# APEX Calculus: for University of Lethbridge

## Section4.5Taylor Polynomials

Consider a function $$y=f(x)$$ and a point $$\big(c,f(c)\big)\text{.}$$ The derivative, $$\fp(c)\text{,}$$ gives the instantaneous rate of change of $$f$$ at $$x=c\text{.}$$ Of all lines that pass through the point $$\big(c,f(c)\big)\text{,}$$ the line that best approximates $$f$$ at this point is the tangent line; that is, the line whose slope (rate of change) is $$\fp(c)\text{.}$$
In Figure 4.5.2, we see a function $$y=f(x)$$ graphed. The table in Table 4.5.3 shows that $$f(0)=2$$ and $$\fp(0) = 1\text{;}$$ therefore, the tangent line to $$f$$ at $$x=0$$ is $$p_1(x) = 1(x-0)+2 = x+2\text{.}$$ The tangent line is also given in the figure. Note that “near” $$x=0\text{,}$$ $$p_1(x) \approx f(x)\text{;}$$ that is, the tangent line approximates $$f$$ well.
One shortcoming of this approximation is that the tangent line only matches the slope of $$f\text{;}$$ it does not, for instance, match the concavity of $$f\text{.}$$ We can find a polynomial, $$p_2(x)\text{,}$$ that does match the concavity near $$0$$ without much difficulty, though. The table in Table 4.5.3 gives the following information:
\begin{equation*} f(0) = 2 \qquad \fp(0) = 1\qquad \fp'(0) = 2\text{.} \end{equation*}
Therefore, we want our polynomial $$p_2(x)$$ to have these same properties. That is, we need
\begin{equation*} p_2(0) = 2 \qquad p_2'(0) = 1 \qquad p_2''(0) = 2\text{.} \end{equation*}
\begin{equation*} p(x) = a_0+a_1x+a_2x^2\text{.} \end{equation*}
We find the following:
\begin{align*} p_2(x) \amp = a_0+a_1x+a_2x^2 \amp p_2(0)\amp=a_0\\ p_2\primeskip '(x) \amp = a_1+2a_2x\amp p_2\primeskip '(0) \amp = a_1\\ p_2\primeskip ''(x) \amp = 2a_2 \amp p_2\primeskip''(0) \amp = 2a_2\text{.} \end{align*}
To get the desired properties above, we must have
\begin{equation*} a_0 = f(0) = 2,\, a_1 = \fp(0) = 1,\, 2a_2 = \fpp(0) = 2\text{,} \end{equation*}
so $$a_0=2\text{,}$$ $$a_1=1\text{,}$$ and $$a_2 = 2/2=1\text{,}$$ giving us the polynomial
\begin{equation*} p_2(x) = 2+x+x^2. \end{equation*}
We can repeat this approximation process by creating polynomials of higher degree that match more of the derivatives of $$f$$ at $$x=0\text{.}$$ In general, a polynomial of degree $$n$$ can be created to match the first $$n$$ derivatives of $$f\text{.}$$ Figure 4.5.4 shows $$p_4(x)= -x^4/2-x^3/6+x^2+x+2\text{,}$$ whose first four derivatives at 0 match those of $$f\text{.}$$
How do we ensure that the derivatives of our polynomial match those of $$f\text{?}$$ We simply begin with a polynomial of the desired degree, compute its derivatives, and compare them to those of $$f\text{!}$$ Recall that each term in a polynomial consists of a power of $$x\text{,}$$ and a coefficient, like so: $$a_nx^n\text{.}$$ Our goal is to determine the value for each coefficient $$a_n$$ so that the derivatives of our polynomial match those of our function $$f\text{.}$$ If we take $$k$$ derivatives of the term $$a_nx^n\text{,}$$ with $$k\leq n\text{,}$$ we obtain
\begin{equation*} \frac{d^k}{dx^k}(a_nx^n)= n(n-1)\cdots (n-k+1)a_nx^{n-k}\text{.} \end{equation*}
For $$k\lt n\text{,}$$ the expression above vanishes when we set $$x=0\text{.}$$ However, for $$n=k\text{,}$$ we obtain the constant value
$$\frac{d^k}{dx^k}(a_kx^k) = k\cdot (k-1)\cdots 2\cdot 1 a_k\text{.}\tag{4.5.1}$$
Consider a polynomial
\begin{equation*} p_n(x) = a_0+a_1x+\cdots +a_kx^k+\cdots +a_nx^n \end{equation*}
of degree $$n\text{.}$$ If we take $$k$$ derivatives, all of the terms involving powers of $$x$$ less than $$k$$ disappear, and when we set $$x=0\text{,}$$ all of the terms involving powers of $$x$$ larger than $$k$$ disappear, leaving us with the single constant given in Equation (4.5.1).
Recalling the notation $$k! = 1\cdot 2\cdot 3\cdots k$$ for the product of the first $$k$$ integers, we have shown that
\begin{equation*} p_n^{(k)}(0) = k!a_k\text{.} \end{equation*}
If we want the derivatives of $$p_n$$ to agree with some unknown function $$f$$ when $$x=0\text{,}$$ then we must have
\begin{equation*} a_k = \frac{f^{(k)}(0)}{k!}\text{.} \end{equation*}
As we use more and more derivatives, our polynomial approximation to $$f$$ gets better and better. In this example, the interval on which the approximation is “good” gets bigger and bigger. Figure 4.5.6 shows $$p_{13}(x)\text{;}$$ we can visually affirm that this polynomial approximates $$f$$ very well on $$[-2,3]\text{.}$$ (The polynomial $$p_{13}(x)$$ is not particularly “nice”. It is
\begin{align*} \amp p_{13}(x)=\frac{16901x^{13}}{6227020800}+\frac{13x^{12}}{1209600}-\frac{1321x^{11}}{39916800}-\frac{779x^{10}}{1814400} -\frac{359x^9}{362880}\\ \amp +\frac{x^8}{240}+\frac{139x^7}{5040}+\frac{11 x^6}{360}-\frac{19x^5}{120}-\frac{x^4}{2}-\frac{x^3}{6}+x^2+x+2\text{.} \end{align*}
The polynomials we have created are examples of Taylor polynomials, named after the British mathematician Brook Taylor who made important discoveries about such functions. In the discussion above, we concentrated on evaluating the derivatives of $$f$$ at 0; however, there is nothing special about this point. Just as we can consider the linear approximation of a function near any point, so too can we determine a polynomial approximation about any value $$c$$ in the domain of $$f\text{.}$$ The only catch is that our polynomial will then be given in terms of powers of $$x-c\text{,}$$ rather than powers of $$x\text{,}$$ as we see in the following definition.

