Introduction to Power Series Power Series As Functions

Section 3.4: Introduction to power series Power Series as Functions

Deﬁnition 70 Deﬁne a “function” by

A power series in the variablex resembles a polynomial, except that it ∞ n 2 may contain inﬁnitely many positive powers ofx. It is an expression of f(x)= x = 1 +x+x + ... the type �n=0 ∞ i 2 ai x =a 0 +a 1x+a 2x + ... , If we try to evaluate this function at x=2 , we get a series of real numbers. �i=0 ∞ n 2 where eacha i is a number. f (2) = 2 = 1 + 2 + 2 + ... �n=0 Example 71 This series is divergent, so our power series does not deﬁne a function that can be evaluated at 2. ∞ xn = 1 +x+x 2 +x 3 + ... �n=0 is a power series.

Question: Can we think of a power series as a function ofx? Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 205 / 171 Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 206 / 171

n 2 Power Series as Functions f(x)= n∞=0 x = 1 +x+x + ... Define a “function” by � If we try evaluating atx= 1 , we get ∞ 2 f(x)= xn = 1 +x+x 2 + ... n 2 1 ∞ 1 1 1 �n=0 f = = 1 + + + ... 2 2 2 2 � � n=0 � � � � If we try to evaluate this function at x=2 , we get a series of real � numbers. This is a geometric series withfirst term a=1 and common ratio 1 ∞ n 2 r= . We know that if r <1 , such a series converges to the f (2) = 2 = 1 + 2 + 2 + ... 2 a | | number . In this case �n=0 1 r This series is divergent, so our power series does not define a − function that can be evaluated at 2. a 1 = 1 = 2, If we try evaluating at 0 (and allow that thefirst termx 0 of the 1 r 1 − − 2 power series is interpreted as 1 for all values ofx), we get 1 and we have f( 2 ) = 2. 2 f (0) = 1 + 0 + 0 + =1 . 1 ··· So we can evaluate our function atx= 2 . So it does make sense to “evaluate” this function atx = 0. Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 206 / 171 Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 207 / 171 n 1 f(x)= n∞=0 x = 1 x , for x <1 Which functions have power series representations? − | | A geometric� series of this sort converges provided that the absolute value of its common ratio is less than 1. In general for any value ofx whose Remark: The power series representation is not particularly useful if you absolute value is less than 1 (i.e. anyx in the interval ( 1, 1)), wefind 1 − 1 want to calculate 1 x for some particular value ofx, because this is that f(x) is a convergent geometric series, converging to . − 1 x easily done directly. However, if we could obtain a power series − representation for a function like sinx and use it to evaluate (or Conclusion: For values of x in the interval ( 1, 1) (i.e. x < 1), the approximate) sin(1) or sin(9) or sin(20), that might be of real practical 1 − | | n function f(x)= 1 x coincides with the power series n∞=0 x . use. These numbers are not easy to obtain directly because the definition − of sinx doesn’t tell us how to calculate sinx for a particularx - you can � 1 ∞ use a calculator of course but how does the calculator do it? = xn, for x <1 . 1 x | | − �n=0 Questions: What functions can be represented by power series, and on The interval ( 1, 1) is called the interval of convergence of the power what sorts of interval or subsets ofR? If a function could be represented − series, and 1 is the radius of convergence. We say that the power series by a power series, how would we calculate the coefficients in this series? 1 n representation of the functionf(x)= 1 x is n∞=0 x , for values ofx in the interval ( 1, 1). − − �

Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 208 / 171 Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 209 / 171

Maclaurin (or Taylor) series Maclaurin (or Taylor) series

Suppose thatf(x) is an infinitely differentiable function (this means that Suppose thatf(x) is an infinitely differentiable function (this means that all the deriviatives off are themselves differentiable), and suppose thatf all the deriviatives off are themselves differentiable), and suppose thatf is represented by the power series is represented by the power series

∞ n ∞ n f(x)= cnx . f(x)= cnx . �n=0 �n=0 We can work out appropriate values for the coefficients cn as follows. We can work out appropriate values for the coefficients cn as follows. n n Put x=0 . Then f (0) =c + ∞ c (0) = f (0) =c . Put x=0 . Then f (0) =c + ∞ c (0) = f (0) =c . 0 n=1 n ⇒ 0 0 n=1 n ⇒ 0 The constant term in the power series is the value off at 0. The constant term in the power series is the value off at 0. � � To calculatec 1, look at the value of the first derivative off at 0, and differentiate the power series term by term. We expect

2 ∞ n 1 f �(x)=c + 2c x+3c x + = nc x − . 1 2 3 ··· n �n=1 Then we should have f (0) =c + 2c 0+3c 0+ =c . Thus � 1 2 × 3 × ··· 1 c1 =f �(0).

Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 210 / 171 Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 210 / 171 n n f(x)= n∞=0 cnx f(x)= n∞=0 cnx

For �c2, look at the second derivative off . We expect For �c2, look at the second derivative off . We expect

2 3 2 3 f ��(x) = 2(1)c2 + 3(2)c3x + 4(3)c4x + 5(4)c5x + ... f ��(x) = 2(1)c2 + 3(2)c3x + 4(3)c4x + 5(4)c5x + ...

