maths.free › Calculus › 6. Power Series › Properties of Power Series

Properties of Power Series

Combine power series by addition or subtraction.

Combining Power Series

If we have two power series with the same interval of convergence, we can add or subtract the two series to create a new power series, also with the same interval of convergence. Similarly, we can multiply a power series by a power of x or evaluate a power series at ${x}^{m}$ for a positive integer m to create a new power series. Being able to do this allows us to find power series representations for certain functions by using power series representations of other functions. For example, since we know the power series representation for $f(x)=\frac{1}{1-x},$ we can find power series representations for related functions, such as

\[y=\frac{3x}{1-{x}^{2}}\ \text{and}\ y=\frac{1}{(x-1)(x-3)}.\]

In we state results regarding addition or subtraction of power series, composition of a power series, and multiplication of a power series by a power of the variable. For simplicity, we state the theorem for power series centered at $x=0.$ Similar results hold for power series centered at $x=a.$

Condensed — the full section is in OpenStax Calculus Volume 2.

Multiplication of Power Series

We can also create new power series by multiplying power series. Being able to multiply two power series provides another way of finding power series representations for functions.

The way we multiply them is similar to how we multiply polynomials. For example, suppose we want to multiply

\[\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\text{\cdots }\]

and

\[\sum _{n=0}^{\infty }{d}_{n}{x}^{n}={d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\text{\cdots }.\]

It appears that the product should satisfy

\[\begin{array}{ll}(\sum _{n=0}^{\infty }{c}_{n}{x}^{n})(\sum _{n=-0}^{\infty }{d}_{n}{x}^{n}) & =({c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\text{\cdots })\cdot ({d}_{0}+{d}_{1}x+{d}_{2}{x}^{2}+\text{\cdots }) \\ & ={c}_{0}{d}_{0}+({c}_{1}{d}_{0}+{c}_{0}{d}_{1})\ x+({c}_{2}{d}_{0}+{c}_{1}{d}_{1}+{c}_{0}{d}_{2})\ {x}^{2}+\text{\cdots }.\end{array}\]

In , we state the main result regarding multiplying power series, showing that if $\sum _{n=0}^{\infty }{c}_{n}{x}^{n}$ and $\sum _{n=0}^{\infty }{d}_{n}{x}^{n}$ converge on a common interval I, then we can multiply the series in this way, and the resulting series also converges on the interval I.

We omit the proof of this theorem, as it is beyond the level of this text and is typically covered in a more advanced course. We now provide an example of this theorem by finding the power series representation for

\[f(x)=\frac{1}{(1-x)(1-{x}^{2})}\]

using the power series representations for

\[y=\frac{1}{1-x}\ \text{and}\ y=\frac{1}{1-{x}^{2}}.\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Differentiating and Integrating Power Series

Consider a power series $\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\text{\cdots }$ that converges on some interval I, and let $f$ be the function defined by this series. Here we address two questions about $f.$

Is $f$ differentiable, and if so, how do we determine the derivative ${f}^{'}?$
How do we evaluate the indefinite integral $\int f(x)\ dx?$

We know that, for a polynomial with a finite number of terms, we can evaluate the derivative by differentiating each term separately. Similarly, we can evaluate the indefinite integral by integrating each term separately. Here we show that we can do the same thing for convergent power series. That is, if

\[f(x)=\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\text{\cdots }\]

converges on some interval I, then

\[{f}^{'}(x)={c}_{1}+2{c}_{2}x+3{c}_{3}{x}^{2}+\text{\cdots }\]

and

\[\int f(x)\ dx=C+{c}_{0}x+{c}_{1}\frac{{x}^{2}}{2}+{c}_{2}\frac{{x}^{3}}{3}+\text{\cdots }\]

converges on I. As noted below, behavior at the endpoints of the interval must be investigated individually.

Evaluating the derivative and indefinite integral in this way is called term-by-term differentiation of a power series and term-by-term integration of a power series, respectively. The ability to differentiate and integrate power series term-by-term also allows us to use known power series representations to find power series representations for other functions. For example, given the power series for $f(x)=\frac{1}{1-x},$ we can differentiate term-by-term to find the power series for ${f}^{'}(x)=\frac{1}{{(1-x)}^{2}}.$ Similarly, using the power series for $g(x)=\frac{1}{1+x},$ we can integrate term-by-term to find the power series for $G(x)=\text{ln}\ (1+x),$ an antiderivative of g. We show how to do this in and . First, we state , which provides the main result regarding differentiation and integration of power series.

