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Taylor polynomials

In Activity, we used f(x) = e^x as a case study to investigate polynomial approximations to f(x) near near a = 0.

Introduction

In Activity, we used \(f(x) = e^x\) as a case study to investigate polynomial approximations to \(f(x)\) near near \(a = 0\). For the degree \(3\) approximation, we chose the conditions \(T_3(0) = f(0)\), \(T_3'(0) = f'(0)\), \(T_3''(0) = f''(0)\), and \(T_3'''(0) = f'''(0)\). Starting with \(T_3(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3\) and \(f(x) = e^x\), we found the first three derivatives of \(f\) and \(T_3\) and evaluated them at \(a = 0\), which led to the results below. \[\begin{aligned}f(x) \amp= e^x \amp T_3(x) \amp= c_0 + c_1 x + c_2 x^2 + c_3 x^3 \\ f'(x) \amp= e^x \amp T_3'(x) \amp= c_1 + 2 c_2 x + 3 c_3 x^2 \\ f''(x) \amp= e^x \amp T_3''(x) \amp= 2 c_2 + 6 c_3 x \\ f'''(x) \amp= e^x \amp T_3'''(x) \amp= 6 c_3 \\ \amp \\ f(0) \amp= 1 \amp T_3(0) \amp= c_0 \\ f'(0) \amp= 1 \amp T_3'(0) \amp= c_1 \\ f''(0) \amp= 1 \amp T_3''(0) \amp= 2 c_2 \\ f'''(0) \amp= 1 \amp T_3'''(0) \amp= 6 c_3\end{aligned}\]

Equating the function and derivative values of \(f\) and \(T_3\) at \(a = 0\), it follows \[\begin{aligned}\end{aligned}\] and therefore \[\begin{aligned}\end{aligned}\] so that \[\begin{aligned}\end{aligned}\]. Moreover, plotting \(f\) and \(T_3\) near \(a = 0\), we see in Figure that the interval of accuracy for a tolerance of \(0.1\) is about \(-1.2 \le x \le 1.2\).

One important pattern we observed in our work with \(f(x) = e^x\) is that for every natural number \(n\), the \(n^{\text{th}}\) derivative's value at \(a=0\) is \(f^{(n)}(0) = 1\). As we will see, this information will ultimately help us find a general formula for \(c_n\), the coefficient of \(x^n\) in the degree \(n\) polynomial approximation of \(f(x) = e^x\).

In this section, we will learn how we can more systematically find degree \(n\) approximations for functions that have at least \(n\) derivatives, as well as how to center the approximation at a value other than \(a = 0\).

Exploration

Taylor polynomials

In our work so far in Chapter, we have found several different approximations of two important functions: \(e^x\) and \(\sin(x)\). In Section, we saw that near \(a = 0\)

\(e^x \approx 1 + x\) (the degree \(1\) approximation);
\(e^x \approx 1 + x + \frac{1}{2}x^2\) (the degree \(2\) approximation);
\(e^x \approx 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3\) (the degree \(3\) approximation); and
\(e^x \approx 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4\) (the degree \(4\) approximation).

In Preview Activity, we saw further that

\(\sin(x) \approx x\) (the degree \(1\) approximation); and
\(\sin(x) \approx x - \frac{1}{6}x^3\) (the degree \(3\) approximation).

We've also observed that as the degree of the approximation increases, the polynomial approximation gets more accurate by being closer to the original function \(f(x)\) at each fixed value of \(x\) as well as on a wider interval. To find better and better approximations of any function with a sufficient number of derivatives, we naturally want to find approximations of arbitrary degree \(n\). We thus define the Taylor polynomial of degree \(n\) centered at \(a = 0\).

By definition, \(T_n\) is the polynomial whose function value and first \(n\) derivative values at \(a = 0\) match the function value and all \(n\) derivative values of \(f\) at \(a = 0\).

