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Power Series and Functions

Identify a power series and provide examples of them.

Form of a Power Series

A series of the form

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

where x is a variable and the coefficients cn are constants, is known as a power series. The series

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

is an example of a power series. Since this series is a geometric series with ratio \(r=x,\) we know that it converges if \(|x|<1\) and diverges if \(|x|\ge 1.\)

To make this definition precise, we stipulate that \({x}^{0}=1\) and \({(x-a)}^{0}=1\) even when \(x=0\) and \(x=a,\) respectively.

The series

\[\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}=1+x+\frac{{x}^{2}}{2\text{!}}+\frac{{x}^{3}}{3\text{!}}+\cdots\]

and

\[\sum _{n=0}^{\infty }n\text{!}{x}^{n}=1+x+2\text{!}{x}^{2}+3\text{!}{x}^{3}+\cdots\]

are both power series centered at \(x=0.\) The series

\[\sum _{n=0}^{\infty }\frac{{(x-2)}^{n}}{(n+1){3}^{n}}=1+\frac{x-2}{2\cdot 3}+\frac{{(x-2)}^{2}}{3\cdot {3}^{2}}+\frac{{(x-2)}^{3}}{4\cdot {3}^{3}}+\cdots\]

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

Convergence of a Power Series

Since the terms in a power series involve a variable x, the series may converge for certain values of x and diverge for other values of x. For a power series centered at \(x=a,\) the value of the series at \(x=a\) is given by \({c}_{0}.\) Therefore, a power series always converges at its center. Some power series converge only at that value of x. Most power series, however, converge for more than one value of x. In that case, the power series either converges for all real numbers x or converges for all x in a finite interval. For example, the geometric series \(\sum _{n=0}^{\infty }{x}^{n}\) converges for all x in the interval \((-1,1),\) but diverges for all x outside that interval. We now summarize these three possibilities for a general power series.

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

Representing Functions as Power Series

Being able to represent a function by an “infinite polynomial” is a powerful tool. Polynomial functions are the easiest functions to analyze, since they only involve the basic arithmetic operations of addition, subtraction, multiplication, and division. If we can represent a complicated function by an infinite polynomial, we can use the polynomial representation to differentiate or integrate it. In addition, we can use a truncated version of the polynomial expression to approximate values of the function. So, the question is, when can we represent a function by a power series?

Consider again the geometric series

\[1+x+{x}^{2}+{x}^{3}+\cdots =\sum _{n=0}^{\infty }{x}^{n}.\]

Recall that the geometric series

\[a+ar+a{r}^{2}+a{r}^{3}+\cdots\]

converges if and only if \(|r|<1.\) In that case, it converges to \(\frac{a}{1-r}.\) Therefore, if \(|x|<1,\) the series in converges to \(\frac{1}{1-x}\) and we write

\[1+x+{x}^{2}+{x}^{3}+\cdots =\frac{1}{1-x}\ \text{for}\ |x|<1.\]

As a result, we are able to represent the function \(f(x)=\frac{1}{1-x}\) by the power series

\[1+x+{x}^{2}+{x}^{3}+\cdots \ \text{when}\ |x|<1.\]

We now show graphically how this series provides a representation for the function \(f(x)=\frac{1}{1-x}\) by comparing the graph of f with the graphs of several of the partial sums of this infinite series.

Example

Try it.

Sketch a graph of \(f(x)=\frac{1}{1-x}\) and the graphs of the corresponding partial sums \({S}_{N}(x)=\sum _{n=0}^{N}{x}^{n}\) for \(N=2,4,6\) on the interval \((-1,1).\) Comment on the approximation \({S}_{N}\) as N increases.

Solution

From the graph in you see that as N increases, \({S}_{N}\) becomes a better approximation for \(f(x)=\frac{1}{1-x}\) for x in the interval \((-1,1).\)

Next we consider functions involving an expression similar to the sum of a geometric series and show how to represent these functions using power series.