The proof of this result is beyond the scope of the text and is omitted. Note that although guarantees the same radius of convergence when a power series is differentiated or integrated term-by-term, it says nothing about what happens at the endpoints. It is possible that the differentiated and integrated power series have different behavior at the endpoints than does the original series. We see this behavior in the next examples.

Condensed — the full section is in OpenStax Calculus Volume 2.

Key Concepts

Given two power series $\sum _{n=0}^{\infty }{c}_{n}{x}^{n}$ and $\sum _{n=0}^{\infty }{d}_{n}{x}^{n}$ that converge to functions f and g on a common interval I, the sum and difference of the two series converge to $f\pm g,$ respectively, on I. In addition, for any real number b and integer $m\ge 0,$ the series $\sum _{n=0}^{\infty }b{x}^{m}{c}_{n}{x}^{n}$ converges to $b{x}^{m}f(x)$ and the series $\sum _{n=0}^{\infty }{c}_{n}{(b{x}^{m})}^{n}$ converges to $f(b{x}^{m})$ whenever bx^m is in the interval I.
Given two power series that converge on an interval $(\text{-}R,R),$ the Cauchy product of the two power series converges on the interval $(\text{-}R,R).$
Given a power series that converges to a function f on an interval $(\text{-}R,R),$ the series can be differentiated term-by-term and the resulting series converges to ${f}^{'}$ on $(\text{-}R,R).$ The series can also be integrated term-by-term and the resulting series converges to $\int f(x)\ dx$ on $(\text{-}R,R).$

Properties of Power Series

In the following exercises, use partial fractions to find the power series of each function.

In the following exercises, express each series as a rational function.

The following exercises explore applications of annuities.

In the following exercises, express the sum of each power series in terms of geometric series, and then express the sum as a rational function.

In the following exercises, find the power series of $f(x)g(x)$ given f and g as defined.

In the following exercises, differentiate the given series expansion of f term-by-term to obtain the corresponding series expansion for the derivative of f.

In the following exercises, integrate the given series expansion of $f$ term-by-term from zero to x to obtain the corresponding series expansion for the indefinite integral of $f.$

Condensed — the full section is in OpenStax Calculus Volume 2.

Practice (40)

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

Suppose that $\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ is a power series whose interval of convergence is $(-1,1),$ and suppose that $\sum _{n=0}^{\infty }{b}_{n}{x}^{n}$ is a power series whose interval of convergence is $(-2,2).$
1. Find the interval of convergence of the series $\sum _{n=0}^{\infty }({a}_{n}{x}^{n}+{b}_{n}{x}^{n}).$
2. Find the interval of convergence of the series $\sum _{n=0}^{\infty }{a}_{n}{3}^{n}{x}^{n}.$
Откриј одговор.
1. Since the interval $(-1,1)$ is a common interval of convergence of the series $\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ and $\sum _{n=0}^{\infty }{b}_{n}{x}^{n},$ the interval of convergence of the series $\sum _{n=0}^{\infty }({a}_{n}{x}^{n}+{b}_{n}{x}^{n})$ is $(-1,1).$
2. Since $\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ is a power series centered at zero with radius of convergence 1, it converges for all x in the interval $(-1,1).$ By , the series
  \[\sum _{n=0}^{\infty }{a}_{n}{3}^{n}{x}^{n}=\sum _{n=0}^{\infty }{a}_{n}{(3x)}^{n}\]
  converges if 3x is in the interval $(-1,1).$ Therefore, the series converges for all x in the interval $(-\frac{1}{3},\frac{1}{3}).$
Suppose that $\sum _{n=0}^{\infty }{a}_{n}{x}^{n}$ has an interval of convergence of $(-1,1).$ Find the interval of convergence of $\sum _{n=0}^{\infty }{a}_{n}{(\frac{x}{2})}^{n}.$

Откриј одговор.