### Definition4.5.7.Taylor Polynomial, Maclaurin Polynomial.

Let $$f$$ be a function whose first $$n$$ derivatives exist at $$x=c\text{.}$$
1. The Taylor polynomial of degree $$n$$ of $$f$$ at $$x=c$$ is
\begin{align*} p_n(x) \amp = f(c) + \fp(c)(x-c) + \frac{\fpp(c)}{2!}(x-c)^2\\ \amp \quad+\frac{\fp''(c)}{3!}(x-c)^3+\cdots+\frac{f\,^{(n)}(c)}{n!}(x-c)^n\text{.} \end{align*}
2. A special case of the Taylor polynomial is the Maclaurin polynomial, where $$c=0\text{.}$$ That is, the Maclaurin polynomial of degree $$n$$ of $$f$$ is
\begin{equation*} p_n(x) = f(0) + \fp(0)x + \frac{\fpp(0)}{2!}x^2+\frac{\fp''(0)}{3!}x^3+\cdots+\frac{f\,^{(n)}(0)}{n!}x^n\text{.} \end{equation*}
We will practice creating Taylor and Maclaurin polynomials in the following examples.

### Example4.5.9.Finding and using Maclaurin polynomials.

1. Find the $$n$$th Maclaurin polynomial for $$f(x) = e^x\text{.}$$
2. Use $$p_5(x)$$ to approximate the value of $$e\text{.}$$
Solution 1. Video solution
Solution 2.
1. We start with creating a table of the derivatives of $$e^x$$ evaluated at $$x=0\text{.}$$ In this particular case, this is relatively simple, as shown in Table 4.5.10.
By the definition of the Maclaurin polynomial, we have
\begin{align*} p_n(x) \amp = f(0) + \fp(0)x + \frac{\fpp(0)}{2!}x^2+\frac{\fp''(0)}{3!}x^3+\cdots+\frac{f\,^{(n)}(0)}{n!}x^n\\ \amp = 1+x+\frac{1}{2}x^2+\frac{1}{6}x^3 + \frac{1}{24}x^4 + \cdots + \frac{1}{n!}x^n\text{.} \end{align*}
2. Using our answer from part 1, we have
\begin{equation*} e^x\approx p_5(x) = 1+x+\frac{1}{2}x^2+\frac{1}{6}x^3 + \frac{1}{24}x^4 + \frac{1}{120}x^5\text{.} \end{equation*}
To approximate the value of $$e\text{,}$$ note that $$e = e^1 = f(1) \approx p_5(1)\text{.}$$ It is very straightforward to evaluate $$p_5(1)\text{:}$$
\begin{equation*} p_5(1) = 1+1+\frac12+\frac16+\frac1{24}+\frac1{120} = \frac{163}{60} \approx 2.71667\text{.} \end{equation*}
A plot of $$f(x)=e^x$$ and $$p_5(x)$$ is given in Figure 4.5.11. To $$5$$ decimal places, the actual value of $$e$$ is $$2.71828\text{.}$$ So this approximation agrees to two decimal places.