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

Key Concepts

  • For a power series centered at \(x=a,\) one of the following three properties hold:
    1. The power series converges only at \(x=a.\) In this case, we say that the radius of convergence is \(R=0.\)
    2. The power series converges for all real numbers x. In this case, we say that the radius of convergence is \(R=\infty .\)
    3. There is a real number R such that the series converges for |x-a|R.\) In this case, the radius of convergence is R.
  • If a power series converges on a finite interval, the series may or may not converge at the endpoints.
  • The ratio test may often be used to determine the radius of convergence.
  • The geometric series \(\sum _{n=0}^{\infty }{x}^{n}=\frac{1}{1-x}\) for \(|x|<1\) allows us to represent certain functions using geometric series.

Key Equations

Power series centered at \(x=0\)\(\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+\cdots\)
Power series centered at \(x=a\)\(\sum _{n=0}^{\infty }{c}_{n}{(x-a)}^{n}={c}_{0}+{c}_{1}(x-a)+{c}_{2}{(x-a)}^{2}+\cdots\)

Power Series and Functions

In the following exercises, state whether each statement is true, or give an example to show that it is false.

In the following exercises, suppose that \(|\frac{{a}_{n+1}}{{a}_{n}}|\to 1\) as \(n\to \infty .\) Find the radius of convergence for each series.

In the following exercises, find the radius of convergence R and interval of convergence for \(\sum {a}_{n}{x}^{n}\) with the given coefficients \({a}_{n}.\)

In the following exercises, find the radius of convergence of each series.

In the following exercises, use the ratio test to determine the radius of convergence of each series.

In the following exercises, given that \(\frac{1}{1-x}=\sum _{n=0}^{\infty }{x}^{n}\) with convergence in \((-1,1),\) find the power series for each function with the given center a, and identify its interval of convergence.

Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.

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

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Praktika (40)

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  1. For each of the following series, find the interval and radius of convergence.

    1. \(\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}\)
    2. \(\sum _{n=0}^{\infty }n\text{!}{x}^{n}\)
    3. \(\sum _{n=0}^{\infty }\frac{{(x-2)}^{n}}{(n+1){3}^{n}}\)
    Atskleisti atsakymą
    1. To check for convergence, apply the ratio test. We have
      \[\begin{array}{ll}\rho & =\underset{n\to \infty }{\text{lim}}|\frac{\frac{{x}^{n+1}}{(n+1)\text{!}}}{\frac{{x}^{n}}{n\text{!}}}| \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{x}^{n+1}}{(n+1)\text{!}}\cdot \frac{n\text{!}}{{x}^{n}}| \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{x}^{n+1}}{(n+1)\cdot n\text{!}}\cdot \frac{n\text{!}}{{x}^{n}}| \\ & =\underset{n\to \infty }{\text{lim}}|\frac{x}{n+1}| \\ & =|x|\underset{n\to \infty }{\text{lim}}\frac{1}{n+1} \\ & =0<1\end{array}\]
      for all values of x. Therefore, the series converges for all real numbers x. The interval of convergence is \((\text{-}\infty ,\infty )\) and the radius of convergence is \(R=\infty .\)
    2. Apply the ratio test. For \(x\ne 0,\) we see that
      \[\begin{array}{ll}\rho & =\underset{n\to \infty }{\text{lim}}|\frac{(n+1)\text{!}{x}^{n+1}}{n\text{!}{x}^{n}}| \\ & =\underset{n\to \infty }{\text{lim}}|(n+1)x| \\ & =|x|\underset{n\to \infty }{\text{lim}}(n+1) \\ & =\infty .\end{array}\]
      Therefore, the series diverges for all \(x\ne 0.\) Since the series is centered at \(x=0,\) it must converge there, so the series converges only for \(x=0.\) The interval of convergence is the single value \(x=0\) and the radius of convergence is \(R=0.\)
    3. In order to apply the ratio test, consider
      \[\begin{array}{ll}\rho & =\underset{n\to \infty }{\text{lim}}|\frac{\frac{{(x-2)}^{n+1}}{(n+2){3}^{n+1}}}{\frac{{(x-2)}^{n}}{(n+1){3}^{n}}}| \\ & =\underset{n\to \infty }{\text{lim}}|\frac{{(x-2)}^{n+1}}{(n+2){3}^{n+1}}\cdot \frac{(n+1){3}^{n}}{{(x-2)}^{n}}| \\ & =\underset{n\to \infty }{\text{lim}}|\frac{(x-2)(n+1)}{3(n+2)}| \\ & =\frac{|x-2|}{3}.\end{array}\]
      The ratio \(\rho <1\) if \(|x-2|<3.\) Since \(|x-2|<3\) implies that -31\) if \(|x-2|>3.\) Therefore, the series diverges if \(x<-1\) or \(x>5.\) The ratio test is inconclusive if \(\rho =1.\) The ratio \(\rho =1\) if and only if \(x=-1\) or \(x=5.\) We need to test these values of x separately. For \(x=-1,\) the series is given by
      \[\sum _{n=0}^{\infty }\frac{{(-1)}^{n}}{n+1}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots .\]
      Since this is the alternating harmonic series, it converges. Thus, the series converges at \(x=-1.\) For \(x=5,\) the series is given by
      \[\sum _{n=0}^{\infty }\frac{1}{n+1}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots .\]
      This is the harmonic series, which is divergent. Therefore, the power series diverges at \(x=5.\) We conclude that the interval of convergence is \([-1,5)\) and the radius of convergence is \(R=3.\)