Interval of convergence is $(-2,2).$
Use the power series representation for $f(x)=\frac{1}{1-x}$ combined with to construct a power series for each of the following functions. Find the interval of convergence of the power series.
1. $f(x)=\frac{3x}{1+{x}^{2}}$
2. $f(x)=\frac{1}{(x-1)(x-3)}$
Откриј одговор.
1. First write $f(x)$ as
  \[f(x)=3x(\frac{1}{1-(\text{-}{x}^{2})}).\]
  Using the power series representation for $f(x)=\frac{1}{1-x}$ and parts ii. and iii. of , we find that a power series representation for f is given by
  \[\sum _{n=0}^{\infty }3x{(\text{-}{x}^{2})}^{n}=\sum _{n=0}^{\infty }3{(-1)}^{n}{x}^{2n+1}.\]
  Since the interval of convergence of the series for $\frac{1}{1-x}$ is $(-1,1),$ the interval of convergence for this new series is the set of real numbers x such that $|{x}^{2}|<1.$ Therefore, the interval of convergence is $(-1,1).$
2. To find the power series representation, use partial fractions to write $f(x)=\frac{1}{(x-1)(x-3)}$ as the sum of two fractions. We have
  \[\begin{array}{ll}\frac{1}{(x-1)(x-3)} & =\frac{\text{-}1\text{/}2}{x-1}+\frac{1\text{/}2}{x-3} \\ & =\frac{1\text{/}2}{1-x}-\frac{1\text{/}2}{3-x} \\ & =\frac{1\text{/}2}{1-x}-\frac{1\text{/}6}{1-\frac{x}{3}}.\end{array}\]
  First, using part ii. of , we obtain
  \[\frac{1\text{/}2}{1-x}=\sum _{n=0}^{\infty }\frac{1}{2}{x}^{n}\ \text{for}\ |x|<1.\]
  Then, using parts ii. and iii. of , we have
  \[\frac{1\text{/}6}{1-x\text{/}3}=\sum _{n=0}^{\infty }\frac{1}{6}{(\frac{x}{3})}^{n}\ \text{for}\ |x|<3.\]
  Since we are combining these two power series, the interval of convergence of the difference must be the smaller of these two intervals. Using this fact and part i. of , we have
  \[\frac{1}{(x-1)(x-3)}=\sum _{n=0}^{\infty }(\frac{1}{2}-\frac{1}{6\cdot {3}^{n}}){x}^{n}\]
  where the interval of convergence is $(-1,1).$
Use the series for $f(x)=\frac{1}{1-x}$ on $|x|<1$ to construct a series for $\frac{1}{(1-x)(x-2)}.$ Determine the interval of convergence.

Откриј одговор.

$\sum _{n=0}^{\infty }(-1+\frac{1}{{2}^{n+1}}){x}^{n}.$ The interval of convergence is $(-1,1).$
Consider the power series $\sum _{n=0}^{\infty }{2}^{n}{x}^{n}.$ Find the function f represented by this series. Determine the interval of convergence of the series.

Откриј одговор.

Writing the given series as
\[\sum _{n=0}^{\infty }{2}^{n}{x}^{n}=\sum _{n=0}^{\infty }{(2x)}^{n},\]
we can recognize this series as the power series for
\[f(x)=\frac{1}{1-2x}.\]
Since this is a geometric series, the series converges if and only if $|2x|<1.$ Therefore, the interval of convergence is $(-\frac{1}{2},\frac{1}{2}).$
Find the function represented by the power series $\sum _{n=0}^{\infty }\frac{1}{{3}^{n}}{x}^{n}.$ Determine its interval of convergence.

Откриј одговор.

