e x = 1 + x + x2 2! + x3

Section 11.10 Taylor Series and the Binomial Series Given a function f (x), we would like to be able to find a power series that represents the funct...
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Section 11.10

Taylor Series and the Binomial Series Given a function f (x), we would like to be able to find a power series that represents the function. For example, in the last section we noted that we can represent ex by the power series ex = 1 + x +

x2 x3 x4 + + + ..., 2! 3! 4!

and the power series converges to ex for all x. In this section we will discuss a method for finding the power series for any function. We begin with a motivating example.

Example. Find the power series centered at x = 0 for f (x) = ln(x + e). Right now we have very little to work with in order to solve this problem; in fact, the only really useful piece of information we have is the form for a power series centered at x = 0: ∞ X

cn xn = c0 + c1 x + c2 x2 + c3 x3 + . . . .

n=0

If ln(x + e) has a power series representation at x = 0, then it should be of this form. In other words, we know that we would like to rewrite ln(x + e) as ln(x + e) = c0 + c1 x + c2 x2 + c3 x3 + . . . ; fortunately for us, the only real unknowns here are the constants cn , which we need to find. It turns out that c0 is quite easy to determine; if we evaluate ln(x + e) at the center x = 0 of the series, the only non-zero term is c0 : 1 = ln(0 + e) = c0 + c1 (0) + c2 (0)2 + c3 (0)3 + . . . = c0 . So c0 = ln(0 + e) = 1. Unfortunately, there are infinitely many constants to find; it would be helpful if we had some method for isolating the constant cn in an analogous way to how c0 was originally isolated. Fortunately for us, we can use calculus to do this: since f (x) = ln(x + e) = c0 + c1 x + c2 x2 + c3 x3 + . . . , we know that f 0 (x) =

1 = c1 + 2c2 x + 3c3 x2 + 4c4 x3 + . . . , x+e

so that

1 1 = . 0+e e Similarly, we can find c2 using the second derivative of f (x): c1 =

f 00 (x) = −

1 = 2c2 + 6c3 x + 12c4 x2 + 20c5 x3 + . . . , (x + e)2 1

Section 11.10

so

1 . 2e2 The remaining constants can be found in a similar fashion; in fact, it is not too difficult to see that f (n) (0) cn = . n! So the power series centered at 0 for ln(x + e) is c2 = −

∞ X f (n) (0)

n!

n=0

xn = 1 +

x x2 2x3 − 2 + 3 + ..., e e e

which converges for all x.

The ideas we used above to find the constants cn in a particular power series representation actually work in general for finding representations of any function f (x): Theorem. If f (x) is a function with a power series expansion at x = a, then the power series is given by ∞ X f (n) (a) f 00 (a) (x − a)n = f (a) + f 0 (a)(x − a) + (x − a)2 + . . . n! 2! n=0

f (n) (a)

where is the nth derivative of f evaluated at x = a; this series is known as the Taylor series generated by f at x = a. The Taylor series generated by f at x = 0, given by ∞ X f (n) (0) n=0

n!

xn = f (0) + f 0 (0)x +

f 00 (0) 2 x + ... 2!

is called the Maclaurin series generated by f . We must be careful to interpret the theorem correctly; it is not always the case that the Taylor series of a function f (x) actually converges to f (x) on its entire interval of convergence. In fact, the function f (x) might not be equal to its Taylor series. We will discuss this problem in more detail in the next section.

Example. Find the Taylor series centered at x = 1 for f (x) = ln(x2 ). The Taylor series has form ∞ X f (n) (1) n=0

n!

(x − 1)n = f (1) + f 0 (1)(x − 1) +

We need to calculate the derivatives f (n) (1): 2

f 00 (1) (x − 1)2 + . . . . 2!