### Example4.5.12.Finding and using Taylor polynomials.

1. Find the $$n$$th Taylor polynomial of $$y=\ln(x)$$ at $$x=1\text{.}$$
2. Use $$p_6(x)$$ to approximate the value of $$\ln(1.5)\text{.}$$
3. Use $$p_6(x)$$ to approximate the value of $$\ln(2)\text{.}$$
Solution 1. Video solution
Solution 2.
1. We begin by creating a table of derivatives of $$\ln(x)$$ evaluated at $$x=1\text{.}$$ While this is not as straightforward as it was in the previous example, a pattern does emerge (for $$n\ge 1$$), as shown in Table 4.5.13. Notice in the table below that each time we take a derivative (starting at the second derivative), we apply the power rule and “bring down” the exponent to multiply by the previous coefficent. So the $$6$$ in the $$4^{th}$$ derivative is actually $$1\cdot 2\cdot 3=3!\text{.}$$
Notice that the coefficients alternate in sign starting at $$n=1\text{.}$$ Using Definition 4.5.7, we have
\begin{align*} p_n(x) \amp = f(c) + \fp(c)(x-c) + \frac{\fpp(c)}{2!}(x-c)^2+\dots\\ \amp \dots\frac{\fp''(c)}{3!}(x-c)^3+\cdots+\frac{f\,^{(n)}(c)}{n!}(x-c)^n\\ \amp = 0 + \frac{0!}{1!}(x-1) - \frac{1!}{2!}(x-1)^2+\dots\\ \amp \dots\frac{2!}{3!}(x-1)^3+\cdots+\frac{(-1)^{n+1}\cdot (n-1)!}{n!}(x-1)^n\\ \amp = (x-1)-\frac12(x-1)^2+\frac13(x-1)^3-\dots\\ \amp \dots \frac14(x-1)^4+\cdots+\frac{(-1)^{n+1}}{n}(x-1)^n\text{.} \end{align*}
Note how the coefficients of the $$(x-1)$$ terms turn out to be “nice.”
2. We can compute $$p_6(x)$$ using our work above:
Since $$p_6(x)$$ approximates $$\ln(x)$$ well near $$x=1\text{,}$$ we approximate $$\ln(1.5) \approx p_6(1.5)\text{:}$$
\begin{align*} p_6(1.5) \amp = (1.5-1)-\frac12(1.5-1)^2+\frac13(1.5-1)^3+\dots\\ \amp \dots -\frac14(1.5-1)^4 +\frac15(1.5-1)^5-\frac16(1.5-1)^6\\ \amp =\frac{259}{640}\\ \amp \approx 0.404688\text{.} \end{align*}
This is a good approximation as a calculator shows that $$\ln(1.5) \approx 0.4055\text{.}$$ Figure 4.5.14 below plots $$y=\ln(x)$$ with $$y=p_6(x)\text{.}$$ We can see that $$\ln(1.5) \approx p_6(1.5)\text{.}$$
3. We approximate $$\ln 2$$ with $$p_6(2)\text{:}$$
\begin{align*} p_6(2) \amp = (2-1)-\frac12(2-1)^2+\frac13(2-1)^3-\frac14(2-1)^4+\cdots\\ \amp \cdots +\frac15(2-1)^5-\frac16(2-1)^6\\ \amp = 1-\frac12+\frac13-\frac14+\frac15-\frac16\\ \amp = \frac{37}{60}\\ \amp \approx 0.616667\text{.} \end{align*}
This approximation is not terribly impressive: a hand held calculator shows that $$\ln(2) \approx 0.693147\text{.}$$ The graph in Figure 4.5.14 shows that $$p_6(x)$$ provides less accurate approximations of $$\ln(x)$$ as $$x$$ gets close to 0 or 2. Surprisingly enough, even the $$20$$th degree Taylor polynomial fails to approximate $$\ln(x)$$ for $$x\gt 2$$ very well, as shown in Figure 4.5.15. We'll soon discuss why this is.
Taylor polynomials are used to approximate functions $$f(x)$$ in mainly two situations:
1. When $$f(x)$$ is known, but perhaps “hard” to compute directly. For instance, we can define the cosine of an angle as either the ratio of sides of a right triangle (“adjacent over hypotenuse”) or using the definition in terms of the unit circle. However, neither of these provides a convenient way of computing $$\cos(2)\text{.}$$ A Taylor polynomial of sufficiently high degree can provide a reasonable method of computing such values using only operations usually hard-wired into a computer ($$+\text{,}$$ $$-\text{,}$$ × and $$\div$$).
2. When $$f(x)$$ is not known, but information about its derivatives is known. This occurs more often than one might think, especially in the study of differential equations.
In both situations, a critical piece of information to have is “How good is my approximation?” If we use a Taylor polynomial to compute $$\cos(2)\text{,}$$ how do we know how accurate the approximation is?
Although much of the content presented in Calculus concerns the search for exact answers to problems such as integration and differentiation, many practical applications of calculus involve attempts to find approximations; for example, using Newton's Method to approximate the zeros of a function, or numerical integration to approximate the value of an integral that cannot be solved exactly. Whenever an approximation is used, one naturally wishes to know how good the approximation is. In other words, we look for a bound on the error introduced by working with an approximation. The following theorem gives bounds on the error introduced in using a Taylor (and hence Maclaurin) polynomial to approximate a function.
The first part of Taylor's Theorem states that $$f(x) = p_n(x) + R_n(x)\text{,}$$ where $$p_n(x)$$ is the $$n$$th order Taylor polynomial and $$R_n(x)$$ is the remainder, or error, in the Taylor approximation. The second part gives bounds on how big that error can be. If the $$(n+1)$$th derivative is large on $$I\text{,}$$ the error may be large; if $$x$$ is far from $$c\text{,}$$ the error may also be large. However, the $$(n+1)!$$ term in the denominator tends to ensure that the error gets smaller as $$n$$ increases.
The following example computes error estimates for the approximations of $$\ln(1.5)$$ and $$\ln(2)$$ made in Example 4.5.12.