$f(x)=\frac{3}{3-x}.$ The interval of convergence is $(-3,3).$
Suppose you win the lottery and are given the following three options: (1) Receive 20 million dollars today; (2) receive 1.5 million dollars per year over the next 20 years; or (3) receive 1 million dollars per year indefinitely (being passed on to your heirs). Which is the best deal, assuming that the annual interest rate is 5%? We answer this by working through the following sequence of questions.
1. How much is the 1.5 million dollars received annually over the course of 20 years worth in terms of today’s dollars, assuming an annual interest rate of 5%?
2. Use the answer to part a. to find a general formula for the present value of payments of C dollars received each year over the next n years, assuming an average annual interest rate r.
3. Find a formula for the present value if annual payments of C dollars continue indefinitely, assuming an average annual interest rate r.
4. Use the answer to part c. to determine the present value of 1 million dollars paid annually indefinitely.
5. Use your answers to parts a. and d. to determine which of the three options is best.
Откриј одговор.
1. Consider the payment of 1.5 million dollars made at the end of the first year. If you were able to receive that payment today instead of one year from now, you could invest that money and earn 5% interest. Therefore, the present value of that money P₁ satisfies ${P}_{1}(1+0.05)=1.5\ \text{million dollars}.$ We conclude that
  \[{P}_{1}=\frac{1.5}{1.05}=\text{\$}1.429\ \text{million dollars}\text{.}\]
  Similarly, consider the payment of 1.5 million dollars made at the end of the second year. If you were able to receive that payment today, you could invest that money for two years, earning 5% interest, compounded annually. Therefore, the present value of that money P₂ satisfies ${P}_{2}{(1+0.05)}^{2}=1.5\ \text{million dollars}.$ We conclude that
  \[{P}_{2}=\frac{1.5}{{(1.05)}^{2}}=\text{\$}1.361\ \text{million dollars}\text{.}\]
  The value of the future payments today is the sum of the present values ${P}_{1},{P}_{2},\text{\ldots },{P}_{20}$ of each of those annual payments. The present value P_k satisfies
  \[{P}_{k}=\frac{1.5}{{(1.05)}^{k}}.\]
  Therefore,
  \[\begin{array}{ll}P & =\frac{1.5}{1.05}+\frac{1.5}{{(1.05)}^{2}}+\text{\cdots }+\frac{1.5}{{(1.05)}^{20}} \\ & =\text{\$}18.693\ \text{million dollars}\text{.}\end{array}\]
2. Using the result from part a. we see that the present value P of C dollars paid annually over the course of n years, assuming an annual interest rate r, is given by
  \[P=\frac{C}{1+r}+\frac{C}{{(1+r)}^{2}}+\text{\cdots }+\frac{C}{{(1+r)}^{n}}\ \text{dollars}\text{.}\]
3. Using the result from part b. we see that the present value of an annuity that continues indefinitely is given by the infinite series
  \[P=\sum _{n=0}^{\infty }\frac{C}{{(1+r)}^{n+1}}.\]
  We can view the present value as a power series in r, which converges as long as $|\frac{1}{1+r}|<1.$ Since $r>0,$ this series converges. Rewriting the series as
  \[P=\frac{C}{(1+r)}\sum _{n=0}^{\infty }{(\frac{1}{1+r})}^{n},\]
  we recognize this series as the power series for
  \[f(r)=\frac{1}{1-(\frac{1}{1+r})}=\frac{1}{(\frac{r}{1+r})}=\frac{1+r}{r}.\]
  We conclude that the present value of this annuity is
  \[P=\frac{C}{1+r}\cdot \frac{1+r}{r}=\frac{C}{r}.\]
4. From the result to part c. we conclude that the present value P of $C=1\ \text{million dollars}$ paid out every year indefinitely, assuming an annual interest rate $r=0.05,$ is given by
  \[P=\frac{1}{0.05}=20\ \text{million dollars}\text{.}\]
5. From part a. we see that receiving $1.5 million dollars over the course of 20 years is worth $18.693 million dollars in today’s dollars. From part d. we see that receiving $1 million dollars per year indefinitely is worth $20 million dollars in today’s dollars. Therefore, either receiving a lump-sum payment of $20 million dollars today or receiving $1 million dollars indefinitely have the same present value.
Multiply the power series representation
\[\begin{array}{ll}\frac{1}{1-x} & =\sum _{n=0}^{\infty }{x}^{n} \\ & =1+x+{x}^{2}+{x}^{3}+\text{\cdots }\end{array}\]
for $|x|<1$ with the power series representation
\[\begin{array}{ll}\frac{1}{1-{x}^{2}} & =\sum _{n=0}^{\infty }{({x}^{2})}^{n} \\ & =1+{x}^{2}+{x}^{4}+{x}^{6}+\text{\cdots }\end{array}\]
for $|x|<1$ to construct a power series for $f(x)=\frac{1}{(1-x)(1-{x}^{2})}$ on the interval $(-1,1).$

Откриј одговор.

We need to multiply
\[(1+x+{x}^{2}+{x}^{3}+\text{\cdots })(1+{x}^{2}+{x}^{4}+{x}^{6}+\text{\cdots }).\]
Writing out the first several terms, we see that the product is given by
\[\begin{array}{l}(1+{x}^{2}+{x}^{4}+{x}^{6}+\text{\cdots })+(x+{x}^{3}+{x}^{5}+{x}^{7}+\text{\cdots })+({x}^{2}+{x}^{4}+{x}^{6}+{x}^{8}+\text{\cdots })+({x}^{3}+{x}^{5}+{x}^{7}+{x}^{9}+\text{\cdots }) \\ =1+x+(1+1){x}^{2}+(1+1){x}^{3}+(1+1+1){x}^{4}+(1+1+1){x}^{5}+\text{\cdots } \\ =1+x+2{x}^{2}+2{x}^{3}+3{x}^{4}+3{x}^{5}+\text{\cdots }.\end{array}\]
Since the series for $y=\frac{1}{1-x}$ and $y=\frac{1}{1-{x}^{2}}$ both converge on the interval $(-1,1),$ the series for the product also converges on the interval $(-1,1).$
Multiply the series $\frac{1}{1-x}=\sum _{n=0}^{\infty }{x}^{n}$ by itself to construct a series for $\frac{1}{(1-x)(1-x)}.$

Откриј одговор.