Next, we evaluate each of the derivatives of \(T_n(x)\) at \(a = 0\) and set each result equal to the corresponding derivative value of \(f\) evaluated at \(a = 0\), which ultimately enables us to determine the coefficients \(c_0, c_1, \ldots, c_n\). These two steps are summarized in Table. Note how we use the index variable, \(k\), to track the various derivatives of \(T_n\) and \(f\).

\(k\)	\(T_n^{(k)}(0)\)	what follows from\(T_n^{(k)}(0) = f^{(k)}(0)\)
\(0\)	\(T_n(0) = c_0\)	\(c_0 = f(0)\)
\(1\)	\(T_n'(0) = c_1\)	\(c_1 = f'(0)\)
\(2\)	\(T_n''(0) = 2c_2\)	\(2c_2 = f''(0)\)
\(3\)	\(T'''(0) = (3 \cdot 2 \cdot 1) c_3\)	\((3 \cdot 2 \cdot 1) c_3 = f'''(0)\)
\(4\)	\(T^{(4)}(0) = (4 \cdot 3 \cdot 2 \cdot 1) c_4\)	\((4 \cdot 3 \cdot 2 \cdot 1) c_4 = f^{(4)}(0)\)
\(\vdots\)	\(\vdots\)	\(\vdots\)
\(n\)	\(T^{(n)}(0) = \bigl[ n (n-1) \cdots 2 \cdot 1 \bigr] c_n\)	\(\bigl[ n (n-1) \cdots 2 \cdot 1 \bigr] c_n = f^{(n)}(0)\)

Condensed — the full section is in Boelkins, Active Calculus.

Taylor polynomial approximations centered at an arbitrary value a

In all of our work so far in Chapter, we have focused on approximating functions such as \(e^x\), \(\sin(x)\), \(\ln(1+x)\), and \(\cos(x)\) near \(a = 0\). But we could instead be interested in the behavior of some function \(f\) near \(a = 5\), or be interested in a function \(f\) that wasn't even defined at \(a = 0\). Thus, we next generalize our earlier work to Taylor polynomial approximations centered at any value \(a\).

From our early studies in Section, we know that at any input value \(x = a\) where a function \(f\) has a first derivative, \(f\) has a tangent line approximation \[\begin{aligned}\end{aligned}\] that satisfies \(f(x) \approx L(x)\) for \(x\) values near \(a\). Provided that \(f\) has a second derivative at \(x=a\), we can build a quadratic approximation near \(a\) for \(f\), similar to the one we found at \(a = 0\) for \(f(x)=e^x\) in Activity. In addition, as long as \(f\) has a third derivative at \(x=a\), we can even find a cubic approximation (just as we did at \(a = 0\) in Activity), and so on.

In developing such approximations centered at any value \(x = a\), our guiding principle is the same as with our work at \(a = 0\): we'll require that at the input value \(a\), the original function's output and its derivatives' outputs match the corresponding approximation's output and derivatives' output.

Building on the form of the tangent line approximation, which we now denote \(T_1(x)\), \[\begin{aligned}\end{aligned}\], it is natural for us to consider quadratic and cubic approximations of form \[\begin{aligned}\end{aligned}\], a cubic approximation of form \[\begin{aligned}\end{aligned}\], and so on. We define the more general degree \(n\) Taylor Polynomial centered at \(a\) as follows.

Similar to the situation when \(a = 0\), it follows that we can find the coefficients \(c_k\) of the Taylor polynomial in terms of the various derivatives of \(f\) evaluated at \(a\).

If \(f\) is a function with at least \(n\) derivatives at \(a\), then the degree \(n\) Taylor polynomial of \(f\) centered at \(a\), \(T_n(x)\), is \[\begin{aligned}\end{aligned}\] where each coefficient \(c_k\) is given by \[\begin{aligned}\end{aligned}\].

Condensed — the full section is in Boelkins, Active Calculus.