Section 11.10

f (x) = ln(x2 )

f (1) = 0

2 x 2 f 00 (x) = − 2 x 4 f 000 (x) = 3 x

f 0 (1) = 2 = 2(0!)

f 0 (x) =

f 00 (1) = −2 = −2(1!) f 000 (1) = 4 = 2(2!) .. .

f (n) (x) = (−1)n+1

2((n − 1)!) xn

f (n) (1) = 2(−1)n+1 (n − 1)!

So the Taylor series for ln(x2 ) is ∞ X n=1



2(−1)n+1 (n − 1)!

(x − 1)n X (x − 1)n = 2(−1)n+1 . n! n n=1

The radius of convergence for the power series is R = 1.

Notice that the partial sums of the Taylor series above, such as T1 = 2(x − 1) = 2x − 2, T2 = 2(x − 1) − (x − 1)2 = −x2 + 4x − 3, and 2 T3 = 2(x − 1) − (x − 1)2 + (x − 1)3 3 are just polynomials. This leads us to a definition: Definition. The Taylor polynomial of order n generated by f at x = a is the polynomial Tn (x) = f (a) + f 0 (a)(x − a) +

f 00 (a) f (n) (a) (x − a)2 + . . . + (x − a)n . 2! n!

Notice that this polynomial (which has a finite number of terms) is simply a truncation of the Taylor series for f (x); we get the kth Taylor polynomial by erasing all terms from the Taylor series so that n > k. We will see momentarily that Taylor polynomials are extremely useful for estimating a function’s value.

3

Section 11.10

Example. The Taylor series for ex is

∞ X xn i=0

n!

;

find the Taylor polynomial of order 4 generated by ex . To find the Taylor polynomial, we simply need to truncate the Taylor series, in this case at n = 4: 1 1 1 T4 (x) = 1 + x + x2 + x3 + x4 . 2 6 24

Convergence of Taylor Series We have a final question left to answer: given a function f (x) and the Taylor series generated by f at x = a, does the Taylor series generated by f actually converge to f (x)? In other words, is the equal sign in ∞ X f (n) (a) f (x) = (x − a)n n! n=0

justified? We can answer this question using Taylor polynomials. We can think of Tn (x) as a tool for estimating the value for the function f (x); the difference f (x) − Tn (x) is also a function, so we set Rn (x) = f (x) − Tn (x) and call R(x) the remainder function. Rewriting the equation, we have f (x) = Tn (x) + Rn (x). We know that lim Tn (x) =

n→∞

∞ X f (n) (a)

n!

n=0

and we want to know if f (x) =

∞ X f (n) (a)

n!

n=0

(x − a)n ,

(x − a)n .

Notice that if lim Rn (x) = 0, then n→∞

f (x) = lim (Tn (x) + Rn (x)) = lim Tn (x) + 0 = lim Tn (x) = n→∞

n→∞

n→∞

This observation leads to the following theorems: 4

∞ X f (n) (a) n=0

n!

(x − a)n .

Section 11.10

Theorem. If f (x) has Taylor polynomials Tn (x) and remainder function Rn (x), and if lim Rn (x) = 0,

n→∞

then the Taylor series generated by f (x) at x = a converges to f for |x − a| < R, and we write f (x) =

∞ X f (n) (a) n=0

n!

(x − a)n

on the interval of convergence of the series. Taylor’s Formula. If |f (n+1) (x)| ≤ M for |x − a| ≤ d, then |Rn (x)| ≤

M |x − a|n+1 (n + 1)!

for |x − a| ≤ d.

Well-known Taylor Series Two Taylor series that we will see quite often are the series for sin x and for cos x. The function f (x) = sin x has Taylor series sin x =

∞ X

(−1)n

n=0

x2n+1 x3 x5 x7 =x− + − ..., (2n + 1)! 3! 5! 7!

and the series converges to sin x for all x. The function f (x) = cos x has Taylor series cos x =

∞ X

(−1)n

n=0

x2n x2 x4 x6 =1− + − ..., (2n)! 2! 4! 6!

and the series converges to cos x for all x.