$1+2x+3{x}^{2}+4{x}^{3}+\text{\cdots }$
1. Use the power series representation
  \[\begin{array}{ll}f(x) & =\frac{1}{1-x} \\ & =\sum _{n=0}^{\infty }{x}^{n} \\ & =1+x+{x}^{2}+{x}^{3}+\text{\cdots }\end{array}\]
  for $|x|<1$ to find a power series representation for
  \[g(x)=\frac{1}{{(1-x)}^{2}}\]
  on the interval $(-1,1).$ Determine whether the resulting series converges at the endpoints.
2. Use the result of part a. to evaluate the sum of the series $\sum _{n=0}^{\infty }\frac{n+1}{{4}^{n}}.$
Откриј одговор.
1. Since $g(x)=\frac{1}{{(1-x)}^{2}}$ is the derivative of $f(x)=\frac{1}{1-x},$ we can find a power series representation for g by differentiating the power series for f term-by-term. The result is
  \[\begin{array}{ll}g(x) & =\frac{1}{{(1-x)}^{2}} \\ & =\frac{d}{dx}(\frac{1}{1-x}) \\ & =\sum _{n=0}^{\infty }\frac{d}{dx}({x}^{n}) \\ & =\frac{d}{dx}(1+x+{x}^{2}+{x}^{3}+\text{\cdots }) \\ & =0+1+2x+3{x}^{2}+4{x}^{3}+\text{\cdots } \\ & =\sum _{n=0}^{\infty }(n+1){x}^{n}\end{array}\]
  for $|x|<1.$ does not guarantee anything about the behavior of this series at the endpoints. Testing the endpoints by using the divergence test, we find that the series diverges at both endpoints $x=\text{\pm }1.$ Note that this is the same result found in .
2. From part a. we know that
  \[\sum _{n=0}^{\infty }(n+1){x}^{n}=\frac{1}{{(1-x)}^{2}}.\]
  Therefore,
  \[\begin{array}{ll}\sum _{n=0}^{\infty }\frac{n+1}{{4}^{n}} & =\sum _{n=0}^{\infty }(n+1){(\frac{1}{4})}^{n} \\ & =\frac{1}{{(1-\frac{1}{4})}^{2}} \\ & =\frac{1}{{(\frac{3}{4})}^{2}} \\ & =\frac{16}{9}.\end{array}\]
Differentiate the series $\frac{1}{{(1-x)}^{2}}=\sum _{n=0}^{\infty }(n+1){x}^{n}$ term-by-term to find a power series representation for $\frac{2}{{(1-x)}^{3}}$ on the interval $(-1,1).$

Откриј одговор.