Summary

Provided that a function \(f(x)\) has \(n\) derivatives at a selected input value \(x = a\), we can find a degree \(n\) polynomial \(T_n(x)\) that approximates \(f(x)\) near \(a\) by requiring that \(T_n(a) = f(a)\), \(T_n'(a) = f'(a)\), \(T_n''(a) = f''(a)\), \(\ldots\), \(T_n^{(n)}(a) = f^{(n)}(a)\).
When \(a = 0\), the degree \(n\) polynomial approximation, \(T_n(x)\), to a function \(f(x)\), centered at \(a = 0\), is a polynomial of the form \[\begin{aligned}\end{aligned}\] and it follows that the coefficients \(c_k\) are determined by the values of the various derivatives of \(f(x)\) evaluated at \(0\) according to the formula \[\begin{aligned}\end{aligned}\] Therefore, for \(x\) near \(a = 0\), \[\begin{aligned}\end{aligned}\].
Just as we can consider any function \(f\) that has \(n\) derivatives at \(a = 0\) and find approximations centered there, we can also consider any input value \(a\) at which those \(n\) derivatives exist, and find a polynomial approximation that satisfies \(T_n(a) = f(a)\), \(T_n'(a) = f'(a)\), \(T_n''(a) = f''(a)\), \(\ldots\), \(T_n^{(n)}(a) = f^{(n)}(a)\).
At such a value \(a\), the degree \(n\) Taylor polynomial of \(f\) centered at \(a\) has form \[\begin{aligned}\end{aligned}\] and it follows that the coefficients \(c_k\) are determined by the values of the various derivatives of \(f(x)\) evaluated at \(a\) according to the formula \[\begin{aligned}\end{aligned}\] Thus, for \(x\) near \(a\), \[\begin{aligned}\end{aligned}\].

Practice (8)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Suppose we know the following information about a function \(f\): \[\begin{aligned}\end{aligned}\].
1. Determine \(T_4(x)\), the degree \(4\) Taylor polynomial of \(f\) that is centered at \(a = 0\).
2. Use \(T_4(x)\) to estimate \(f(0.5)\).
3. State each of the Taylor polynomials \(T_3(x)\), \(T_2(x)\), and \(T_1(x)\).
Zbulo përgjigjen
1. Using the given information that \(f(0) = 2, f'(0) = -3, f''(0) = -1, f'''(0) = 0, f^{(4)}(0) = -3\) and the fact that the coeffficients of the Taylor polynomial centered at \(a = 0\) are \(c_k = \frac{f^{k}(0)}{k!}\), we find that \[\begin{aligned}\end{aligned}\].
2. Since \(0.5\) is near \(0\), we have \[\begin{aligned}\end{aligned}\].
3. Since each lower degree Taylor polynomial is embedded within a higher degree one, now that we know \(T_4(x)\), it follows that \[\begin{aligned}T_3(x) =\mathstrut \amp 2 - 3x - \frac{1}{2!}x^2 + 0x^3 \\ T_2(x) =\mathstrut \amp 2 - 3x - \frac{1}{2!}x^2 \\ T_1(x) =\mathstrut \amp 2 - 3x\end{aligned}\]. We note that \(T_3(x) = T_2(x)\) since \(f'''(0) = 0\).
In an earlier exercise, we found that the degree 2 Taylor polynomial centered at \(a = 0\) of a quadratic function is the quadratic function itself. In this exercise, we explore how changing the center of the approximation offers additional insight into the function.
Let \(f(x) = \frac{1}{2}x^2 - 2x + 5\), and let \(a = 2\) be the center at which we will find a degree \(2\) Taylor polynomial approximation of \(f\).
1. By finding \(f'(x)\), \(f''(x)\), \(f(2)\), \(f'(2)\), and \(f''(2)\), determine \(T_2(x)\), the degree \(2\) Taylor polynomial approximation of \(f\) that is centered at \(a = 2\).
2. Plot both \(f(x)\) and \(T_2(x)\) on the same axes. What do you observe?
3. What does the algebraic form of \(T_2(x)\) tell you about the original function \(f(x)\)?
4. What is the tangent line approximation to \(f(x)\) at \(a = 2\)? What is special about the function's behavior at this input value?
Zbulo përgjigjen
1. Computing the various derivatives of \(f\) and evaluating them at \(a = 2\), we see the results in Table.
  \(f(x)=\) \(\frac{1}{2}x^2 - 2x + 5\) \(f(2)=\) \(3\)
  \(f'(x)=\) \(x - 2\) \(f'(2)=\) \(0\)
  \(f''(x)=\) \(1\) \(f''(2)=\) \(1\)
  Using the fact that \(c_k = \frac{f^{k}(2)}{k!}\), we find that \[\begin{aligned}\end{aligned}\].
2. When we plot both \(f(x)\) and \(T_2(x)\) on the same axes we observe that they are the same function: \(f(x) = T_2(x)\).
3. Because \(T_2(x) = 3 + \frac{1}{2}(x-2)^2\) is a quadratic function in vertex form, we see that \((2,3)\) is the vertex of the function \(f(x)\).
4. The tangent line approximation to \(f(x)\) at \(a = 2\) is \(L(x) = T_1(x) = 3\), which indicates that \(f\) has a horizontal tangent line at \(x = 2\). This corresponds to the fact that \(f\) has its global minimum at \(x = 2\).