### Example4.5.18.Finding error bounds of a Taylor polynomial.

Use Theorem 4.5.16 to find error bounds when approximating $$\ln(1.5)$$ and $$\ln(2)$$ with $$p_6(x)\text{,}$$ the Taylor polynomial of degree 6 of $$f(x)=\ln(x)$$ at $$x=1\text{,}$$ as calculated in Example 4.5.12.
Solution 1. Video solution
Solution 2.
1. We start with the approximation of $$\ln(1.5)$$ with $$p_6(1.5)\text{.}$$ The theorem references an open interval $$I$$ that contains both $$x$$ and $$c\text{.}$$ The smaller the interval we use the better; it will give us a more accurate (and smaller!) approximation of the error. We let $$I = (0.9,1.6)\text{,}$$ as this interval contains both $$c=1$$ and $$x=1.5\text{.}$$ The theorem references $$\max\abs{f^{(n+1)}(z)}\text{.}$$ In our situation, this is asking “How big can the $$7$$th derivative of $$y=\ln(x)$$ be on the interval $$(0.9,1.6)\text{?}$$” The seventh derivative is $$y = -6!/x^7\text{.}$$ The largest absolute value it attains on $$I$$ is about 1506. (There are no critical numbers of $$f^{(7)}$$ in the interval so we evaluate the endpoints: $$f^{(7)}(0.9)\approx 1506$$ and $$f^{(7)}(1.6)\approx 27\text{.}$$) In particular, we are evaluating at $$x=1.5\text{,}$$ so we let $$x=1.5\text{.}$$ Thus we can bound the error as:
\begin{align*} \abs{R_6(1.5)} \amp \leq \frac{\max\abs{f^{(7)}(z)}}{7!}\abs{(1.5-1)^7}\\ \amp \leq \frac{1506}{5040}\cdot\frac1{2^7}\\ \amp \approx 0.0023\text{.} \end{align*}
We computed $$p_6(1.5) = 0.404688\text{;}$$ using a calculator, we find $$\ln(1.5) \approx 0.405465\text{,}$$ so the actual error is about $$0.000778\text{,}$$ which is less than our bound of $$0.0023\text{.}$$ This affirms Taylor's Theorem; the theorem states that our approximation would be within about 2 thousandths of the actual value, whereas the approximation was actually closer. Taylor's Theorem only gives an upper bound on the error.
2. We again find an interval $$I$$ that contains both $$c=1$$ and $$x=2\text{;}$$ we choose $$I = (0.9,2.1)\text{.}$$ The maximum value of the seventh derivative of $$f$$ on this interval is again about 1506 (as the largest values come near $$x=0.9$$). Thus
\begin{align*} \abs{ R_6(2)} \amp \leq \frac{\max\abs{f^{(7)}(z)}}{7!}\abs{(2-1)^7}\\ \amp \leq \frac{1506}{5040}\cdot1^7\\ \amp \approx 0.30\text{.} \end{align*}
This bound is not as nearly as good as before. Using the degree 6 Taylor polynomial at $$x =1$$ will bring us within 0.3 of the correct answer. As $$p_6(2)\approx 0.61667\text{,}$$ our error estimate guarantees that the actual value of $$\ln(2)$$ is somewhere between $$0.31667$$ and $$0.91667\text{.}$$ These bounds are not particularly useful. In reality, our approximation was only off by about 0.07. However, we are approximating ostensibly because we do not know the real answer. In order to be assured that we have a good approximation, we would have to resort to using a polynomial of higher degree.
We practice again. This time, we use Taylor's theorem to find $$n$$ that guarantees our approximation is within a certain amount.

