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1.3: Taylor and Maclaurin Series - Mathematics


In the previous two sections we discussed how to find power series representations for certain types of functions––specifically, functions related to geometric series. In particular, we address the following questions: Which functions can be represented by power series and how do we find such representations? If we can find a power series representation for a particular function (f) and the series converges on some interval, how do we prove that the series actually converges to (f)?

Overview of Taylor/Maclaurin Series

Consider a function (f) that has a power series representation at (x=a). Then the series has the form

[sum_{n=0}^∞c_n(x−a)^n=c_0+c_1(x−a)+c_2(x−a)^2+ dots. label{eq1}]

What should the coefficients be? For now, we ignore issues of convergence, but instead focus on what the series should be, if one exists. We return to discuss convergence later in this section. If the series Equation ef{eq1} is a representation for (f) at (x=a), we certainly want the series to equal (f(a)) at (x=a). Evaluating the series at (x=a), we see that

[sum_{n=0}^∞c_n(x−a)^n=c_0+c_1(a−a)+c_2(a−a)^2+dots=c_0.label{eq2}]

Thus, the series equals (f(a)) if the coefficient (c_0=f(a)). In addition, we would like the first derivative of the power series to equal (f′(a)) at (x=a). Differentiating Equation ef{eq2} term-by-term, we see that

[dfrac{d}{dx}left( sum_{n=0}^∞c_n(x−a)^n ight)=c_1+2c_2(x−a)+3c_3(x−a)^2+dots.label{eq3}]

Therefore, at (x=a,) the derivative is

[dfrac{d}{dx}left( sum_{n=0}^∞c_n(x−a)^n ight)=c_1+2c_2(a−a)+3c_3(a−a)^2+dots=c_1.label{eq4}]

Therefore, the derivative of the series equals (f′(a)) if the coefficient (c_1=f′(a).) Continuing in this way, we look for coefficients (c_n) such that all the derivatives of the power series Equation ef{eq4} will agree with all the corresponding derivatives of (f) at (x=a). The second and third derivatives of Equation ef{eq3} are given by

[dfrac{d^2}{dx^2} left(sum_{n=0}^∞c_n(x−a)^n ight)=2c_2+3⋅2c_3(x−a)+4⋅3c_4(x−a)^2+dotslabel{eq5}]

and

[dfrac{d^3}{dx^3} left( sum_{n=0}^∞c_n(x−a)^n ight)=3⋅2c_3+4⋅3⋅2c_4(x−a)+5⋅4⋅3c_5(x−a)^2+⋯.label{eq6}]

Therefore, at (x=a), the second and third derivatives

[dfrac{d^2}{dx^2}(sum_{n=0}^∞c_n(x−a)^n)=2c_2+3⋅2c_3(a−a)+4⋅3c_4(a−a)^2+dots=2c_2label{eq7}]

and

[dfrac{d^3}{dx^3} left(sum_{n=0}^∞c_n(x−a)^n ight)=3⋅2c_3+4⋅3⋅2c_4(a−a)+5⋅4⋅3c_5(a−a)^2+dots =3⋅2c_3label{eq8}]

equal (f''(a)) and (f'''(a)), respectively, if (c_2=dfrac{f''(a)}{2}) and (c_3=dfrac{f'''(a)}{3}⋅2). More generally, we see that if (f) has a power series representation at (x=a), then the coefficients should be given by (c_n=dfrac{f^{(n)}(a)}{n!}). That is, the series should be

[sum_{n=0}^∞dfrac{f^{(n)}(a)}{n!}(x−a)^n=f(a)+f′(a)(x−a)+dfrac{f''(a)}{2!}(x−a)^2+dfrac{f'''(a)}{3!}(x−a)^3+⋯]

This power series for (f) is known as the Taylor series for (f) at (a.) If (x=0), then this series is known as the Maclaurin series for (f).

Definition (PageIndex{1}): Maclaurin and Taylor series

If (f) has derivatives of all orders at (x=a), then theTaylor series for the function (f) at (a) is

[sum_{n=0}^∞dfrac{f^{(n)}(a)}{n!}(x−a)^n=f(a)+f′(a)(x−a)+dfrac{f''(a)}{2!}(x−a)^2+⋯+dfrac{f^{(n)}(a)}{n!}(x−a)^n+⋯]

The Taylor series for (f) at 0 is known as the Maclaurin series for (f).

Later in this section, we will show examples of finding Taylor series and discuss conditions under which the Taylor series for a function will converge to that function. Here, we state an important result. Recall that power series representations are unique. Therefore, if a function (f) has a power series at (a), then it must be the Taylor series for (f) at (a).

Uniqueness of Taylor Series

If a function (f) has a power series at a that converges to (f) on some open interval containing (a), then that power series is the Taylor series for (f) at (a).

The proof follows directly from that discussed previously.

To determine if a Taylor series converges, we need to look at its sequence of partial sums. These partial sums are finite polynomials, known as Taylor polynomials.

Taylor Polynomials

The nth partial sum of the Taylor series for a function (f) at (a) is known as the nth Taylor polynomial. For example, the 0th, 1st, 2nd, and 3rd partial sums of the Taylor series are given by

[egin{align*} p_0(x) &=f(a) [5pt] p_1(x) &=f(a)+f′(a)(x−a) [5pt]p_2(x) &=f(a)+f′(a)(x−a)+dfrac{f''(a)}{2!}(x−a)^2 [5pt]p_3(x)&=f(a)+f′(a)(x−a)+dfrac{f''(a)}{2!}(x−a)^2+dfrac{f'''(a)}{3!}(x−a)^3 end{align*}]

respectively. These partial sums are known as the 0th, 1st, 2nd, and 3rd Taylor polynomials of (f) at (a), respectively. If (x=a), then these polynomials are known as Maclaurin polynomials for (f). We now provide a formal definition of Taylor and Maclaurin polynomials for a function (f).

Definition (PageIndex{2}): Maclaurin polynomial

If (f) has n derivatives at (x=a), then the nth Taylor polynomial for (f) at (a) is

[p_n(x)=f(a)+f′(a)(x−a)+dfrac{f''(a)}{2!}(x−a)^2+dfrac{f'''(a)}{3!}(x−a)^3+⋯+dfrac{f^{(n)}(a)}{n!}(x−a)^n.]

The nth Taylor polynomial for (f) at 0 is known as the nth Maclaurin polynomial for (f).

We now show how to use this definition to find several Taylor polynomials for (f(x)=ln x) at (x=1).

Example (PageIndex{1}): Finding Taylor Polynomials

Find the Taylor polynomials (p_0,p_1,p_2) and (p_3) for (f(x)=ln x) at (x=1). Use a graphing utility to compare the graph of (f) with the graphs of (p_0,p_1,p_2) and (p_3).