Recall that we found in Preview and subsequent work that \(\sin(x) \approx x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 - \frac{1}{7!}x^7\), which is the degree \(7\) Taylor approximation centered at \(0\). And in Activity, we found that the degree \(6\) Taylor approximation centered at \(a = 0\) for \(\cos(x)\) is \(\cos(x) \approx 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6\).

In this exercise, we investigate Taylor polynomial approximations of \(f(x) = \sin(x)\) centered at \(a = \frac{\pi}{2}\).

By finding the appropriate derivatives of \(f(x) = \sin(x)\) and evaluating them at \(a = \frac{\pi}{2}\), determine the degree \(6\) Taylor polynomial approximation of \(\sin(x)\) centered at \(a = \frac{\pi}{2}\).
How is your result in (a) similar to the degree \(6\) Taylor polynomial of \(\cos(x)\) that is centered at \(a = 0\)?
Recall the trigonometric identity that states \(\sin(x) = \cos(x - \frac{\pi}{2})\). How does this identity help explain what you found in (a) and (b)?

Zbulo përgjigjen

Finding the first \(6\) derivatives of \(f(x) = \sin(x)\) and evaluating them at \(a = \frac{\pi}{2}\), we see the results in Table.

\(f(x) =\)	\(\sin(x)\)	\(f(\frac{pi}{2}) =\)	\(\sin(\frac{pi}{2}) = 1\)
\(f'(x) =\)	\(\cos(x)\)	\(f'(\frac{pi}{2}) =\)	\(0\)
\(f''(x) =\)	\(-\sin(x)\)	\(f''(\frac{pi}{2})) =\)	\(-\sin(\frac{pi}{2}) = -1\)
\(f'''(x) =\)	\(-\cos(x)\)	\(f'''(\frac{pi}{2})) =\)	\(0\)
\(f^{(4)}(x) =\)	\(\sin(x)\)	\(f^{(4)}(\frac{pi}{2}) =\)	\(\sin(\frac{pi}{2}) = 1\)
\(f^{(5)}(x) =\)	\(\cos(x)\)	\(f^{(5)}(\frac{pi}{2}) =\)	\(0\)
\(f^{(6)}(x) =\)	\(-\sin(x)\)	\(f^{(6)}(\frac{pi}{2})) =\)	\(-\sin(\frac{pi}{2}) = -1\)

Using the fact that \(c_k = \frac{f^{k}(\frac{pi}{2})}{k!}\), we find that the degree \(6\) Taylor polynomial approximation of \(\sin(x)\) centered at \(a = \frac{\pi}{2}\) is \[\begin{aligned}\end{aligned}\]