  2. Find the interval and radius of convergence for the series \(\sum _{n=1}^{\infty }\frac{{x}^{n}}{\sqrt{n}}.\)

    Atskleisti atsakymą

    The interval of convergence is \([-1,1).\) The radius of convergence is \(R=1.\)

  3. Sketch a graph of \(f(x)=\frac{1}{1-x}\) and the graphs of the corresponding partial sums \({S}_{N}(x)=\sum _{n=0}^{N}{x}^{n}\) for \(N=2,4,6\) on the interval \((-1,1).\) Comment on the approximation \({S}_{N}\) as N increases.

    Atskleisti atsakymą

    From the graph in you see that as N increases, \({S}_{N}\) becomes a better approximation for \(f(x)=\frac{1}{1-x}\) for x in the interval \((-1,1).\)

  4. Sketch a graph of \(f(x)=\frac{1}{1-{x}^{2}}\) and the corresponding partial sums \({S}_{N}(x)=\sum _{n=0}^{N}{x}^{2n}\) for \(N=2,4,6\) on the interval \((-1,1).\)

  5. Use a power series to represent each of the following functions \(f.\) Find the interval of convergence.

    1. \(f(x)=\frac{1}{1+{x}^{3}}\)
    2. \(f(x)=\frac{{x}^{2}}{4-{x}^{2}}\)
    Atskleisti atsakymą
    1. You should recognize this function f as the sum of a geometric series, because
      \[\frac{1}{1+{x}^{3}}=\frac{1}{1-(\text{-}{x}^{3})}.\]
      Using the fact that, for \(|r|<1,\frac{a}{1-r}\) is the sum of the geometric series
      \[\sum _{n=0}^{\infty }a{r}^{n}=a+ar+a{r}^{2}+\cdots ,\]
      we see that, for \(|\text{-}{x}^{3}|<1,\)
      \[\begin{array}{ll}\frac{1}{1+{x}^{3}} & =\frac{1}{1-(\text{-}{x}^{3})} \\ & =\sum _{n=0}^{\infty }{(\text{-}{x}^{3})}^{n} \\ & =1-{x}^{3}+{x}^{6}-{x}^{9}+\cdots .\end{array}\]
      Since this series converges if and only if \(|\text{-}{x}^{3}|<1,\) the interval of convergence is \((-1,1),\) and we have
      \[\frac{1}{1+{x}^{3}}=1-{x}^{3}+{x}^{6}-{x}^{9}+\cdots \ \text{for}\ |x|<1.\]
    2. This function is not in the exact form of a sum of a geometric series. However, with a little algebraic manipulation, we can relate f to a geometric series. By factoring 4 out of the two terms in the denominator, we obtain
      \[\begin{array}{ll}\frac{{x}^{2}}{4-{x}^{2}} & =\frac{{x}^{2}}{4(1-\frac{{x}^{2}}{4})} \\ & =\frac{{x}^{2}}{4(1-{(\frac{x}{2})}^{2})}.\end{array}\]
      Therefore, we have
      \[\begin{array}{ll}\frac{{x}^{2}}{4-{x}^{2}} & =\frac{{x}^{2}}{4(1-{(\frac{x}{2})}^{2})} \\ & =\frac{\frac{{x}^{2}}{4}}{1-{(\frac{x}{2})}^{2}} \\ & =\sum _{n=0}^{\infty }\frac{{x}^{2}}{4}{(\frac{x}{2})}^{2n}.\end{array}\]
      The series converges as long as \(|{(\frac{x}{2})}^{2}|<1\) (note that when \(|{(\frac{x}{2})}^{2}|=1\) the series does not converge). Solving this inequality, we conclude that the interval of convergence is \((-2,2)\) and
      \[\begin{array}{ll}\frac{{x}^{2}}{4-{x}^{2}} & =\sum _{n=0}^{\infty }\frac{{x}^{2n+2}}{{4}^{n+1}} \\ & =\frac{{x}^{2}}{4}+\frac{{x}^{4}}{{4}^{2}}+\frac{{x}^{6}}{{4}^{3}}+\cdots \end{array}\]
      for \(|x|<2.\)

