumed the addition formula for sine and managed to show that sine is continuous and differentiable of all orders for all x. From such results we have d x dx sin x = cos x = sin + π ), 2 d 2 x dx sin x = sin x = sin + 2 π )

25 And in general, for n N, So d n dx sin x n = sin x=0 d n x dx sin x = sin + n π ). n 2 n π ) = 2 0 if n even 1 if n = 1, 5, 9,... 1 if n = 3, 7, 11,... = { 0 if n = 2r 1) r if n = 2r + 1 Thus the Taylor polynomial for sin x at 0 of degree n is T n,0 sin x) = x x3 3! + x5 5! x7 7! )R x 2R+1 2R + 1)! = R r=0 1) r x 2r+1, 2r + 1)! where 2R + 1 n < 2R + 1) Find Lagrange s form for R n,0 sin x) and show that R n,0 sin x) x n+1 n + 1)! Solution By Lagrange s bound on the error, R n,0 sin x) = dn+1 sin x x n+1 dxn+1 x=c n + 1)! for some c between 0 and x. But this derivative is either sin c or cos c, both of which in modulus are 1. Hence result. 16. Prove that sin 0.1) T 5,0 sin 0.1)) Hence give sin 0.1 to 8 decimal places. 25

26 Solution By definition of the Remainder we have sin x) T 5,0 sin x)) = R 5,0 sin x) x 6 6! from above. Apply this with x = 0.1 to get i.e. sin 0.1) T 5,0 sin 0.1)) ! = , T 5,0 sin 0.1)) sin 0.1) T 5,0 sin 0.1)) Yet T 5,0 sin 0.1) = ) )5 = So sin 0.1) that is, sin 0.1) Looking for digits in common between the upper and lower bounds we see that to 8 decimal places sin 0.1 is In fact sin 0.1 = ) 17. An example in the notes is not as strong as it could be. Example Use Lagrange s form for the error to show that cos2 x 1 x ) 3 x x 6. Hence show that cos Thus to 6 decimal places. cos =

27 In fact cos = Solution The observation to make is that the polynomial of degree 4 is, in fact, the Taylor polynomial of degree 5. This is because 1 x x4 = 1 + 0x x 2 + 0x x4 + 0x 5 = T 5,0 cos 2 x ). Thus cos2 x 1 x ) 3 x4 = R5,0 cos 2 x ) f 6) c) = x 6 6! = 25 6! sin 2c x x 6. The Taylor polynomial approximation to cos 2 x at x = 0.1 is T 5,0 cos 2 x ) x=0.1 = 1 0.1) )4 = The error in this approximation is Hence while 2 45 x 6 = )6 = cos = cos =