Puttingx = 0 givesf ��(0) = 2(1)c2 or Puttingx = 0 givesf ��(0) = 2(1)c2 or f (0) f (0) c = �� . c = �� . 2 2(1) 2 2(1)

(3) Forc 3, look at the third derivativef (x). We have

(3) 2 f (x) = 3(2)(1)c 3 + 4(3)(2)c4x + 5(4)(3)c5x + ...

(3) Settingx = 0 givesf (0) = 3(2)(1)c3 or

f (3)(0) c = 3 3(2)(1)

Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 211 / 171 Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 211 / 171

Coeﬃcients of the Maclaurin Series Power series representation of sinx

Continuing this process, we obtain the following general formula forc n: n Write f(x) = sinx , and write ∞ c x for the Maclaurin series of 1 n=0 n c = f (n)(0). sinx. Then n n! � f (0) = sin 0 = 0 = c = 0 ⇒ 0 Deﬁnition 72 For a positive integern, the number n factorial, denoted n! is deﬁned by

n!=n (n 1) (n 2) ... 3 2 1 . × − × − × × × The number 0! (zero factorial) is deﬁned to be 1.

Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 212 / 171 Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 213 / 171 Power series representation of sinx Power series representation of sinx

Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 213 / 171 Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 213 / 171

Power series representation of sinx Power series representation of sinx

n n Write f(x) = sinx , and write n∞=0 cnx for the Maclaurin series of Write f(x) = sinx , and write n∞=0 cnx for the Maclaurin series of sinx. Then sinx. Then � � f (0) = sin 0 = 0 = c = 0 f (0) = sin 0 = 0 = c = 0 ⇒ 0 ⇒ 0 f (0) = cos 0 = 1 = c = 1 f (0) = cos 0 = 1 = c = 1 � ⇒ 1 � ⇒ 1 f (0) = sin 0 = 0 = c = 0 = 0 f (0) = sin 0 = 0 = c = 0 = 0 �� − ⇒ 2 2! �� − ⇒ 2 2! (3) 1 1 (3) 1 1 f (0) = cos 0 = 1 = c = − = f (0) = cos 0 = 1 = c = − = − − ⇒ 3 3! − 6 − − ⇒ 3 3! − 6 f (4)(0) = sin 0 = 0 = c = 0 = 0 ⇒ 4 4!

Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 213 / 171 Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 213 / 171 Power series representation of sinx Power series representation of sinx

This pattern continues : This pattern continues : If k is even then f (k) (0) = sin 0 = 0 , so c = 0. If k is even then f (k) (0) = sin 0 = 0 , so c = 0. ± k ± k Ifk is odd and k 1 mod 4 then f (k) (0) = cos 0 = 1 and c = 1 . ≡ k k!

Thus the Maclaurin series for sinx is given by Thus the Maclaurin series for sinx is given by

∞ ( 1)k 1 1 1 ∞ ( 1)k 1 1 1 − x2k+1 =x x3 + x5 x7 + ... − x2k+1 =x x3 + x5 x7 + ... (2k + 1)! − 3! 5! − 7! (2k + 1)! − 3! 5! − 7! �k=0 �k=0 Note that this series only involves odd powers ofx - this is not surprising Note that this series only involves odd powers ofx - this is not surprising because sin is an odd function; it satisﬁes sin( x)= sinx . because sin is an odd function; it satisﬁes sin( x)= sinx . − − − −

Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 214 / 171 Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 214 / 171

Power series representation of sinx Power series representations of sinx and cosx

Theorem 73 This pattern continues : For every real number x, the above series converges to sinx. If k is even then f (k) (0) = sin 0 = 0 , so c = 0. ± k Thus computing partial sums of this series gives us an eﬀective way of Ifk is odd and k 1 mod 4 then f (k) (0) = cos 0 = 1 and c = 1 . ≡ k k! approximating sinx for any real numberx. Ifk is odd and k 3 mod 4 then f (k) (0) = cos 0 = 1 and ≡ − − c = 1 . k − k! Exercise 74 Thus the Maclaurin series for sinx is given by Show that the Maclaurin series for cos x is given by

∞ ( 1)k 1 1 1 k − x2k+1 =x x3 + x5 x7 + ... ∞ ( 1) 2k (2k + 1)! − 3! 5! − 7! − x . k=0 (2k!) � �k=0 Note that this series only involves odd powers ofx - this is not surprising because sin is an odd function; it satisﬁes sin( x)= sinx . (Note that this can be obtained by diﬀerentiating term-by-term the series − − d for sinx, as we would expect since (sinx) = cosx.) dx

Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 214 / 171 Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 215 / 171 Learning outcomes for Section 3.4

After studying this section you should be able to State the meaning of the term power series, Explain the concept of the radius of convergence of a power series, Calculate the coeﬃcients in (an initial segment of) the Maclaurin series representation of a given function.

Dr Rachel Quinlan MA180/MA186/MA190 Calculus Power series 216 / 171