Solution

To find these Taylor polynomials, we need to evaluate (f) and its first three derivatives at (x=1).

(f(x)=ln x) (f(1)=0)

(f′(x)=dfrac{1}{x}) (f′(1)=1)

(f''(x)=−dfrac{1}{x^2}) (f''(1)=−1)

(f'''(x)=dfrac{2}{x^3}) (f'''(1)=2)

Therefore,

[egin{align*} p_0(x) &=f(1)=0,[5pt]p_1(x)&=f(1)+f′(1)(x−1) =x−1,[5pt]p_2(x) &=f(1)+f′(1)(x−1)+dfrac{f''(1)}{2}(x−1)^2 = (x−1)−dfrac{1}{2}(x−1)^2 [5pt]p_3(x)& =f(1)+f′(1)(x−1)+dfrac{f''(1)}{2}(x−1)^2+dfrac{f'''(1)}{3!}(x−1)^3=(x−1)−dfrac{1}{2}(x−1)^2+dfrac{1}{3}(x−1)^3 end{align*}]

The graphs of (y=f(x)) and the first three Taylor polynomials are shown in Figure (PageIndex{1}).

Exercise (PageIndex{1})

Find the Taylor polynomials (p_0,p_1,p_2) and (p_3) for (f(x)=dfrac{1}{x^2}) at (x=1).

Hint

Find the first three derivatives of (f) and evaluate them at (x=1.)

Answer

[ p_0(x)=1]

[p_1(x)=1−2(x−1)]

[p_2(x)=1−2(x−1)+3(x−1)^2]

[p_3(x)=1−2(x−1)+3(x−1)^2−4(x−1)^3]

We now show how to find Maclaurin polynomials for (e^x, sin x,) and (cos x). As stated above, Maclaurin polynomials are Taylor polynomials centered at zero.

Example (PageIndex{2}): Finding Maclaurin Polynomials

For each of the following functions, find formulas for the Maclaurin polynomials (p_0,p_1,p_2) and (p_3). Find a formula for the nth Maclaurin polynomial and write it using sigma notation. Use a graphing utility to compare the graphs of (p_0,p_1,p_2) and (p_3) with (f).

  1. (f(x)=e^x)
  2. (f(x)=sin x)
  3. (f(x)=cos x)

Solution

Since (f(x)=e^x),we know that (f(x)=f′(x)=f''(x)=⋯=f^{(n)}(x)=e^x) for all positive integers n. Therefore,

[f(0)=f′(0)=f''(0)=⋯=f^{(n)}(0)=1 onumber]

for all positive integers n. Therefore, we have

(p_0(x)=f(0)=1,)

(p_1(x)=f(0)+f′(0)x=1+x,)

(p_2(x)=f(0)+f′(0)x+dfrac{f''(0)}{2!}x^2=1+x+dfrac{1}{2}x^2),

(p_3(x)=f(0)+f′(0)x+dfrac{f''(0)}{2}x^2+dfrac{f'''(0)}{3!}x^3=1+x+dfrac{1}{2}x^2+dfrac{1}{3!}x^3),

(p_n(x)=f(0)+f′(0)x+dfrac{f''(0)}{2}x^2+dfrac{f'''(0)}{3!}x^3+⋯+dfrac{f^{(n)}(0)}{n!}x^n=1+x+dfrac{x^2}{2!}+dfrac{x^3}{3!}+⋯+dfrac{x^n}{n!}=sum_{k=0}^ndfrac{x^k}{k!}).

The function and the first three Maclaurin polynomials are shown in Figure 2.

b. For (f(x)=sin x), the values of the function and its first four derivatives at (x=0) are given as follows:

(f(x)=sin x) (f(0)=0)

(f′(x)=cos x) (f′(0)=1)

(f''(x)=−sin x) (f''(0)=0)

(f'''(x)=−cos x) (f'''(0)=−1)

(f^{(4)}(x)=sin x) (f^{(4)}(0)=0).

Since the fourth derivative is (sin x,) the pattern repeats. That is, (f^{(2m)}(0)=0) and (f^{(2m+1)}(0)=(−1)^m) for (m≥0.) Thus, we have

(p_0(x)=0,)

(p_1(x)=0+x=x,)

(p_2(x)=0+x+0=x,)

(p_3(x)=0+x+0−dfrac{1}{3!}x^3=x−dfrac{x^3}{3!},)

(p_4(x)=0+x+0−dfrac{1}{3!}x^3+0=x−dfrac{x^3}{3!}),

(p_5(x)=0+x+0−dfrac{1}{3!}x^3+0+dfrac{1}{5!}x^5=x−dfrac{x^3}{3!}+dfrac{x^5}{5!}),

and for (m≥0),

(p_{2m+1}(x)=p_{2m+2}(x)=x−dfrac{x^3}{3!}+dfrac{x^5}{5!}−⋯+(−1)^mdfrac{x^{2m+1}}{(2m+1)!}=sum_{k=0}^m(−1)^kdfrac{x^{2k+1}}{(2k+1)!}).

Graphs of the function and its Maclaurin polynomials are shown in Figure 3.

c. For (f(x)=cos x), the values of the function and its first four derivatives at (x=0) are given as follows:

(f(x)=cos x) (f(0)=1)

(f′(x)=−sin x) (f′(0)=0)

(f''(x)=−cos x) (f''(0)=−1)

(f'''(x)=sin x) (f'''(0)=0)

(f^{(4)}(x)=cos x) (f^{(4)}(0)=1.)

Since the fourth derivative is (sin x), the pattern repeats. In other words, (f^{(2m)}(0)=(−1)^m) and (f^{(2m+1)}=0) for (m≥0). Therefore,

(p_0(x)=1,)

(p_1(x)=1+0=1,)

(p_2(x)=1+0−dfrac{1}{2!}x^2=1−dfrac{x^2}{2!}),

(p_3(x)=1+0−dfrac{1}{2!}x^2+0=1−dfrac{x^2}{2!}),

(p_4(x)=1+0−dfrac{1}{2!}x^2+0+dfrac{1}{4!}x^4=1−dfrac{x^2}{2!}+dfrac{x^4}{4!}),

(p_5(x)=1+0−dfrac{1}{2!}x^2+0+dfrac{1}{4!}x^4+0=1−dfrac{x^2}{2!}+dfrac{x^4}{4!}),

and for (n≥0),

(p_{2m}(x)=p_{2m+1}(x)=1−dfrac{x^2}{2!}+dfrac{x^4}{4!}−⋯+(−1)^mdfrac{x^{2m}}{(2m)!}=sum_{k=0}^m(−1)^kdfrac{x^{2k}}{(2k)!}).