Earlier, we found that the degree \(6\) Taylor polynomial of \(\cos(x)\) centered at \(a = 0\) is given by \[\begin{aligned}\end{aligned}\], so the degree \(6\) Taylor polynomial of \(\sin(x)\) centered at \(a = \frac{\pi}{2}\), \(T_6(x)\) appears to be the same as \(P_6(x-\frac{\pi}{2})\), the degree \(6\) Taylor polynomial of \(\cos(x)\) centered at \(a = 0\), shifted \(\frac{\pi}{2}\) units to the right.
Since \(\sin(x) = \cos(x - \frac{\pi}{2})\), this tells us that the sine function is simply a shifted version of the cosine function (the cosine function shifted \(\frac{\pi}{2}\) units to the right), so it makes sense that their Taylor polynomials align as we found in (c), provided that we center the sine function's approximations at \(a = \frac{\pi}{2}\).

In Example, we found that \[\begin{aligned}\end{aligned}\], where \(T_5\) is the degree \(5\) Taylor approximation of \(\ln(1+x)\) centered at \(a = 0\).
In Activity, we found that \[\begin{aligned}\end{aligned}\], where \(P_5\) is the degree \(5\) Taylor approximation of \(\ln(x)\) centered at \(a = 1\). (Here we are using \(T_5\) and \(P_5\) to distinguish between these two degree \(5\) polynomial approximations of the two different functions \(f(x) = \ln(1+x)\) and \(g(x) = \ln(x)\), centered at two different values.)
1. Note that \(f(0.5) = \ln(1.5)\). Use \(T_5(x)\) appropriately to find an estimate of \(\ln(1.5)\).
2. Observe that \(g(1.5) = \ln(1.5)\). Use \(P_5(x)\) to estimate \(\ln(1.5)\).
3. Are the estimates of \(\ln(1.5)\) generated by \(T_5(x)\) and \(P_5(x)\) the same or different? Why do you think this happens?
Zbulo përgjigjen
1. We have that \[\begin{aligned}\end{aligned}\].
2. Here, we find that \[\begin{aligned}\end{aligned}\]. Since \((1.5 - 1) = 0.5\), we see that \(P_5(1.5) = T_5(0.5) = 0.4072916\overline{6}\).
3. In (b), we just observed that the estimates of \(\ln(1.5)\) generated by \(T_5(x)\) and \(P_5(x)\) are the same. This occurs because \(f(x) = \ln(1+x)\) is just a shifted version of \(g(x) = \ln(x)\) (\(f(x) = g(1+x)\), so \(f\) is \(g\) shifted 1 unit left), so by changing where their respective Taylor polynomials are centered, we are using the same structure to generate each approximation.
Suppose we know the following information about a function \(f\): \[\begin{aligned}\end{aligned}\].
1. Determine \(T_4(x)\), the degree \(4\) Taylor polynomial of \(f\) that is centered at \(a = 0\).
2. Use \(T_4(x)\) to estimate \(f(0.5)\).
3. State each of the Taylor polynomials \(T_3(x)\), \(T_2(x)\), and \(T_1(x)\).
Zbulo përgjigjen
1. Using the given information that \(f(0) = 2, f'(0) = -3, f''(0) = -1, f'''(0) = 0, f^{(4)}(0) = -3\) and the fact that the coeffficients of the Taylor polynomial centered at \(a = 0\) are \(c_k = \frac{f^{k}(0)}{k!}\), we find that \[\begin{aligned}\end{aligned}\].
2. Since \(0.5\) is near \(0\), we have \[\begin{aligned}\end{aligned}\].
3. Since each lower degree Taylor polynomial is embedded within a higher degree one, now that we know \(T_4(x)\), it follows that \[\begin{aligned}T_3(x) =\mathstrut \amp 2 - 3x - \frac{1}{2!}x^2 + 0x^3 \\ T_2(x) =\mathstrut \amp 2 - 3x - \frac{1}{2!}x^2 \\ T_1(x) =\mathstrut \amp 2 - 3x\end{aligned}\]. We note that \(T_3(x) = T_2(x)\) since \(f'''(0) = 0\).
In an earlier exercise, we found that the degree 2 Taylor polynomial centered at \(a = 0\) of a quadratic function is the quadratic function itself. In this exercise, we explore how changing the center of the approximation offers additional insight into the function.
Let \(f(x) = \frac{1}{2}x^2 - 2x + 5\), and let \(a = 2\) be the center at which we will find a degree \(2\) Taylor polynomial approximation of \(f\).
1. By finding \(f'(x)\), \(f''(x)\), \(f(2)\), \(f'(2)\), and \(f''(2)\), determine \(T_2(x)\), the degree \(2\) Taylor polynomial approximation of \(f\) that is centered at \(a = 2\).
2. Plot both \(f(x)\) and \(T_2(x)\) on the same axes. What do you observe?
3. What does the algebraic form of \(T_2(x)\) tell you about the original function \(f(x)\)?
4. What is the tangent line approximation to \(f(x)\) at \(a = 2\)? What is special about the function's behavior at this input value?
Zbulo përgjigjen
1. Computing the various derivatives of \(f\) and evaluating them at \(a = 2\), we see the results in Table.
  \(f(x)=\) \(\frac{1}{2}x^2 - 2x + 5\) \(f(2)=\) \(3\)
  \(f'(x)=\) \(x - 2\) \(f'(2)=\) \(0\)
  \(f''(x)=\) \(1\) \(f''(2)=\) \(1\)
  Using the fact that \(c_k = \frac{f^{k}(2)}{k!}\), we find that \[\begin{aligned}\end{aligned}\].
2. When we plot both \(f(x)\) and \(T_2(x)\) on the same axes we observe that they are the same function: \(f(x) = T_2(x)\).
3. Because \(T_2(x) = 3 + \frac{1}{2}(x-2)^2\) is a quadratic function in vertex form, we see that \((2,3)\) is the vertex of the function \(f(x)\).
4. The tangent line approximation to \(f(x)\) at \(a = 2\) is \(L(x) = T_1(x) = 3\), which indicates that \(f\) has a horizontal tangent line at \(x = 2\). This corresponds to the fact that \(f\) has its global minimum at \(x = 2\).