### Example4.5.19.Finding sufficiently accurate Taylor polynomials.

Find $$n$$ such that the $$n$$th Taylor polynomial of $$f(x)=\cos(x)$$ at $$x=0$$ approximates $$\cos(2)$$ to within $$0.001$$ of the actual answer. What is $$p_n(2)\text{?}$$
Solution 1. Video solution
Solution 2.
Following Taylor's theorem, we need bounds on the size of the derivatives of $$f(x)=\cos(x)\text{.}$$ In the case of this trigonometric function, this is easy. All derivatives of cosine are $$\pm \sin(x)$$ or $$\pm \cos(x)\text{.}$$ In all cases, these functions are never greater than 1 in absolute value. We want the error to be less than $$0.001\text{.}$$ To find the appropriate $$n\text{,}$$ consider the following inequalities:
\begin{align*} \frac{\max\abs{f^{(n+1)}(z)}}{(n+1)!}\abs{(2-0)^{(n+1)}} \amp \leq 0.001\\ \frac1{(n+1)!}\cdot2^{(n+1)} \amp \leq 0.001\text{.} \end{align*}
We find an $$n$$ that satisfies this last inequality with trial-and-error. When $$n=8\text{,}$$ we have $$\ds \frac{2^{8+1}}{(8+1)!} \approx 0.0014\text{;}$$ when $$n=9\text{,}$$ we have $$\ds \frac{2^{9+1}}{(9+1)!} \approx 0.000282 \lt 0.001\text{.}$$ Thus we want to approximate $$\cos(2)$$ with $$p_9(2)\text{.}$$
We now set out to compute $$p_9(x)\text{.}$$ We again need a table of the derivatives of $$f(x)=\cos(x)$$ evaluated at $$x=0\text{.}$$ A table of these values is given in Table 4.5.20.
Notice how the derivatives, evaluated at $$x=0\text{,}$$ follow a certain pattern. All the odd powers of $$x$$ in the Taylor polynomial will disappear as their coefficient is $$0\text{.}$$ While our error bounds state that we need $$p_9(x)\text{,}$$ our work shows that this will be the same as $$p_8(x)\text{.}$$
Since we are forming our polynomial at $$x=0\text{,}$$ we are creating a Maclaurin polynomial, and:
\begin{align*} p_8(x) \amp = f(0) + \fp(0)x + \frac{\fpp(0)}{2!}x^2 + \frac{\fp''(0)}{3!}x^3 + \cdots +\frac{f^{(8)}(0)}{8!}x^8\\ \amp = 1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\frac{1}{6!}x^6+\frac{1}{8!}x^8\text{.} \end{align*}
We finally approximate $$\cos(2)\text{:}$$
\begin{equation*} \cos(2) \approx p_8(2) = -\frac{131}{315} \approx -0.41587\text{.} \end{equation*}
Our error bound guarantee that this approximation is within $$0.001$$ of the correct answer. Technology shows us that our approximation is actually within about $$0.0003$$ of the correct answer.
Figure 4.5.21 shows a graph of $$y=p_8(x)$$ and $$y=\cos(x)\text{.}$$ Note how well the two functions agree on about $$(-\pi,\pi)\text{.}$$