Graphs of the function and the Maclaurin polynomials appear in Figure 4.

Exercise (PageIndex{2})

Find formulas for the Maclaurin polynomials (p_0,p_1,p_2) and (p_3) for (f(x)=dfrac{1}{1+x}).

Find a formula for the nth Maclaurin polynomial. Write your answer using sigma notation.

Hint

Evaluate the first four derivatives of (f) and look for a pattern.

Answer

(p_0(x)=1;p_1(x)=1−x;p_2(x)=1−x+x^2;p_3(x)=1−x+x^2−x^3;p_n(x)=1−x+x^2−x^3+⋯+(−1)^nx^n=_{k=0}^n(−1)^kx^k)

Taylor’s Theorem with Remainder

Recall that the nth Taylor polynomial for a function (f) at a is the nth partial sum of the Taylor series for (f) at a. Therefore, to determine if the Taylor series converges, we need to determine whether the sequence of Taylor polynomials ({p_n}) converges. However, not only do we want to know if the sequence of Taylor polynomials converges, we want to know if it converges to (f). To answer this question, we define the remainder (R_n(x)) as

[R_n(x)=f(x)−p_n(x).]

For the sequence of Taylor polynomials to converge to (f), we need the remainder (R_n) to converge to zero. To determine if (R_n) converges to zero, we introduce Taylor’s theorem with remainder. Not only is this theorem useful in proving that a Taylor series converges to its related function, but it will also allow us to quantify how well the nth Taylor polynomial approximates the function.

Here we look for a bound on (|R_n|.) Consider the simplest case: (n=0). Let (p_0) be the 0th Taylor polynomial at a for a function (f). The remainder (R_0) satisfies

(R_0(x)=f(x)−p_0(x)=f(x)−f(a).)

If (f) is differentiable on an interval I containing (a) and (x), then by the Mean Value Theorem there exists a real number c between a and x such that (f(x)−f(a)=f′(c)(x−a)). Therefore,

[R_0(x)=f′(c)(x−a).]

Using the Mean Value Theorem in a similar argument, we can show that if (f) is n times differentiable on an interval I containing a and x, then the nth remainder (R_n) satisfies

[R_n(x)=dfrac{f^{(n+1)}(c)}{(n+1)!}(x−a)^{n+1}]

for some real number c between a and x. It is important to note that the value c in the numerator above is not the center a, but rather an unknown value c between a and x. This formula allows us to get a bound on the remainder (R_n). If we happen to know that (∣f^{(n+1)}(x)∣) is bounded by some real number M on this interval I, then

[|R_n(x)|≤dfrac{M}{(n+1)!}|x−a|^{n+1}]

for all x in the interval I.

We now state Taylor’s theorem, which provides the formal relationship between a function (f) and its nth degree Taylor polynomial (p_n(x)). This theorem allows us to bound the error when using a Taylor polynomial to approximate a function value, and will be important in proving that a Taylor series for (f) converges to (f).

Taylor’s Theorem with Remainder

Let (f) be a function that can be differentiated (n+1) times on an interval I containing the real number a. Let (p_n) be the nth Taylor polynomial of (f) at a and let

[R_n(x)=f(x)−p_n(x)]

be the nth remainder. Then for each x in the interval I, there exists a real number c between a and x such that

[R_n(x)=dfrac{f^{(n+1)}(c)}{(n+1)!}(x−a)^{n+1}].

If there exists a real number (M) such that (∣f^{(n+1)}(x)∣≤M) for all (x∈I), then

[|R_n(x)|≤dfrac{M}{(n+1)!}|x−a|^{n+1}]

for all (x) in (I).

Proof

Fix a point (x∈I) and introduce the function g such that

[g(t)=f(x)−f(t)−f′(t)(x−t)−dfrac{f''(t)}{2!}(x−t)^2−⋯−dfrac{f^{(n)}(t)}{n!}(x−t)^n−R_n(x)dfrac{(x−t)^{n+1}}{(x−a)^{n+1}}.]

We claim that (g) satisfies the criteria of Rolle’s theorem. Since (g) is a polynomial function (in t), it is a differentiable function. Also, g is zero at (t=a) and (t=x) because

[ egin{align} g(a)&=f(x)−f(a)−f′(a)(x−a)−dfrac{f''(a)}{2!}(x−a)^2+⋯+dfrac{f^{(n)}(a)}{n!}(x−a)^n−R_n(x) [5pt] &=f(x)−p_n(x)−R_n(x) [5pt] &=0, [5pt] g(x) &=f(x)−f(x)−0−⋯−0 [5pt] &=0. end{align}]

Therefore, g satisfies Rolle’s theorem, and consequently, there exists c between a and x such that (g′(c)=0.) We now calculate (g′). Using the product rule, we note that

[dfrac{d}{dt}[dfrac{f^{(n)}(t)}{n!}(x−t)^n]=−dfrac{f^{(n)}(t)}{(n−1)!}(x−t)^{n−1}+dfrac{f^{(n+1)}(t)}{n!}(x−t)^n.]

Consequently,

[g′(t)=−f′(t)+[f′(t)−f''(t)(x−t)]+[f''(t)(x−t)−dfrac{f'''(t)}{2!}(x−t)^2]+⋯+[dfrac{f^{(n)}(t)}{(n−1)!}(x−t)^{n−1}−dfrac{f^{(n+1)}(t)}{n!}(x−t)^n]+(n+1)R_n(x)dfrac{(x−t)^n}{(x−a)^{n+1}}].

Notice that there is a telescoping effect. Therefore,

[g′(t)=−dfrac{f^{(n+1)}(t)}{n!}(x−t)^n+(n+1)R_n(x)dfrac{(x−t)^n}{(x−a)^{n+1}}].

By Rolle’s theorem, we conclude that there exists a number c between a and x such that (g′(c)=0.) Since

[g′(c)=−dfrac{f^{(n+1})(c)}{n!}(x−c)^n+(n+1)R_n(x)dfrac{(x−c)^n}{(x−a)^{n+1}}]

we conclude that

[−dfrac{f^{(n+1)}(c)}{n!}(x−c)^n+(n+1)R_n(x)dfrac{(x−c)^n}{(x−a)^{n+1}}=0.]

Adding the first term on the left-hand side to both sides of the equation and dividing both sides of the equation by (n+1,) we conclude that

[R_n(x)=dfrac{f^{(n+1)}(c)}{(n+1)!}(x−a)^{n+1}]

as desired. From this fact, it follows that if there exists M such that (∣f^{(n+1)}(x)∣≤M) for all x in I, then

[|R_n(x)|≤dfrac{M}{(n+1)!}|x−a|^{n+1}].

