11.10

11.10 Taylor and Maclaurin

Series

2

Taylor and Maclaurin Series

We start by supposing that f is any function that can be represented by a power series

f(x) = c₀ + c₁(x – a) + c₂(x – a)² + c₃(x – a)³ + c₄(x – a)⁴ + ^{. . .} | x – a | < R

Let’s try to determine what the coefficients c_n must be in terms of f.

To begin, notice that if we put x = a in Equation 1, then all terms after the first one are 0 and we get

f(a) = c₀

Taylor and Maclaurin Series

We can differentiate the series in Equation 1 term by term:

f′(x) = c₁ + 2c₂(x – a) + 3c₃(x – a)² + 4c₄(x – a)³ + ^{. . .}

|x – a | < R

and substitution of x = a in Equation 2 gives f′(a) = c₁

Now we differentiate both sides of Equation 2 and obtain f″(x) = 2c₂ + 2  3c₃(x – a) + 3  4c₄(x – a)² + ^{. . .}

| x – a | < R

Again we put x = a in Equation 3. The result is

4

Taylor and Maclaurin Series

Let’s apply the procedure one more time. Differentiation of the series in Equation 3 gives

f'''(x) = 2  3c₃ + 2  3  4c₄(x – a) + 3  4  5c₅(x – a)² +

. . . | x – a | < R

and substitution of x = a in Equation 4 gives f'''(a) = 2  3c₃ = 3!c₃

By now you can see the pattern. If we continue to differentiate and substitute x = a, we obtain

f ⁽ⁿ⁾ (a) = 2  3  4  ^{. . .}  nc_n = n!c_n

Taylor and Maclaurin Series

Solving this equation for the nth coefficient c_n, we get

This formula remains valid even for n = 0 if we adopt the conventions that 0! = 1 and f⁽⁰⁾ = f. Thus we have proved the following theorem.

6

Taylor and Maclaurin Series

Substituting this formula for c_n back into the series, we see that if f has a power series expansion at a, then it must be of the following form.

The series in Equation 6 is called the Taylor series of the function f at a (or about a or centered at a).

Taylor and Maclaurin Series

For the special case a = 0 the Taylor series becomes

This case arises frequently enough that it is given the special name Maclaurin series.

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Example 1

Find the Maclaurin series of the function f(x) = e^x and its radius of convergence.

Solution:

If f(x) = e^x, then f⁽ⁿ⁾(x) = e^x, so f⁽ⁿ⁾(0) = e⁰ = 1 for all n.

Therefore the Taylor series for f at 0 (that is, the Maclaurin series) is

Example 1 – Solution

To find the radius of convergence we let a_n = xⁿ/n!.

Then

so, by the Ratio Test, the series converges for all x and the radius of convergence is R = .

cont’d

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Taylor and Maclaurin Series

The conclusion we can draw from Theorem 5 and Example 1 is that if e^x has a power series expansion at 0, then

So how can we determine whether e^x does have a power series representation?

Taylor and Maclaurin Series

Let’s investigate the more general question: under what circumstances is a function equal to the sum of its Taylor series?

In other words, if f has derivatives of all orders, when is it true that

As with any convergent series, this means that f(x) is the limit of the sequence of partial sums.

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Taylor and Maclaurin Series

In the case of the Taylor series, the partial sums are

Notice that T_n is a polynomial of degree n called the n th-degree Taylor polynomial of f at a.

Taylor and Maclaurin Series

For instance, for the exponential function f(x) = e^x, the

result of Example 1 shows that the Taylor polynomials at 0 (or Maclaurin polynomials) with n = 1, 2, and 3 are

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Taylor and Maclaurin Series

The graphs of the exponential function and these three Taylor polynomials are drawn in Figure 1.

Figure 1

As n increases, T_n(x) appears to approach e^x in Figure 1. This suggests that e^x is equal to the sum of its Taylor series.

Taylor and Maclaurin Series

In general, f(x) is the sum of its Taylor series if

If we let

R_n(x) = f(x) – T_n(x) so that f(x) = T_n(x) + R_n(x) then R_n(x) is called the remainder of the Taylor series. If we can somehow show that lim_n→ R_n(x) = 0, then it

follows that

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Taylor and Maclaurin Series

We have therefore proved the following.

In trying to show that lim_n→ R_n(x) = 0 for a specific function f, we usually use the following Theorem.