$\sum _{n=0}^{\infty }(n+2)(n+1){x}^{n}$
For each of the following functions f, find a power series representation for f by integrating the power series for ${f}^{'}$ and find its interval of convergence.
1. $f(x)=\text{ln}\ (1+x)$
2. $f(x)={\text{tan}}^{-1}x$
Откриј одговор.
1. For $f(x)=\text{ln}\ (1+x),$ the derivative is ${f}^{'}(x)=\frac{1}{1+x}.$ We know that
  \[\begin{array}{ll}\frac{1}{1+x} & =\frac{1}{1-(\text{-}x)} \\ & =\sum _{n=0}^{\infty }{(\text{-}x)}^{n} \\ & =1-x+{x}^{2}-{x}^{3}+\text{\cdots }\end{array}\]
  for $|x|<1.$ To find a power series for $f(x)=\text{ln}\ (1+x),$ we integrate the series term-by-term.
  \[\begin{array}{ll}\int {f}^{'}(x)\ dx & =\int (1-x+{x}^{2}-{x}^{3}+\text{\cdots })\ dx \\ & =C+x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-\frac{{x}^{4}}{4}+\text{\cdots }\end{array}\]
  Since $f(x)=\text{ln}\ (1+x)$ is an antiderivative of $\frac{1}{1+x},$ it remains to solve for the constant C. Since $\text{ln}\ (1+0)=0,$ we have $C=0.$ Therefore, a power series representation for $f(x)=\text{ln}\ (1+x)$ is
  \[\begin{array}{ll}\text{ln}\ (1+x) & =x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-\frac{{x}^{4}}{4}+\text{\cdots } \\ & =\sum _{n=1}^{\infty }{(-1)}^{n+1}\frac{{x}^{n}}{n}\end{array}\]
  for $|x|<1.$ does not guarantee anything about the behavior of this power series at the endpoints. However, checking the endpoints, we find that at $x=1$ the series is the alternating harmonic series, which converges. Also, at $x=-1,$ the series is the harmonic series, which diverges. It is important to note that, even though this series converges at $x=1,$ does not guarantee that the series actually converges to $\text{ln}\ (2).$ In fact, the series does converge to $\text{ln}\ (2),$ but showing this fact requires more advanced techniques. (Abel’s theorem, covered in more advanced texts, deals with this more technical point.) The interval of convergence is $(-1,1].$
2. The derivative of $f(x)={\text{tan}}^{-1}x$ is ${f}^{'}(x)=\frac{1}{1+{x}^{2}}.$ We know that
  \[\begin{array}{ll}\frac{1}{1+{x}^{2}} & =\frac{1}{1-(\text{-}{x}^{2})} \\ & =\sum _{n=0}^{\infty }{(\text{-}{x}^{2})}^{n} \\ & =1-{x}^{2}+{x}^{4}-{x}^{6}+\text{\cdots }\end{array}\]
  for $|x|<1.$ To find a power series for $f(x)={\text{tan}}^{-1}x,$ we integrate this series term-by-term.
  \[\begin{array}{ll}\int {f}^{'}(x)\ dx & =\int (1-{x}^{2}+{x}^{4}-{x}^{6}+\text{\cdots })\ dx \\ & =C+x-\frac{{x}^{3}}{3}+\frac{{x}^{5}}{5}-\frac{{x}^{7}}{7}+\text{\cdots }\end{array}\]
  Since ${\text{tan}}^{-1}(0)=0,$ we have $C=0.$ Therefore, a power series representation for $f(x)={\text{tan}}^{-1}x$ is
  \[\begin{array}{ll}{\text{tan}}^{-1}x & =x-\frac{{x}^{3}}{3}+\frac{{x}^{5}}{5}-\frac{{x}^{7}}{7}+\text{\cdots } \\ & =\sum _{n=0}^{\infty }{(-1)}^{n}\frac{{x}^{2n+1}}{2n+1}\end{array}\]
  for $|x|<1.$ Again, does not guarantee anything about the convergence of this series at the endpoints. However, checking the endpoints and using the alternating series test, we find that the series converges at $x=1$ and $x=-1.$ As discussed in part a., using Abel’s theorem, it can be shown that the series actually converges to ${\text{tan}}^{-1}(1)$ and ${\text{tan}}^{-1}(-1)$ at $x=1$ and $x=-1,$ respectively. Thus, the interval of convergence is $[-1,1].$
Integrate the power series $\text{ln}\ (1+x)=\sum _{n=1}^{\infty }{(-1)}^{n+1}\frac{{x}^{n}}{n}$ term-by-term to evaluate $\int \text{ln}\ (1+x)\ dx.$

Откриј одговор.

$\sum _{n=2}^{\infty }\frac{{(-1)}^{n}{x}^{n}}{n(n-1)}$
If $f(x)=\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}$ and $g(x)=\sum _{n=0}^{\infty }{(-1)}^{n}\frac{{x}^{n}}{n\text{!}},$ find the power series of $\frac{1}{2}(f(x)+g(x))$ and of $\frac{1}{2}(f(x)-g(x)).$

Откриј одговор.

$\frac{1}{2}(f(x)+g(x))=\sum _{n=0}^{\infty }\frac{{x}^{2n}}{(2n)\text{!}}$ and $\frac{1}{2}(f(x)-g(x))=\sum _{n=0}^{\infty }\frac{{x}^{2n+1}}{(2n+1)\text{!}}.$
If $C(x)=\sum _{n=0}^{\infty }\frac{{x}^{2n}}{(2n)\text{!}}$ and $S(x)=\sum _{n=0}^{\infty }\frac{{x}^{2n+1}}{(2n+1)\text{!}},$ find the power series of $C(x)+S(x)$ and of $C(x)-S(x).$
$\frac{4}{(x-3)(x+1)}$

Откриј одговор.