Not only does Taylor’s theorem allow us to prove that a Taylor series converges to a function, but it also allows us to estimate the accuracy of Taylor polynomials in approximating function values. We begin by looking at linear and quadratic approximations of (f(x)=dfrac[3]{x}) at (x=8) and determine how accurate these approximations are at estimating (dfrac[3]{11}).

Example (PageIndex{3}): Using Linear and Quadratic Approximations to Estimate Function Values

Consider the function (f(x)=sqrt[3]{x}).

  1. Find the first and second Taylor polynomials for (f) at (x=8). Use a graphing utility to compare these polynomials with (f) near (x=8.)
  2. Use these two polynomials to estimate (sqrt[3]{11}).
  3. Use Taylor’s theorem to bound the error.

Solution:

a. For (f(x)=sqrt[3]{x}), the values of the function and its first two derivatives at (x=8) are as follows:

(f(x)=sqrt[3]{x}) (f(8)=2)

(f′(x)=dfrac{1}{3x^{2/3}}) (f′(8)=dfrac{1}{12})

(f''(x)=dfrac{−2}{9x^{5/3}}) (f''(8)=−dfrac{1}{144.})

Thus, the first and second Taylor polynomials at (x=8) are given by

(p_1(x)=f(8)+f′(8)(x−8))

(=2+dfrac{1}{12}(x−8))

(p_2(x)=f(8)+f′(8)(x−8)+dfrac{f''(8)}{2!}(x−8)^2)

(=2+dfrac{1}{12}(x−8)−dfrac{1}{288}(x−8)^2).

The function and the Taylor polynomials are shown in Figure 4.

b. Using the first Taylor polynomial at (x=8), we can estimate

[dfrac[3]{11}≈p_1(11)=2+dfrac{1}{12}(11−8)=2.25.]

Using the second Taylor polynomial at (x=8), we obtain

[sqrt[3]{11}≈p_2(11)=2+dfrac{1}{12}(11−8)−dfrac{1}{288}(11−8)^2=2.21875.]

c. By Note, there exists a c in the interval ((8,11)) such that the remainder when approximating (sqrt[3]{11}) by the first Taylor polynomial satisfies

[R_1(11)=dfrac{f''(c)}{2!}(11−8)^2.]

We do not know the exact value of c, so we find an upper bound on (R_1(11)) by determining the maximum value of (f'') on the interval ((8,11)). Since (f''(x)=−dfrac{2}{9x^{5/3}}), the largest value for (|f''(x)|) on that interval occurs at (x=8). Using the fact that (f''(8)=−dfrac{1}{144}), we obtain

(|R_1(11)|≤dfrac{1}{144⋅2!}(11−8)^2=0.03125.)

Similarly, to estimate (R_2(11)), we use the fact that

(R_2(11)=dfrac{f'''(c)}{3!}(11−8)^3).

Since (f'''(x)=dfrac{10}{27x^{8/3}}), the maximum value of (f''') on the interval ((8,11)) is (f'''(8)≈0.0014468). Therefore, we have

(|R_2(11)|≤dfrac{0.0011468}{3!}(11−8)^3≈0.0065104.)

Exercise (PageIndex{3}):

Find the first and second Taylor polynomials for (f(x)=sqrt{x}) at (x=4). Use these polynomials to estimate (sqrt{6}). Use Taylor’s theorem to bound the error.

Hint

Evaluate (f(4),f′(4),) and (f''(4).)

Answer

(p_1(x)=2+dfrac{1}{4}(x−4);p_2(x)=2+dfrac{1}{4}(x−4)−dfrac{1}{64}(x−4)^2;p_1(6)=2.5;p_2(6)=2.4375;)

(|R_1(6)|≤0.0625;|R_2(6)|≤0.015625)

Example (PageIndex{4}): Approximating (sin x) Using Maclaurin Polynomials

From Example b., the Maclaurin polynomials for (sin x) are given by

[p_{2m+1}(x)=p_{2m+2}(x)=x−dfrac{x^3}{3!}+dfrac{x^5}{5!}−dfrac{x^7}{7!}+⋯+(−1)^mdfrac{x^{2m+1}}{(2m+1)!} onumber]

for (m=0,1,2,….)

  1. Use the fifth Maclaurin polynomial for (sin x) to approximate (sin(dfrac{π}{18})) and bound the error.
  2. For what values of (x) does the fifth Maclaurin polynomial approximate (sin x) to within (0.0001)?

Solution

a.

The fifth Maclaurin polynomial is

[p_5(x)=x−dfrac{x^3}{3!}+dfrac{x^5}{5!}].

Using this polynomial, we can estimate as follows:

[sin(dfrac{π}{18})≈p_5(dfrac{π}{18})=dfrac{π}{18}−dfrac{1}{3!}(dfrac{π}{18})^3+dfrac{1}{5!}(dfrac{π}{18})^5≈0.173648.]

To estimate the error, use the fact that the sixth Maclaurin polynomial is (p_6(x)=p_5(x)) and calculate a bound on (R_6(dfrac{π}{18})). By Note, the remainder is

[R_6(dfrac{π}{18})=dfrac{f^{(7)}(c)}{7!}(dfrac{π}{18})^7]

for some c between 0 and (dfrac{π}{18}). Using the fact that (∣f^{(7)}(x)∣≤1) for all (x), we find that the magnitude of the error is at most

[dfrac{1}{7!}⋅(dfrac{π}{18})^7≤9.8×10^{−10}.]

b.

We need to find the values of (x) such that

[dfrac{1}{7}!|x|^7≤0.0001.]

Solving this inequality for (x), we have that the fifth Maclaurin polynomial gives an estimate to within (0.0001) as long as (|x|<0.907.)

Exercise (PageIndex{4})

Use the fourth Maclaurin polynomial for (cos x) to approximate (cos(dfrac{π}{12}).)

Hint

The fourth Maclaurin polynomial is (p_4(x)=1−dfrac{x^2}{2!}+dfrac{x^4}{4!}).

Answer

0.96593

Now that we are able to bound the remainder (R_n(x)), we can use this bound to prove that a Taylor series for (f) at a converges to (f).

Representing Functions with Taylor and Maclaurin Series

We now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.

Example (PageIndex{5}): Finding a Taylor Series

Find the Taylor series for (f(x)=dfrac{1}{x}) at (x=1). Determine the interval of convergence.

Solution

For (f(x)=dfrac{1}{x},) the values of the function and its first four derivatives at (x=1) are

(f(x)=dfrac{1}{x}) (f(1)=1)

(f′(x)=−dfrac{1}{x^2}) (f′(1)=−1)

(f''(x)=dfrac{2}{x^3}) (f''(1)=2!)