Taylor and Maclaurin Series

To see why this is true for n = 1, we assume that

| f″(x)| ≤ M. In particular, we have f″(x) ≤ M, so for a ≤ x ≤ a + d we have

An antiderivative of f″ is f′, so by Part 2 of the Fundamental Theorem of Calculus, we have

f′(x) – f′(a) ≤ M(x – a) or f′(x) ≤ f′(a) + M(x – a)

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Taylor and Maclaurin Series

Thus

But R₁(x) = f(x) – T₁(x) = f(x) – f(a) – f′(a)(x – a). So

Taylor and Maclaurin Series

A similar argument, using f″(x) ≥ –M, shows that

So

Although we have assumed that x > a, similar calculations show that this inequality is also true for x < a.

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Taylor and Maclaurin Series

This proves Taylor’s Inequality for the case where n = 1.

The result for any n is proved in a similar way by integrating n + 1 times.

In applying Theorems 8 and 9 it is often helpful to make use of the following fact.

This is true because we know from Example 1 that the series Σ xⁿ/n! converges for all x and so its nth term approaches 0.

Example 2

Prove that e^x is equal to the sum of its Maclaurin series.

Solution:

If f(x) = e^x, then f^{(n + 1)}(x) = e^x for all n. If d is any positive number and | x | ≤ d, then |f^{(n + 1)}(x) | = e^x ≤ e^d.

So Taylor’s Inequality, with a = 0 and M = e^d, says that

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Example 2 – Solution

Notice that the same constant M = e^d works for every value of n. But, from Equation 10, we have

It follows from the Squeeze Theorem that

lim_{n →} | R_n(x) | = 0 and therefore lim_{n →} R_n(x) = 0 for all values of x. By Theorem 8, e^x is equal to the sum of its Maclaurin series, that is,

cont’d

Taylor and Maclaurin Series

In particular, if we put x = 1 in Equation 11, we obtain the following expression for the number e as a sum of an

infinite series:

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Example 8

Find the Maclaurin series for f(x) = (1 + x)^k, where k is any real number.

Solution:

Arranging our work in columns, we have

f(x) = (1 + x)^k f(0) = 1

f′(x) = k(1 + x)^{k – 1} f′(0) = k

f″(x) = k(k – 1)(1 + x)^{k – 2} f″(0) = k(k – 1)

f'''(x) = k(k – 1)(k – 2)(1 + x)k – 3 f'''(0) = k(k – 1)(k – 2)

. . . .

. .

f⁽ⁿ⁾ (x) = k(k – 1) ^{. . .} (k – n + 1) f⁽ⁿ⁾(0) = k(k – 1) ^{. . .} (1 + x)^{k – n} (k – n + 1)

Example 8 – Solution

Therefore the Maclaurin series of f(x) = (1 + x)^k is

This series is called the binomial series.

Notice that if k is a nonnegative integer, then the terms are eventually 0 and so the series is finite. For other values of k none of the terms is 0 and so we can try the Ratio Test.

cont’d

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Example 8 – Solution

If its nth term is a_n, then

Thus, by the Ratio Test, the binomial series converges if

| x | < 1 and diverges if | x| > 1.

cont’d

Taylor and Maclaurin Series

The traditional notation for the coefficients in the binomial series is

and these numbers are called the binomial coefficients.

The following theorem states that (1 + x)^k is equal to the sum of its Maclaurin series.

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Taylor and Maclaurin Series

It is possible to prove this by showing that the remainder term R_n(x) approaches 0, but that turns out to be quite difficult.

Taylor and Maclaurin Series

Although the binomial series always converges when

| x | < 1, the question of whether or not it converges at the endpoints, ±1, depends on the value of k.

It turns out that the series converges at 1 if –1 < k ≤ 0 and at both endpoints if k ≥ 0.

Notice that if k is a positive integer and n > k, then the

expression for contains a factor (k – k), so for n > k.

This means that the series terminates and reduces to the

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Taylor and Maclaurin Series

We collect in the following table, for future reference, some

important Maclaurin series that we have derived in this section and the preceding one.

Table 1

Important Maclaurin Series and their Radii of Convergence

Multiplication and Division of

Power Series

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Example 13

Find the first three nonzero terms in the Maclaurin series for (a) e^x sin x and (b) tan x.

Solution:

(a) Using the Maclaurin series for e^x and sin x in Table 1, we have

Example 13 – Solution

We multiply these expressions, collecting like terms just as for polynomials:

cont’d

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Example 13 – Solution

Thus

(b) Using the Maclaurin series in Table 1, we have

cont’d

Example 13 – Solution

We use a procedure like long division:

cont’d