### Example4.5.22.Finding and using Taylor polynomials.

1. Find the degree 4 Taylor polynomial, $$p_4(x)\text{,}$$ for $$f(x)=\sqrt{x}$$ at $$x=4\text{.}$$
2. Use $$p_4(x)$$ to approximate $$\sqrt{3}\text{.}$$
3. Find bounds on the error when approximating $$\sqrt{3}$$ with $$p_4(3)\text{.}$$
Solution.
1. We begin by evaluating the derivatives of $$f$$ at $$x=4\text{.}$$ This is done in Table 4.5.23.
These values allow us to form the Taylor polynomial $$p_4(x)\text{:}$$
\begin{align*} p_4(x) = 2 \amp + \frac14(x-4) +\frac{-1/32}{2!}(x-4)^2+\dots\\ \amp \dots \frac{3/256}{3!}(x-4)^3+\frac{-15/2048}{4!}(x-4)^4\text{.} \end{align*}
2. As $$p_4(x) \approx \sqrt{x}$$ near $$x=4\text{,}$$ we approximate $$\sqrt{3}$$ with $$p_4(3) = 1.73212\text{.}$$
3. To find a bound on the error, we need an open interval that contains $$x=3$$ and $$x=4\text{.}$$ We set $$I = (2.9,4.1)\text{.}$$ The largest value the fifth derivative of $$f(x)=\sqrt{x}$$ takes on this interval is near $$x=2.9\text{,}$$ at about $$0.0273\text{.}$$ (We often graph the $$(n+1)^{th}$$ derivative to find its extrema. In this case is $$f^{(5)}(x)=105/(32x^{9/2})$$ is always decreasing, so the maximum occurs at $$2.9\text{.}$$) Thus
\begin{equation*} \abs{R_4(3)} \leq \frac{0.0273}{5!}\abs{(3-4)^5} \approx 0.00023\text{.} \end{equation*}
This shows our approximation is accurate to at least the first 2 places after the decimal. (It turns out that our approximation is actually accurate to 4 places after the decimal.) A graph of $$f(x)=\sqrt x$$ and $$p_4(x)$$ is given in Figure 4.5.24. Note how the two functions are nearly indistinguishable on $$(2,7)\text{.}$$
Our final example gives a brief introduction to using Taylor polynomials to solve differential equations.

### Example4.5.25.Approximating an unknown function.

A function $$y=f(x)$$ is unknown save for the following two facts.
1. $$y(0) = f(0) = 1\text{,}$$ and
2. $$\displaystyle y'= y^2$$
(This second fact says that amazingly, the derivative of the function is actually the function squared!)
Find the degree 3 Maclaurin polynomial $$p_3(x)$$ of $$y=f(x)\text{.}$$
Solution.
One might initially think that not enough information is given to find $$p_3(x)\text{.}$$ However, note how the second fact above actually lets us know what $$y'(0)$$ is:
\begin{equation*} y' = y^2 \Rightarrow y'(0) = y^2(0)\text{.} \end{equation*}
Since $$y(0) = 1\text{,}$$ we conclude that $$y'(0) = 1\text{.}$$
Now we find information about $$y''\text{.}$$ Starting with $$y'=y^2\text{,}$$ take derivatives of both sides, with respect to $$x$$. That means we must use implicit differentiation.
\begin{align*} y' \amp = y^2\\ \frac{d}{dx}\big(y'\big) \amp = \frac{d}{dx}\big(y^2\big)\\ y'' \amp = 2y\cdot y'.\\ \end{align*}

Now evaluate both sides at $$x=0\text{:}$$

\begin{align*} y''(0) \amp = 2y(0)\cdot y'(0)\\ y''(0) \amp = 2\text{.} \end{align*}
We repeat this once more to find $$y'''(0)\text{.}$$ We again use implicit differentiation; this time the Product Rule is also required.
\begin{align*} \frac{d}{dx}\big(y''\big) \amp = \frac{d}{dx} \big(2yy'\big)\\ y''' \amp = 2y'\cdot y' + 2y\cdot y''.\\ \end{align*}