(f'''(x)=−dfrac{3⋅2}{x^4}) (f'''(1)=−3!)

(f^{(4)}(x)=dfrac{4⋅3⋅2}{x^5}) (f^{(4)}(1)=4!).

That is, we have (f^{(n)}(1)=(−1)^nn!) for all (n≥0). Therefore, the Taylor series for (f) at (x=1) is given by

(displaystyle sum_{n=0}^∞dfrac{f^{(n)}(1)}{n!}(x−1)^n=sum_{n=0}^∞(−1)^n(x−1)^n).

To find the interval of convergence, we use the ratio test. We find that

(dfrac{|a_{n+1}|}{|a_n|}=dfrac{∣(−1)^{n+1}(x−1)n^{+1}∣}{|(−1)^n(x−1)^n|}=|x−1|).

Thus, the series converges if (|x−1|<1.) That is, the series converges for (0

(displaystyle sum_{n=0}^∞(−1)^n(2−1)^n=sum_{n=0}^∞(−1)^n)

diverges by the divergence test. Similarly, at (x=0,)

(displaystyle sum_{n=0}^∞(−1)^n(0−1)^n=sum_{n=0}^∞(−1)^{2n}=sum_{n=0}^∞1)

diverges. Therefore, the interval of convergence is ((0,2)).

Exercise (PageIndex{5})

Find the Taylor series for (f(x)=dfrac{1}{2}) at (x=2) and determine its interval of convergence.

Hint

(f^{(n)}(2)=dfrac{(−1)^nn!}{2^{n+1}})

Answer

(dfrac{1}{2}displaystyle sum_{n=0}^∞(dfrac{2−x}{2})^n). The interval of convergence is ((0,4)).

We know that the Taylor series found in this example converges on the interval ((0,2)), but how do we know it actually converges to (f)? We consider this question in more generality in a moment, but for this example, we can answer this question by writing

[ f(x)=dfrac{1}{x}=dfrac{1}{1−(1−x)}.]

That is, (f) can be represented by the geometric series (displaystyle sum_{n=0}^∞(1−x)^n). Since this is a geometric series, it converges to (dfrac{1}{x}) as long as (|1−x|<1.) Therefore, the Taylor series found in Example does converge to (f(x)=dfrac{1}{x}) on ((0,2).)

We now consider the more general question: if a Taylor series for a function (f) converges on some interval, how can we determine if it actually converges to (f)? To answer this question, recall that a series converges to a particular value if and only if its sequence of partial sums converges to that value. Given a Taylor series for (f) at a, the nth partial sum is given by the nth Taylor polynomial pn. Therefore, to determine if the Taylor series converges to (f), we need to determine whether

(displaystyle lim_{n→∞}p_n(x)=f(x)).

Since the remainder (R_n(x)=f(x)−p_n(x)), the Taylor series converges to (f) if and only if

(displaystyle lim_{n→∞}R_n(x)=0.)

We now state this theorem formally.

Convergence of Taylor Series

Suppose that (f) has derivatives of all orders on an interval (I) containing (a). Then the Taylor series

[sum_{n=0}^∞dfrac{f^{(n)}(a)}{n!}(x−a)^n]

converges to (f(x)) for all (x) in (I) if and only if

[lim_{n→∞}R_n(x)=0]

for all (x) in (I).

With this theorem, we can prove that a Taylor series for (f) at a converges to (f) if we can prove that the remainder (R_n(x)→0). To prove that (R_n(x)→0), we typically use the bound

[|R_n(x)|≤dfrac{M}{(n+1)!}|x−a|^{n+1}]

from Taylor’s theorem with remainder.

In the next example, we find the Maclaurin series for (e^x) and (sin x) and show that these series converge to the corresponding functions for all real numbers by proving that the remainders (R_n(x)→0) for all real numbers (x).

Example (PageIndex{6}): Finding Maclaurin Series

For each of the following functions, find the Maclaurin series and its interval of convergence. Use Note to prove that the Maclaurin series for (f) converges to (f) on that interval.

  1. (e^x)
  2. (sin x)

Solution

a. Using the nth Maclaurin polynomial for (e^x) found in Example a., we find that the Maclaurin series for (e^x) is given by

(displaystyle sum_{n=0}^∞dfrac{x^n}{n!}).

To determine the interval of convergence, we use the ratio test. Since

(dfrac{|a_{n+1}|}{|a_n|}=dfrac{|x|^{n+1}}{(n+1)!}⋅dfrac{n!}{|x|^n}=dfrac{|x|}{n+1}),

we have

(displaystyle lim_{n→∞}dfrac{|a_{n+1}|}{|a_n|}=lim_{n→∞}dfrac{|x|}{n+1}=0)

for all (x). Therefore, the series converges absolutely for all (x), and thus, the interval of convergence is ((−∞,∞)). To show that the series converges to (e^x) for all (x), we use the fact that (f^{(n)}(x)=e^x) for all (n≥0) and (e^x) is an increasing function on ((−∞,∞)). Therefore, for any real number (b), the maximum value of (e^x) for all (|x|≤b) is (e^b). Thus,

(|R_n(x)|≤dfrac{e^b}{(n+1)!}|x|^{n+1}).

Since we just showed that

(displaystyle sum_{n=0}^∞dfrac{|x|^n}{n!})

converges for all x, by the divergence test, we know that

(displaystyle lim_{n→∞}dfrac{|x|^{n+1}}{(n+1)!}=0)

for any real number x. By combining this fact with the squeeze theorem, the result is (lim_{n→∞}R_n(x)=0.)

b. Using the nth Maclaurin polynomial for (sin x) found in Example b., we find that the Maclaurin series for (sin x) is given by

(displaystyle sum_{n=0}^∞(−1)^ndfrac{x^{2n+1}}{(2n+1)!}).

In order to apply the ratio test, consider

(dfrac{|a_{n+1}|}{|a_n|}=dfrac{|x|^{2n+3}}{(2n+3)!}⋅dfrac{(2n+1)!}{|x|^{2n+1}}=dfrac{|x|^2}{(2n+3)(2n+2)}).

Since

(displaystyle lim_{n→∞}dfrac{|x|^2}{(2n+3)(2n+2)}=0)

for all (x), we obtain the interval of convergence as ((−∞,∞).) To show that the Maclaurin series converges to (sin x), look at (R_n(x)). For each (x) there exists a real number (c) between (0) and (x) such that

(R_n(x)=dfrac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}).

Since (∣f^{(n+1)}(c)∣≤1) for all integers (n) and all real numbers (c), we have

(|R_n(x)|≤dfrac{|x|^{n+1}}{(n+1)!})

for all real numbers (x). Using the same idea as in part a., the result is (displaystyle lim_{n→∞}R_n(x)=0) for all (x), and therefore, the Maclaurin series for (sin x) converges to (sin x) for all real (x).