$\frac{4}{(x-3)(x+1)}=\frac{1}{x-3}-\frac{1}{x+1}=-\frac{1}{3(1-\frac{x}{3})}-\frac{1}{1-(\text{-}x)}=-\frac{1}{3}\sum _{n=0}^{\infty }{(\frac{x}{3})}^{n}-\sum _{n=0}^{\infty }{(-1)}^{n}{x}^{n}=\sum _{n=0}^{\infty }({(-1)}^{n+1}-\frac{1}{{3}^{n+1}}){x}^{n}$
$\frac{3}{(x+2)(x-1)}$
$\frac{5}{({x}^{2}+4)({x}^{2}-1)}$

Откриј одговор.

$\frac{5}{({x}^{2}+4)({x}^{2}-1)}=\frac{1}{{x}^{2}-1}-\frac{1}{4}\ \frac{1}{1+{(\frac{x}{2})}^{2}}=\text{-}\sum _{n=0}^{\infty }{x}^{2n}-\frac{1}{4}\sum _{n=0}^{\infty }{(-1)}^{n}{(\frac{x}{2})}^{2n}=\sum _{n=0}^{\infty }(-1+(-1){\cdot }^{n+1}\frac{1}{{2}^{n+2}}){x}^{2n}$
$\frac{30}{({x}^{2}+1)({x}^{2}-9)}$
$\sum _{n=1}^{\infty }\frac{1}{{x}^{n}}$

Откриј одговор.

$\frac{1}{x}\sum _{n=0}^{\infty }\frac{1}{{x}^{n}}=\frac{1}{x}\ \frac{1}{1-\frac{1}{x}}=\frac{1}{x-1}$
$\sum _{n=1}^{\infty }\frac{1}{{x}^{2n}}$
$\sum _{n=1}^{\infty }\frac{1}{{(x-3)}^{2n-1}}$

Откриј одговор.

$\frac{1}{x-3}\ \frac{1}{1-\frac{1}{{(x-3)}^{2}}}=\frac{x-3}{{(x-3)}^{2}-1}$
$\sum _{n=1}^{\infty }(\frac{1}{{(x-3)}^{2n-1}}-\frac{1}{{(x-2)}^{2n-1}})$
Calculate the present values P of an annuity in which $10,000 is to be paid out annually for a period of 20 years, assuming interest rates of $r=0.03,r=0.05,$ and $r=0.07.$

Откриј одговор.

$P={P}_{1}+\text{\cdots }+{P}_{20}$ where ${P}_{k}=10,000\frac{1}{{(1+r)}^{k}}.$ Then $P=10,000\sum _{k=1}^{20}\frac{1}{{(1+r)}^{k}}=10,000\frac{1-{(1+r)}^{-20}}{r}.$ When $r=0.03,P\approx 10,000\ \times \ 14.8775=148,775.$ When $r=0.05,P\approx 10,000\ \times \ 12.4622=124,622.$ When $r=0.07,P\approx 105,940.$
Calculate the present values P of annuities in which $9,000 is to be paid out annually perpetually, assuming interest rates of $r=0.03,r=0.05$ and $r=0.07.$
Calculate the annual payouts C to be given for 20 years on annuities having present value $100,000 assuming respective interest rates of $r=0.03,r=0.05,$ and $r=0.07.$

Откриј одговор.

In general, $P=\frac{C(1-{(1+r)}^{\text{-}N})}{r}$ for N years of payouts, or $C=\frac{Pr}{1-{(1+r)}^{\text{-}N}}.$ For $N=20$ and $P=100,000,$ one has $C=6721.57$ when $r=0.03;C=8024.26$ when $r=0.05;$ and $C\approx 9439.29$ when $r=0.07.$
Calculate the annual payouts C to be given perpetually on annuities having present value $100,000 assuming respective interest rates of $r=0.03,r=0.05,$ and $r=0.07.$
Suppose that an annuity has a present value $P=1\ \text{million dollars}.$ What interest rate r would allow for perpetual annual payouts of $50,000?

Откриј одговор.

In general, $P=\frac{C}{r}.$ Thus, $r=\frac{C}{P}=5\ \times \ \frac{{10}^{4}}{{10}^{6}}=0.05.$
Suppose that an annuity has a present value $P=10\ \text{million dollars}\text{.}$ What interest rate r would allow for perpetual annual payouts of $100,000?
$x+{x}^{2}-{x}^{3}+{x}^{4}+{x}^{5}-{x}^{6}+\text{\cdots }$ (Hint: Group powers x^3k, ${x}^{3k-1},$ and ${x}^{3k-2}.)$

Откриј одговор.