Now evaluate both sides at $$x=0\text{:}$$

\begin{align*} y'''(0) \amp = 2y'(0)^2 + 2y(0)y''(0)\\ y'''(0) \amp = 2+4=6\text{.} \end{align*}
In summary, we have:
\begin{equation*} y(0) = 1 \qquad y'(0) = 1 \qquad y''(0) = 2 \qquad y'''(0) = 6\text{.} \end{equation*}
We can now form $$p_3(x)\text{:}$$
\begin{align*} p_3(x) \amp = 1 + x + \frac{2}{2!}x^2 + \frac{6}{3!}x^3\\ \amp = 1+x+x^2+x^3\text{.} \end{align*}
It turns out that the differential equation we started with, $$y'=y^2\text{,}$$ where $$y(0)=1\text{,}$$ can be solved without too much difficulty:
\begin{equation*} y = \frac{1}{1-x}\text{.} \end{equation*}
Figure 4.5.26 shows this function plotted with $$p_3(x)\text{.}$$ Note how similar they are near $$x=0\text{.}$$
It is beyond the scope of this text to pursue error analysis when using Taylor polynomials to approximate solutions to differential equations. This topic is often broached in introductory Differential Equations courses and usually covered in depth in Numerical Analysis courses. Such an analysis is very important; one needs to know how good their approximation is. We explored this example simply to demonstrate the usefulness of Taylor polynomials.
We first learned of the derivative in the context of instantaneous rates of change and slopes of tangent lines. We furthered our understanding of the power of the derivative by studying how it relates to the graph of a function (leading to ideas of increasing/decreasing and concavity).

### ExercisesExercises

#### Terms and Concepts

##### 1.
What is the difference between a Taylor polynomial and a Maclaurin polynomial?
##### 2.
True or False? In general, $$p_n(x)$$ approximates $$f(x)$$ better and better as $$n$$ gets larger.
• True
• False
##### 3.
For some function $$f(x)\text{,}$$ the Maclaurin polynomial of degree 4 is $$p_4(x) = 6+3x-4x^2+5x^3-7x^4\text{.}$$ What is $$p_2(x)\text{?}$$
$$p_2(x)=$$
##### 4.
For some function $$f(x)\text{,}$$ the Maclaurin polynomial of degree 4 is $$p_4(x) = 6+3x-4x^2+5x^3-7x^4\text{.}$$ What is $$\fpp'(0)\text{?}$$
$$\fpp'(0)=$$