Exercise (PageIndex{6})

Find the Maclaurin series for (f(x)=cos x). Use the ratio test to show that the interval of convergence is ((−∞,∞)). Show that the Maclaurin series converges to (cos x) for all real numbers (x).

Hint

Use the Maclaurin polynomials for (cos x.)

Answer

(sum_{n=0}^∞dfrac{(−1)^nx^{2n}}{(2n)!})

By the ratio test, the interval of convergence is ((−∞,∞).) Since (|R_n(x)|≤dfrac{|x|^{n+1}}{(n+1)!}), the series converges to (cos x) for all real (x).

Proving that E is Irrational

In this project, we use the Maclaurin polynomials for (e^x) to prove that e is irrational. The proof relies on supposing that e is rational and arriving at a contradiction. Therefore, in the following steps, we suppose (e=r/s) for some integers r and s where (s≠0.)

  1. Write the Maclaurin polynomials (p_0(x),p_1(x),p_2(x),p_3(x),p_4(x)) for (e^x). Evaluate (p_0(1),p_1(1),p_2(1),p_3(1),p_4(1)) to estimate e.
  2. Let (R_n(x)) denote the remainder when using (p_n(x)) to estimate (e^x). Therefore, (R_n(x)=e^x−p_n(x)), and (R_n(1)=e−p_n(1)). Assuming that (e=dfrac{r}{s}) for integers r and s, evaluate (R_0(1),R_1(1),R_2(1),R_3(1),R_4(1).)
  3. Using the results from part 2, show that for each remainder (R_0(1),R_1(1),R_2(1),R_3(1),R_4(1),) we can find an integer k such that (kR_n(1)) is an integer for (n=0,1,2,3,4.)
  4. Write down the formula for the nth Maclaurin polynomial (p_n(x)) for (e^x) and the corresponding remainder (R_n(x).) Show that (sn!R_n(1)) is an integer.
  5. Use Taylor’s theorem to write down an explicit formula for (R_n(1)). Conclude that (R_n(1)≠0), and therefore, (sn!R_n(1)≠0).
  6. Use Taylor’s theorem to find an estimate on (R_n(1)). Use this estimate combined with the result from part 5 to show that (|sn!R_n(1)|n is large enough, then (|sn!R_n(1)|<1). Therefore, (sn!R_n(1)) is an integer with magnitude less than 1. Thus, (sn!R_n(1)=0). But from part 5, we know that (sn!R_n(1)≠0). We have arrived at a contradiction, and consequently, the original supposition that e is rational must be false.

Key Concepts

  • Taylor polynomials are used to approximate functions near a value (x=a). Maclaurin polynomials are Taylor polynomials at (x=0).
  • The nth degree Taylor polynomials for a function (f) are the partial sums of the Taylor series for (f).
  • If a function (f) has a power series representation at (x=a), then it is given by its Taylor series at (x=a).
  • A Taylor series for (f) converges to (f) if and only if (displaystyle lim_{n→∞}R_n(x)=0) where (R_n(x)=f(x)−p_n(x)).
  • The Taylor series for (e^x, sin x), and (cos x) converge to the respective functions for all real x.

Key Equations

  • Taylor series for the function (f) at the point (x=a)

(sum_{n=0}^∞dfrac{f^{(n)}(a)}{n!}(x−a)^n=f(a)+f′(a)(x−a)+dfrac{f''(a)}{2!}(x−a)^2+⋯+dfrac{f^{(n)}(a)}{n!}(x−a)^n+⋯)

Glossary

Maclaurin polynomial
a Taylor polynomial centered at 0; the nth Taylor polynomial for (f) at 0 is the nth Maclaurin polynomial for (f)
Maclaurin series
a Taylor series for a function (f) at (x=0) is known as a Maclaurin series for (f)
Taylor polynomials
the nth Taylor polynomial for (f) at (x=a) is (p_n(x)=f(a)+f′(a)(x−a)+dfrac{f''(a)}{2!}(x−a)^2+⋯+dfrac{f^{(n)}(a)}{n!}(x−a)^n)
Taylor series
a power series at a that converges to a function (f) on some open interval containing a
Taylor’s theorem with remainder

for a function (f) and the nth Taylor polynomial for (f) at (x=a), the remainder (R_n(x)=f(x)−p_n(x)) satisfies (R_n(x)=dfrac{f^{(n+1)}(c)}{(n+1)!}(x−a)^{n+1})

for some c between x and a; if there exists an interval I containing a and a real number M such that (∣f^{(n+1)}(x)∣≤M) for all x in I, then (|R_n(x)|≤dfrac{M}{(n+1)!}|x−a|^{n+1})

Contributors

  • Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.


1.3: Taylor and Maclaurin Series - Mathematics

Lecture Description

This video lecture, part of the series Calculus Videos: Series and Sequences by Prof. , does not currently have a detailed description and video lecture title. If you have watched this lecture and know what it is about, particularly what Mathematics topics are discussed, please help us by commenting on this video with your suggested description and title. Many thanks from,

- The CosmoLearning Team

Course Index

  1. Sequence: Convergence and Divergence (Part 1)
  2. Sequences: Converging or Diverging (Part 2)
  3. Series: Geometric Series and the Test for Divergence
  4. Geometric Series: Fraction Representation
  5. Examples of Geometric Series and the Test for Divergence (Part 1)
  6. Examples of Geometric Series and the Test for Divergence (Part 2)
  7. Telescoping Series
  8. Series Diverges: Using Partial Sums
  9. Integral Test for Series
  10. Using the Integral Test for Series: Example 1
  11. Using the Integral Test for Series: Example 2
  12. Using the Integral Test for Series: Example 3
  13. Series: Limit and Direct Comparison Test
  14. Series: Limit and Direct Comparison Test Examples
  15. Series: Limit and Direct Comparison Test Examples (cont.)
  16. Alternating Series (Part 1)
  17. Alternating Series (Part 2)
  18. Alternating Series Estimation Theorem
  19. Ratio Test to Determine if a Series Converges (Ex. 1)
  20. Ratio Test to Determine if a Series Converges (Ex. 2)
  21. Root Test for Series
  22. Power Series Representation of Functions
  23. Power Series: Interval of Convergence
  24. Differentiating and Integrating Power Series
  25. Power Series: Multiplying and Dividing
  26. Taylor and MacLaurin Series (Ex. 1)
  27. Taylor and MacLaurin Series (Ex. 2)
  28. Taylor's Remainder Theorem or Taylor's Inequality
  29. The Binomial Series (Ex. 1)
  30. The Binomial Series (Ex. 2)

Course Description


In this course, Calculus Instructor Patrick gives 30 video lessons on Series and Sequences. Some of the Topics covered are: Convergence and Divergence, Geometric Series, Test for Divergence, Telescoping Series, Integral Test, Limit and Direct Comparison Test, Alternating Series, Alternating Series Estimation Theorem, Ratio Test, Power Series, Taylor and MacLaurin Series, Taylor's Remainder Theorem (Taylor's Inequality), Binomial Series, and many more.