$(x+{x}^{2}-{x}^{3})(1+{x}^{3}+{x}^{6}+\text{\cdots })=\frac{x+{x}^{2}-{x}^{3}}{1-{x}^{3}}$
$x+{x}^{2}-{x}^{3}-{x}^{4}+{x}^{5}+{x}^{6}-{x}^{7}-{x}^{8}+\text{\cdots }$ (Hint: Group powers x^4k, ${x}^{4k-1},$ etc.)
$x-{x}^{2}-{x}^{3}+{x}^{4}-{x}^{5}-{x}^{6}+{x}^{7}-\text{\cdots }$ (Hint: Group powers x^3k, ${x}^{3k-1},$ and ${x}^{3k-2}.)$

Откриј одговор.

$(x-{x}^{2}-{x}^{3})(1+{x}^{3}+{x}^{6}+\text{\cdots })=\frac{x-{x}^{2}-{x}^{3}}{1-{x}^{3}}$
$\frac{x}{2}+\frac{{x}^{2}}{4}-\frac{{x}^{3}}{8}+\frac{{x}^{4}}{16}+\frac{{x}^{5}}{32}-\frac{{x}^{6}}{64}+\text{\cdots }$ (Hint: Group powers ${(\frac{x}{2})}^{3k},{(\frac{x}{2})}^{3k-1},$ and ${(\frac{x}{2})}^{3k-2}.)$
$f(x)=2\sum _{n=0}^{\infty }{x}^{n},g(x)=\sum _{n=0}^{\infty }n{x}^{n}$

Откриј одговор.

${a}_{n}=2,{b}_{n}=n$ so ${c}_{n}=\sum _{k=0}^{n}{b}_{k}{a}_{n-k}=2\sum _{k=0}^{n}k=(n)(n+1)$ and $f(x)g(x)=\sum _{n=1}^{\infty }n(n+1){x}^{n}$
$f(x)=\sum _{n=1}^{\infty }{x}^{n},g(x)=\sum _{n=1}^{\infty }\frac{1}{n}{x}^{n}.$ Express the coefficients of $f(x)g(x)$ in terms of ${H}_{n}=\sum _{k=1}^{n}\frac{1}{k}.$
$f(x)=g(x)=\sum _{n=1}^{\infty }{(\frac{x}{2})}^{n}$

Откриј одговор.

${a}_{n}={b}_{n}={2}^{\text{-}n}$ so ${c}_{n}=\sum _{k=1}^{n-1}{b}_{k}{a}_{n-k}={2}^{\text{-}n}\sum _{k=1}^{n-1}1=\frac{n-1}{{2}^{n}}$ and $f(x)g(x)=\sum _{n=2}^{\infty }(n-1){(\frac{x}{2})}^{n}$
$f(x)=g(x)=\sum _{n=1}^{\infty }n{x}^{n}$
$f(x)=\frac{1}{1+x}=\sum _{n=0}^{\infty }{(-1)}^{n}{x}^{n}$

Откриј одговор.

The derivative of $f$ is $-\frac{1}{{(1+x)}^{2}}=\text{-}\sum _{n=0}^{\infty }{(-1)}^{n}(n+1){x}^{n}.$
$f(x)=\frac{1}{1-{x}^{2}}=\sum _{n=0}^{\infty }{x}^{2n}$
$f(x)=\frac{2x}{{(1+{x}^{2})}^{2}}=\sum _{n=1}^{\infty }{(-1)}^{n}(2n){x}^{2n-1}$

Откриј одговор.

The indefinite integral of $f$ is $-\frac{1}{1+{x}^{2}}=\sum _{n=0}^{\infty }{(-1)}^{n}{x}^{2n}.$

Symbols used here

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

summation
Add a_k for k = 1 up to n.

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

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

\infty

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

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

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

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pi

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

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

\sin,\ \cos,\ \tan

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

\log_b x,\ \ln x

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

\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.

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: Properties of Power Series

Combine power series by addition or subtraction.
Create a new power series by multiplication by a power of the variable or a constant, or by substitution.
Multiply two power series together.
Differentiate and integrate power series term-by-term.
The power series
For any integer
For any integer
Find the interval of convergence of the series

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.

Покушај сам.

Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Више у 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