  6. Represent the function \(f(x)=\frac{{x}^{3}}{2-x}\) using a power series and find the interval of convergence.

    Atskleisti atsakymą

    \(\sum _{n=0}^{\infty }\frac{{x}^{n+3}}{{2}^{n+1}}\) with interval of convergence \((-2,2)\)

  7. If \(\sum _{n=1}^{\infty }{a}_{n}{x}^{n}\) converges, then \({a}_{n}{x}^{n}\to 0\) as \(n\to \infty .\)

    Atskleisti atsakymą

    True. If a series converges then its terms tend to zero.

  8. \(\sum _{n=1}^{\infty }{a}_{n}{x}^{n}\) converges at \(x=0\) for any real numbers \({a}_{n}.\)

  9. Given any sequence \({a}_{n},\) there is always some \(R>0,\) possibly very small, such that \(\sum _{n=1}^{\infty }{a}_{n}{x}^{n}\) converges on \((\text{-}R,R).\)

    Atskleisti atsakymą

    False. It would imply that \({a}_{n}{x}^{n}\to 0\) for |x|

  10. If \(\sum _{n=1}^{\infty }{a}_{n}{x}^{n}\) has radius of convergence \(R>0\) and if \(|{b}_{n}|\le |{a}_{n}|\) for all n, then the radius of convergence of \(\sum _{n=1}^{\infty }{b}_{n}{x}^{n}\) is greater than or equal to R.

  11. Suppose that \(\sum _{n=0}^{\infty }{a}_{n}{(x-3)}^{n}\) converges at \(x=6.\) At which of the following points might the series diverge? Use the fact that if \(\sum {a}_{n}{(x-c)}^{n}\) converges at x, then it converges at any point closer to c than x.

    1. \(x=1\)
    2. \(x=2\)
    3. \(x=3\)
    4. \(x=0\)
    5. \(x=5.99\)
    6. \(x=0.000001\)
    Atskleisti atsakymą

    It must converge on \((0,6]\) and hence at: a. \(x=1;\) b. \(x=2;\) c. \(x=3;\) d. \(x=0;\) e. \(x=5.99;\) and f. \(x=0.000001.\)

  12. Suppose that \(\sum _{n=0}^{\infty }{a}_{n}{(x+1)}^{n}\) converges at \(x=-2.\) At which of the following points must the series also converge? Use the fact that if \(\sum {a}_{n}{(x-c)}^{n}\) converges at x, then it converges at any point closer to c than x.