Differences Between the Taylor and Maclaurin Series

Aside from flying cockroaches, here is another thing that most people detest – math. We are often stricken with fear when we are facing math. The numbers seem like they are rattling our head, and it seems that math is eating up all of our life force. No matter what we do, we can’t escape the clutches of math. From counting to complex equations, we are always dealing with math. Nevertheless, we have to deal with it. Face your fear and learn to handle it. We have to meet Taylor and Maclaurin. Who are these people? These are not people. These are mathematical series.

In the field of mathematics, a Taylor series is defined as the representation of a function as an infinite sum of terms that are calculated from the values of the function’s derivatives at a single point. The Taylor series got its name from Brook Taylor. Brook Taylor was an English mathematician in 1715. It is all right to approximate the value of a function through utilizing the finite number of terms in the Taylor series. Approximating the value is already a common practice. In this approximation process, the Taylor series can yield quantitative estimates on the error. A Taylor polynomial is the term used to represent the finite number of the Taylor series’ initial function terms.

According to wikipedia.org, there are other uses of the Taylor series for determining analytic functions. The Taylor series can be used in obtaining the partial sums or the Taylor polynomials through using approximation techniques in the entire function. Another usage of the Taylor series is the differentiation and integration of the power series which can be done with each term. The Taylor series can also provide a complex analysis through integrating the analytic function with a holomorphic function in a complex plane. It can also be used to obtain and compute values numerically in a truncated series. This is done by applying the Chebyshev formula and Clenshaw algorithm. Another thing is that you can use the Taylor series in algebraic operations. An example of this is applying the Euler’s formula connecting with the Taylor series for the expansion of trigonometric and exponential functions. This can be used in the field of harmonic analysis. You can also use the Taylor series in the field of physics.

A Taylor series becomes a Maclaurin series if the Taylor series is centered at the point of zero. The Maclaurin series is named after Colin Maclaurin. Colin Maclaurin was a Scottish mathematician who had greatly used the Taylor series during the 18th century. A Maclaurin series is the expansion of the Taylor series of a function about zero. According to mathworld.wolfram.com, the Maclaurin series is a type of series expansion in which all terms are non-negative integer powers of the variable. Other more general types of series include the Laurent series and the Puiseux series. The Taylor and Maclaurin series have many uses in the mathematical field including the sciences.

In the field of mathematics, a Taylor series is defined as the representation of a function as an infinite sum of terms that are calculated from the values of the function’s derivatives at a single point.

A Taylor series becomes a Maclaurin series if the Taylor series is centered at the point of zero. A Maclaurin series is the expansion of the Taylor series of a function about zero.

The Taylor series got its name from Brook Taylor. Brook Taylor was an English mathematician in 1715. The Maclaurin series is named after Colin Maclaurin. Colin Maclaurin was a Scottish mathematician who had greatly used the Taylor series during the 18th century.


Maclaurin series of $sin x $ and $cos x $

Let $f(x) = sin x $. Then $f(0) = 0$. Going through the derivatives

$ egin f'(x) &= cos x Rightarrow f'(0) = 1 f''(x) &= -sin x Rightarrow f''(0) = 0 f'''(x) &= -cos x Rightarrow f'''(0) = -1 end $

And so on. The formula for a Maclaurin series is

Therefore the Maclaurin series of $sin x $ is

By similar reasoning the Maclaurin series of $cos x $ is

Here's a graph showing truncated version of the Maclaurin series compared to the actual $sin x $ graph.

  • $y=sin x $ is in blue
  • $y=x$ is in red
  • $y=x - frac<3!>$ is in orange
  • $y=x - frac<3!>+ frac<5!>$ is in purple
  • $y=x - frac<3!>+ frac<5!>- frac<7!>$ is in green

See how each progressively longer truncation gets closer to the $sin x $ graph.

Maclaurin series for trigonometric functions are particularly useful because many of them are periodic over $x$, and longer truncations provide extremely close approximations for relatively small $x$.


How do you find the maclaurin series expansion of #1/(x+3)#?

Another way of finding the Maclaurin series is basically to write it out term by term as follows.

If the Maclaurin series for #1/(x+3)# is #sum_(n=0)^oo a_n x^n# , then #1 = (3+x)sum_(n=0)^oo a_n x^n#

So examining each power of #x# in turn, we can start writing:

since we require #a_0 = 1/3# in order that the constant term (i.e. the term in #x^0# ) is #1# when multiplied by #(3+x)# .

Note that this will result in an unwanted term #x * 1/3# . So to cancel that out, the next term must be #-x/9# to give #-x/3# when multiplied by #3# . So our product looks like this:

This will result in an unwanted term #-x^2/9# . To cancel that out, the next term must be #x^2/27# .

Repeating, it soon becomes clear that the series we want is a geometric series with common ratio #-x/3# , looking like this:


1.3: Taylor and Maclaurin Series - Mathematics

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Mathematical Expression Editor

We find the Maclaurin series representation of a function.

Maclaurin Series

Maclaurin series are a special case of Taylor series with center . In this section we will develop the Maclaurin series for and and use these to create Maclaurin series of other, related functions.

Recall the formula for the coefficients of a Taylor series centered at is . Substituting , we get the formula for the coefficients of a Maclaurin series: We now use this to create the Maclaurin series for .

We now turn to two examples of finding Maclauirin series by making modifications to the previous example.

We can now write the Maclaurin series representation as This can be writtien in summation notation as The ratio test can be used to verify that this representation is valid on the interval .

We now consider an example of a Maclaurin series obtained by making modifications to the previous example.

We now consider an example of a Maclaurin series obtained by making modifications to the previous example.

We will use the Maclaurin series for with replacing and then multiply the result by . This gives This representation is valid in the interval .


The Basic Idea

Ok, so how do we come up with a power series to represent a function? In other words, start with the following (just using the general form of a power series: f ( x ) = ∑ n = 0 ∞ c n ( x − a ) n f(x)=sum_^ c_n(x−a)^n f ( x ) = n = 0 ∑ ∞ ​ c n ​ ( x − a ) n

It may help to try the following demo: (click to follow the link to Desmos)


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Example 1

Reprove that if $f(x) = sum_^ a_n(x - c)^n$ on the interval $(c - R, c + R)$ then $a_n = frac(c)>$ for each $n = 0, 1, 2, . $ .