In this exercise, we investigate Taylor polynomial approximations of \(f(x) = \sin(x)\) centered at \(a = \frac{\pi}{2}\).

By finding the appropriate derivatives of \(f(x) = \sin(x)\) and evaluating them at \(a = \frac{\pi}{2}\), determine the degree \(6\) Taylor polynomial approximation of \(\sin(x)\) centered at \(a = \frac{\pi}{2}\).
How is your result in (a) similar to the degree \(6\) Taylor polynomial of \(\cos(x)\) that is centered at \(a = 0\)?
Recall the trigonometric identity that states \(\sin(x) = \cos(x - \frac{\pi}{2})\). How does this identity help explain what you found in (a) and (b)?

Zbulo përgjigjen

Finding the first \(6\) derivatives of \(f(x) = \sin(x)\) and evaluating them at \(a = \frac{\pi}{2}\), we see the results in Table.

\(f(x) =\)	\(\sin(x)\)	\(f(\frac{pi}{2}) =\)	\(\sin(\frac{pi}{2}) = 1\)
\(f'(x) =\)	\(\cos(x)\)	\(f'(\frac{pi}{2}) =\)	\(0\)
\(f''(x) =\)	\(-\sin(x)\)	\(f''(\frac{pi}{2})) =\)	\(-\sin(\frac{pi}{2}) = -1\)
\(f'''(x) =\)	\(-\cos(x)\)	\(f'''(\frac{pi}{2})) =\)	\(0\)
\(f^{(4)}(x) =\)	\(\sin(x)\)	\(f^{(4)}(\frac{pi}{2}) =\)	\(\sin(\frac{pi}{2}) = 1\)
\(f^{(5)}(x) =\)	\(\cos(x)\)	\(f^{(5)}(\frac{pi}{2}) =\)	\(0\)
\(f^{(6)}(x) =\)	\(-\sin(x)\)	\(f^{(6)}(\frac{pi}{2})) =\)	\(-\sin(\frac{pi}{2}) = -1\)