Example. Use the 4th order Taylor polynomial generated by cos x at x = 0 to estimate cos(1). Since x2 x4 x6 cos x = 1 − + − + ... 2! 4! 6! for all x, we could evaluate cos(1) by “plugging in” 1 into either side of the equation. Fortunately, the 4th Taylor polynomial for cos x, T4 (x) = 1 − 5

x2 x4 + 2! 4!

Section 11.10

can be evaluated quite easily at x = 1 (unlike the infinite series), and 1−

x2 x4 x6 x2 x4 + − ... ≈ 1 − + . 2! 4! 6! 2! 4!

Thus our approximation is 12 14 + 2! 4! 1 1 1− + 2 24 24 − 12 + 1 24 13 24 .542.

cos(1) ≈ 1 − = = = ≈

Mathematica estimates cos(1) ≈ .540, so our estimate was reasonably accurate.

The Binomial Series Another important type of Taylor series, called the binomial series, will allow us to calculate the power series representation for an entire class of functions: every function of the form (1 + x)m can be represented by the binomial series. Before we look at this series in detail we need a bit of new notation.  Definition. Given a real number k and nonnegative integer n, the number nk , read “k choose n”, is given by k(k − 1)(k − 2) · . . . · (k − n + 1) n!  k if n ≥ 1; 0 is defined to be 1.

 Example. Find 74 . In this example, we have k = 7 and n = 4. Then k − n + 1 = 4   7 7·6·5·4 = 4! 4 = 35.

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Section 11.10

Example. Calculate

− 13  . 3

With k = − 13 and n = 3, we compute

− 13  3

by noting that

1 7 k−n+1=− −3+1=− ; 3 3 so

− 31  3

is given by 

− 31 3



(− 13 ) · (− 43 ) · (− 73 ) 3! 14 =− . 81 =

Theorem. The function (1 + x)k , where k is any real number and |x| < 1, has power series representation ∞   X k n k (1 + x) = 1 + x . n n=1

This power series is called the binomial series, and converges to (1 + x)k when −1 < x < 1.

Two different cases emerge depending on the value for k. If k ≥ 0 is an integer, then for any n so that n + 1 ≥ k,   k k · (k − 1) · (k − 2) · . . . · (k − k) · . . . · (k − n + 1) = = 0. n n! In other words, if k is an integer and k ≤ n + 1, then the binomial series will have finitely many terms. For any value of k other than a nonnegative integer, the binomial series will have an infinite number of terms.

Example. Find the power series for

√ 3

8 + x.

If we rewrite the function in the form √ 3

r

x 8(1 + ) 8 r x = 23 1+ 8 x 1 = 2(1 + ) 3 8 3

8+x =

7

Section 11.10

it is clear that it may be represented with the binomial series. Since k

(1 + x) =

∞   X k n=0

n

xn

with k = 13 , we have √ 3

x 1 )3 8  n ∞  X 1/3 x =2 n 8

8 + x = 2(1 +

n=0

1

x + 1 8 3

= 2(1 + = 2(1 +

1 3

1! 1 = 2(1 + · 3 = 2(1 + =2+

·

1 1 3 ( x )2 + 3 ( x )3 + . . .) 2 8 3 8

1 −2 1 −2 −5 · · · 3 x x x + 3 3 · ( )2 + 3 3 · ( )3 + . . .) 8 2! 8 3! 8 x 1 x 2 5 x − ·( ) + · ( )3 + . . .) 8 9 8 81 8

x x2 5x3 − + + . . .) 24 576 41472

x x2 5x3 − + + ... 12 288 20736

Applications of Taylor Series Taylor series behave very well on their intervals of convergence, as indicated by the following theorems: Theorem. If the Taylor series Tf (x) converges to f (x) on the interval I and the Taylor series Tg (x) converges to g(x) on the same interval, then the Taylor series for the function f (x) ± g(x) is given by Tf (x) ± Tg (x), and this Taylor series converges to f (x) ± g(x) on I. In essence, the theorem says that we can add or subtract Taylor series on a common interval of convergence, and the result is again a converging Taylor series. We may also multiply a Taylor series by a constant without changing it too much: Theorem. If the Taylor series Tf (x) converges to f (x) on the interval I and c is any constant, then the Taylor series for the function cf (x) is given by cTf (x), and this Taylor series converges to cf (x) on I.