Suppose that $f(x) = sum_^ a_n(x - c)^n$ on $(c - R, c + R)$ . Then we have that:

Plugging in $x = c$ and we have that $a_0 = frac(c)> <0!>= f(c)$ .

If we differentiate the equation from above, we have that:

Plugging in $x = c$ once again and we have that $a_1 = frac(c)> <1!>= f'(c)$ .

If we differentiate the equative above yet again, then we have that:

Plugging in $x = c$ and we see that $2a_2 = f^<(2)>(c)$ and so $a_2 = frac(c)><2!>$ . We can repeat this process inductive and have that for $n = 0, 1, 2, . $ :


Section 3.7 Taylor and Maclaurin Series ¶ permalink

In this section we will derive a way to represent any smooth (infinitely differentiable) function by a power series.

Suppose that a function (f(x)) can be represented by a power series centered on (a ext<,>) i.e. assume (f(x)) can be written as power series egin f(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + c_4(x-a)^4 + cdots label ag <3.7.1>end where the coefficients, (c_n) are to be determined. We know from theorem <<thm-differentiating-and-integrating-power-series>> that power series are differentiable term by term, thus upon differentiating serveral times we get: egin f'(x) & = 1 c_1 + 2 c_2(x-a) + 3 c_3(x-a)^2 + 4 c_4(x-a)^3 + 5 c_5(x-a)^4 + cdots f''(x) & = (2cdot 1) c_2 + (3cdot 2) c_3(x-a) + (4cdot 3) c_4(x-a)^2 + (5cdot 4) c_5(x-a)^3 + cdots f'''(x) & = (3 cdot 2cdot 1) c_3 + (4cdot 3cdot 2) c_4(x-a) + (5cdot 4 cdot 3) c_5(x-a)^2 + (6cdot 5 cdot 4) c_6(x-a)^3 + cdots & , vdots f^<(n)>(x) & = n! c_n + frac<(n+1)!> <1!>c_(x-a) + frac<(n+2)!> <2!>c_(x-a)^2 +cdots end

Evaluating (f(x)) and its derivatives at (x=a) yields: egin f(a) & = c_0 f'(a) & = 1cdot c_1 f''(a) & = 2cdot 1cdot c_2 f'''(a) & = 3cdot 2cdot 1 cdot c_3 & : , vdots f^<(n)>(a) & = n! c_n end The final line above allows us to determine the needed coefficients. egin oxed(a)>> label ag <3.7.2>end Where we define (0! equiv 1) and (f^<(0)>(x) equiv f(x) ext<.>)

Theorem 3.7.1 Taylor Series Representation of a Function

If (f(x)) has a power series representation (expansion) at (a ext<,>) i.e. if egin f(x) = sum_^ c_n(x-a)^n qquad abs lt R, end then the coefficients, (c_n ext<,>) are egin c_n = frac(a)> end

When the point that we center on, (a ext<,>) is zero, then we call the series a Maclaurin series. Let's compute the Maclaurin series for an especially important function (f(x) = e^x ext<.>) Since, ( fracleft( e^x ight) = e^x ext<,>) we have egin f(x) = f'(x) = f''(x) = cdots = f^<(n)>(x) = e^x, end and therefore egin f(0) = f'(0) = f''(0) = cdots = f^<(n)>(0) = 1. end Thus egin f(x) & = c_0 x^0 + c_1 x^1 + c_2 x^2 + c_3 x^3 + cdots f(x) & = frac <0!>x^0 + frac <1!>x^1 + frac<2!>x^2 + frac<3!>x^3 + cdots e^x & = 1 + x + frac<1><2!>x^2 + frac<1><3!>x^3 + frac<1><4!>x^4 + cdots end

Definition 3.7.2 Taylor Polynomial

The degree (n) Taylor polynomial for the function (f(x)) is given by: egin T_n(x) = sum_^ c_k (x-a)^k qquad c_k = frac(a)> label ag <3.7.3>end

Example 3.7.3 Taylor Polynomial

Here we will compute the Taylor polynomial of degree 3 for the function (f(x) = sin 2x) centered on (a = pi/2 ext<.>)

egin f(x) & = sin 2x & f(pi/2) &= 0 f'(x) & = 2cos 2x & f'(pi/2) &= -2 f''(x) & = -4sin 2x & f''(pi/2) &= 0 f'''(x) & = -8cos 2x & f'''(pi/2) &= 8 end egin c_0 = frac <0!>= 0 quad c_1 = frac <1!>= -2 quad c_2 = frac <2!>= 0 quad c_3 = frac <3!>= frac<8> <6>= frac<4> <3>end Thus the degree three Taylor polynomial which approximates (f(x) = sin 2x) near (pi/2) is: egin T_3(x) = -2 left( x- frac <2> ight) + frac<4> <3>left( x- frac <2> ight)^3 end

The preceding work can be checked with Sage.

The degree 3 Taylor polynomial, (T_3(x) ext<,>) as well as several other Taylor polynomials for the function (sin 2x) are plotted in figure 3.7.4.

Figure 3.7.4 Taylor polynomials for (sin 2x) centered on (pi/2 ext<.>)

Example 3.7.5 Maclaurin Series for (cos x) and (sin x)

The table below summarizes the condition that a function must satisfy in order to be even or odd.

Even: (f(-x) = f(x)) for all (x)
Odd: (f(-x) = -f(x)) for all (x)

The function (f(x) = cos(x)) is even because (cos(-x) = cos(x)) for all (x ext<.>) When a function is even its Maclaurin series will only contain even powers of (x ext<.>) Similarly, when a function is odd, its Maclaurin series will only contain odd powers of (x ext<.>) In this example we will compute the Maclaurin series for (cos x) to demonstrate this phenomenon.

Next we compute the Maclaurin series for (g(x) = sin x) which is and odd function because (sin(-x) = -sin(x)) for all (x ext<.>) egin g(x) &= sin x & g(0) & = 0 g'(x) &= cos x & g'(0) & = 1 g''(x) &= -sin x & g''(0) & = 0 g'''(x) &= -cos x & g'''(0) & = -1 g^<(4)>(x) &= sin x & g^<(4)>(0) & = 0 g^<(5)>(x) &= cos x & g^<(5)>(0) & = 1 end Thus the Maclaurin series for (g(x)=sin x) is: egin g(x) = x - frac <3!>+ frac <5!>- frac <7!>+ cdots = sum_^ (-1)^n frac> <(2n+1)!> ext <.>end


Watch the video: Taylor Series and Maclaurin Series - Calculus 2 (October 2021).