Earlier, we found that the degree \(6\) Taylor polynomial of \(\cos(x)\) centered at \(a = 0\) is given by \[\begin{aligned}\end{aligned}\], so the degree \(6\) Taylor polynomial of \(\sin(x)\) centered at \(a = \frac{\pi}{2}\), \(T_6(x)\) appears to be the same as \(P_6(x-\frac{\pi}{2})\), the degree \(6\) Taylor polynomial of \(\cos(x)\) centered at \(a = 0\), shifted \(\frac{\pi}{2}\) units to the right.
Since \(\sin(x) = \cos(x - \frac{\pi}{2})\), this tells us that the sine function is simply a shifted version of the cosine function (the cosine function shifted \(\frac{\pi}{2}\) units to the right), so it makes sense that their Taylor polynomials align as we found in (c), provided that we center the sine function's approximations at \(a = \frac{\pi}{2}\).

In Example, we found that \[\begin{aligned}\end{aligned}\], where \(T_5\) is the degree \(5\) Taylor approximation of \(\ln(1+x)\) centered at \(a = 0\).
In Activity, we found that \[\begin{aligned}\end{aligned}\], where \(P_5\) is the degree \(5\) Taylor approximation of \(\ln(x)\) centered at \(a = 1\). (Here we are using \(T_5\) and \(P_5\) to distinguish between these two degree \(5\) polynomial approximations of the two different functions \(f(x) = \ln(1+x)\) and \(g(x) = \ln(x)\), centered at two different values.)
1. Note that \(f(0.5) = \ln(1.5)\). Use \(T_5(x)\) appropriately to find an estimate of \(\ln(1.5)\).
2. Observe that \(g(1.5) = \ln(1.5)\). Use \(P_5(x)\) to estimate \(\ln(1.5)\).
3. Are the estimates of \(\ln(1.5)\) generated by \(T_5(x)\) and \(P_5(x)\) the same or different? Why do you think this happens?
Zbulo përgjigjen
1. We have that \[\begin{aligned}\end{aligned}\].
2. Here, we find that \[\begin{aligned}\end{aligned}\]. Since \((1.5 - 1) = 0.5\), we see that \(P_5(1.5) = T_5(0.5) = 0.4072916\overline{6}\).
3. In (b), we just observed that the estimates of \(\ln(1.5)\) generated by \(T_5(x)\) and \(P_5(x)\) are the same. This occurs because \(f(x) = \ln(1+x)\) is just a shifted version of \(g(x) = \ln(x)\) (\(f(x) = g(1+x)\), so \(f\) is \(g\) shifted 1 unit left), so by changing where their respective Taylor polynomials are centered, we are using the same structure to generate each approximation.

Symbols used here

factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\approx

approximately equal
Equal to the precision shown, not exactly.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Taylor polynomials

For functions such as \sin(x), \cos(x), and \ln(1+x) that, like f(x) = e^x, have n derivatives at a = 0 for any choice of n, how can we use patterns in those derivatives to find a general formula for the degree n polynomial approximation to each?
How are the coefficients of the polynomial approximation to a function f(x) near a = 0 completely determined by the values of the various derivatives of f, evaluated at a = 0?
How do the polynomial approximations to a given function f change when we center the approximation at a point other than a = 0?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

Provo timen.

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

Më shumë në Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests

\(f(x)=\)	\(\frac{1}{2}x^2 - 2x + 5\)	\(f(2)=\)	\(3\)
\(f'(x)=\)	\(x - 2\)	\(f'(2)=\)	\(0\)
\(f''(x)=\)	\(1\)	\(f''(2)=\)	\(1\)