8

Section 11.10

One more important point to note: if a Taylor series converges absolutely on an interval, then we may rearrange infinitely many of its terms without changing the value for the series. This is not true for series that only converges conditionally.

ex − 1 − x . x→0 x2

Example. Evaluate lim

Notice that the limit above yields the indeterminate form 00 , so we could use L’Hopital’s rule to evaluate it. However, we can use series to do the same thing: we know that the Taylor series for ex is given by x2 x3 ex = 1 + x + + + ...; 2! 3! in addition, it is easy to see that the Taylor series for −1 − x is just −1 − x. So ex − 1 − x can be rewritten as ex − 1 − x = (1 + x +

x2 x3 x4 x2 x3 + + . . .) − 1 − x = + + + ... 2! 3! 2! 3! 4!

Then x2 2!

x3 3!

4

+ x4! + . . . x2 1 x x2 = lim + + + ... x→0 2 3! 4! 1 = . 2

ex − 1 − x lim = lim x→0 x→0 x2

+

Example. Find the Taylor series for cos2 x. Since

1 + cos 2x 1 1 = + cos 2x, 2 2 2 it will not be difficult to find the required series. Since cos2 x =

cos x =

∞ X

(−1)n

n=0

x2n , (2n)!

we can replace x with 2x to see that cos 2x =

∞ X

(−1)n

n=0

9

(2x)2n . (2n)!

Section 11.10

So 1 + cos 2x 1 1 = + cos 2x 2 2 2 ∞ (2x)2n 1 1X = + (−1)n 2 2 (2n)! n=0

1 1 4x2 16x4 64x6 + (1 − + − . . .) 2 2 2! 4! 6! 1 1 2x2 8x4 32x6 = +( − + − . . .) 2 2 2! 4! 6! 2x2 8x4 32x6 + − ... =1− 2! 4! 6! ∞ X 22n−1 x2n =1+ (−1)n . (2n)! =

n=1

x iθ Example. Use the Taylor √ series for e to calculate e , where θ is a (real) variable and i is the imaginary number i = −1. √ Before we make the calculation, note that since i = −1, powers of i are given by

i0 = 1, i1 = i, i2 = −1, i3 = −i, i4 = 1, i5 = i, . . . ; notice that the pattern repeats after the third power. Since x2 x3 x4 x5 x6 + + + + + ..., ex = 1 + x + 2! 3! 4! 5! 6! we see that eiθ = 1 + iθ +

(iθ)2 (iθ)3 (iθ)4 (iθ)5 (iθ)6 + + + + + ... 2! 3! 4! 5! 6!

θ2 θ3 θ4 θ5 θ6 −i + +i − − .... 2! 3! 4! 5! 6! Now the Taylor series converges absolutely for all θ, so we may rearrange terms as we wish; in particular, we write = 1 + iθ −

eiθ = 1 + iθ − 1−

θ2 2!

1−

θ2 2!

 =  =

θ2 θ3 θ4 θ5 θ6 −i + +i − − ... 2! 3! 4! 5! 6!    θ4 θ3 θ5 + − . . . + iθ − i + i − . . . 4! 3! 5!    θ4 θ3 θ5 + − ... + i θ − + − ... 4! 3! 5!

= cos θ + i sin θ

10

Section 11.10

using the Taylor series formulas for cos θ and sin θ. We have just shown that eiθ = cos θ + i sin θ; this remarkable formula is know as Euler’s formula, and leads to one of the most beautiful identities in mathematics: since cos π = −1 and sin π = 0, −eiπ = 1. This identity relates the five most important constants in mathematics, namely 1, − 1, e, i, and π.

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