#### Problems

##### Exercise Group.
In the following exercises, find the Maclaurin polynomial of degree $$n$$ for the given function.
###### 5.
Find the Maclaurin polynomial of degree $$n=3$$ for $$f(x) = e^{-x}\text{.}$$
$$p_3(x)=$$
###### 6.
Find the Maclaurin polynomial of degree $$n=8$$ for $$f(x) = \sin(x)\text{.}$$
$$p_8(x)=$$
###### 7.
Find the Maclaurin polynomial of degree $$n=5$$ for $$f(x) = x\cdot e^x\text{.}$$
$$p_5(x)=$$
###### 8.
Find the Maclaurin polynomial of degree $$n=6$$ for $$f(x) = \tan(x)\text{.}$$
$$p_6(x)=$$
###### 9.
Find the Maclaurin polynomial of degree $$n=4$$ for $$f(x) = e^{2x}\text{.}$$
$$p_4(x)=$$
###### 10.
Find the Maclaurin polynomial of degree $$n=4$$ for $$\ds f(x) = \frac1{1-x}\text{.}$$
$$p_4(x)=$$
###### 11.
Find the Maclaurin polynomial of degree $$n=4$$ for $$\ds f(x) = \frac1{1+x}\text{.}$$
$$p_4(x)=$$
###### 12.
Find the Maclaurin polynomial of degree $$n=7$$ for $$\ds f(x) = \frac1{1+x}\text{.}$$
$$p_7(x)=$$
##### Exercise Group.
In the following exercises, find the Taylor polynomial of degree $$n\text{,}$$ at $$x=c\text{,}$$ for the given function.
###### 13.
Find the Taylor polynomial for $$f(x) = \sqrt x$$ of degree $$n=4\text{,}$$ at $$c=1\text{.}$$
$$p_4(x)=$$
###### 14.
Find the degree $$n=4$$ Taylor polynomial for $$f(x) = \ln(x+1)\text{,}$$ at $$c=1\text{.}$$
$$p_4(x)=$$
###### 15.
Find the degree $$n=6$$ Taylor polynomial for $$f(x) = \cos(x)\text{,}$$ at $$c=\pi/4\text{.}$$
$$p_6(x)=$$
###### 16.
Find the degree $$n=5$$ Taylor poplynomial for $$f(x) = \sin(x)\text{,}$$ at $$c=\pi/6\text{.}$$
$$p_5(x)=$$
###### 17.
Find the degree $$n=5$$ Taylor poplynomial for $$f(x) = \frac1x\text{,}$$ at $$c=2\text{.}$$
$$p_5(x)=$$
###### 18.
Find the degree $$n=8$$ Taylor poplynomial for $$\ds f(x) = \frac{1}{x^2}\text{,}$$ at $$c=1\text{.}$$
$$p_8(x)=$$
###### 19.
Find the degree $$n=3$$ Taylor poplynomial for $$\ds f(x) = \frac{1}{x^2+1}\text{,}$$ at $$c=-1\text{.}$$
$$p_3(x)=$$
###### 20.
Find the degree $$n=2$$ Taylor polynomial for $$f(x) = x^2\cos(x)\text{,}$$ at $$c=\pi\text{.}$$
$$p_2(x)=$$
##### Exercise Group.
In the following exercises, approximate the function value with the indicated Taylor polynomial and give approximate bounds on the error.
###### 21.
Approximate $$\sin(0.1)$$ with the Maclaurin polynomial of degree 3.
###### 22.
Approximate $$\cos(1)$$ with the Maclaurin polynomial of degree 4.
###### 23.
Approximate $$\sqrt{10}$$ with the Taylor polynomial of degree 2 centered at $$x=9\text{.}$$
###### 24.
Approximate $$\ln(1.5)$$ with the Taylor polynomial of degree 3 centered at $$x=1\text{.}$$
##### Exercise Group.
The following exercises ask for an $$n$$ to be found such that $$p_n(x)$$ approximates $$f(x)$$ within a certain bound of accuracy.
###### 25.
Find $$n$$ such that the Maclaurin polynomial of degree $$n$$ of $$f(x)= e^x$$ approximates $$e$$ within $$0.0001$$ of the actual value.
###### 26.
Find $$n$$ such that the Taylor polynomial of degree $$n$$ of $$f(x)= \sqrt x\text{,}$$ centered at $$x=4\text{,}$$ approximates $$\sqrt 3$$ within $$0.0001$$ of the actual value.
###### 27.
Find $$n$$ such that the Maclaurin polynomial of degree $$n$$ of $$f(x)= \cos(x)$$ approximates $$\cos(\pi/3)$$ within $$0.0001$$ of the actual value.
###### 28.
Find $$n$$ such that the Maclaurin polynomial of degree $$n$$ of $$f(x)= \sin(x)$$ approximates $$\cos(\pi)$$ within $$0.0001$$ of the actual value.
##### Exercise Group.
In the following exercises, find the $$n$$th term of the indicated Taylor polynomial.
###### 29.
Find a formula for the $$n$$th term of the Maclaurin polynomial for $$f(x)=e^x\text{.}$$
###### 30.
Find a formula for the $$n$$th term of the Maclaurin polynomial for $$f(x)=\cos(x)\text{.}$$
###### 31.
Find a formula for the $$n$$th term of the Maclaurin polynomial for $$f(x)=\sin x\text{.}$$
###### 32.
Find a formula for the $$n$$th term of the Maclaurin polynomial for $$\ds f(x)=\frac{1}{1-x}\text{.}$$
###### 33.
Find a formula for the $$n$$th term of the Maclaurin polynomial for $$\ds f(x)=\frac{1}{1+x}\text{.}$$
###### 34.
Find a formula for the $$n$$th term of the Taylor polynomial for $$\ds f(x)=\ln(x)$$ centered at $$x=1\text{.}$$
##### Exercise Group.
In the following exercises, approximate the solution to the given differential equation with a degree 4 Maclaurin polynomial.
###### 35.
$$y'=y\text{,}$$ $$y(0) = 1$$
###### 36.
$$y'=5y\text{,}$$ $$y(0) = 3$$
###### 37.
$$\ds y'=\frac2y\text{,}$$ $$y(0) = 1$$
en.wikipedia.org/wiki/Colin_Maclaurin