    1. \(x=2\)
    2. \(x=-1\)
    3. \(x=-3\)
    4. \(x=0\)
    5. \(x=0.99\)
    6. \(x=0.000001\)

  13. \(\sum _{n=0}^{\infty }{a}_{n}{2}^{n}{x}^{n}\)

    Atskleisti atsakymą

    \(|\frac{{a}_{n+1}{2}^{n+1}{x}^{n+1}}{{a}_{n}{2}^{n}{x}^{n}}|=2|x||\frac{{a}_{n+1}}{{a}_{n}}|\to 2|x|\) so \(R=\frac{1}{2}\)

  14. \(\sum _{n=0}^{\infty }\frac{{a}_{n}{x}^{n}}{{2}^{n}}\)

  15. \(\sum _{n=0}^{\infty }\frac{{a}_{n}{\pi }^{n}{x}^{n}}{{e}^{n}}\)

    Atskleisti atsakymą

    \(|\frac{{a}_{n+1}{(\frac{\pi }{e})}^{n+1}{x}^{n+1}}{{a}_{n}{(\frac{\pi }{e})}^{n}{x}^{n}}|=\frac{\pi |x|}{e}|\frac{{a}_{n+1}}{{a}_{n}}|\to \frac{\pi |x|}{e}\) so \(R=\frac{e}{\pi }\)

  16. \(\sum _{n=0}^{\infty }\frac{{a}_{n}{(-1)}^{n}{x}^{n}}{{10}^{n}}\)

  17. \(\sum _{n=0}^{\infty }{a}_{n}{(-1)}^{n}{x}^{2n}\)

    Atskleisti atsakymą

    \(|\frac{{a}_{n+1}{(-1)}^{n+1}{x}^{2n+2}}{{a}_{n}{(-1)}^{n}{x}^{2n}}|=|{x}^{2}||\frac{{a}_{n+1}}{{a}_{n}}|\to |{x}^{2}|\) so \(R=1\)

  18. \(\sum _{n=0}^{\infty }{a}_{n}{(-4)}^{n}{x}^{2n}\)

  19. \(\sum _{n=1}^{\infty }\frac{{(2x)}^{n}}{n}\)

    Atskleisti atsakymą

    \({a}_{n}=\frac{{2}^{n}}{n}\) so \(\frac{{a}_{n+1}x}{{a}_{n}}\to 2x.\) so \(R=\frac{1}{2}.\) When \(x=\frac{1}{2}\) the series is harmonic and diverges. When \(x=-\frac{1}{2}\) the series is alternating harmonic and converges. The interval of convergence is \(I=[-\frac{1}{2},\frac{1}{2}).\)

  20. \(\sum _{n=1}^{\infty }{(-1)}^{n}\frac{{x}^{n}}{\sqrt{n}}\)

  21. \(\sum _{n=1}^{\infty }\frac{n{x}^{n}}{{2}^{n}}\)

    Atskleisti atsakymą

    \({a}_{n}=\frac{n}{{2}^{n}}\) so \(\frac{{a}_{n+1}x}{{a}_{n}}\to \frac{x}{2}\) so \(R=2.\) When \(x=\pm 2\) the series diverges by the divergence test. The interval of convergence is \(I=(-2,2).\)

  22. \(\sum _{n=1}^{\infty }\frac{n{x}^{n}}{{e}^{n}}\)

  23. \(\sum _{n=1}^{\infty }\frac{{n}^{2}{x}^{n}}{{2}^{n}}\)

    Atskleisti atsakymą

    \({a}_{n}=\frac{{n}^{2}}{{2}^{n}}\) so \(R=2.\) When \(x=\pm 2\) the series diverges by the divergence test. The interval of convergence is \(I=(-2,2).\)

  24. \(\sum _{k=1}^{\infty }\frac{{k}^{e}{x}^{k}}{{e}^{k}}\)

  25. \(\sum _{k=1}^{\infty }\frac{{\pi }^{k}{x}^{k}}{{k}^{\pi }}\)

    Atskleisti atsakymą

    \({a}_{k}=\frac{{\pi }^{k}}{{k}^{\pi }}\) so \(R=\frac{1}{\pi }.\) When \(x=\pm \frac{1}{\pi }\) the series is an absolutely convergent p-series. The interval of convergence is \(I=[-\frac{1}{\pi },\frac{1}{\pi }].\)

  26. \(\sum _{n=1}^{\infty }\frac{{x}^{n}}{n\text{!}}\)

  27. \(\sum _{n=1}^{\infty }\frac{{10}^{n}{x}^{n}}{n\text{!}}\)

    Atskleisti atsakymą

    \({a}_{n}=\frac{{10}^{n}}{n\text{!}},\frac{{a}_{n+1}x}{{a}_{n}}=\frac{10x}{n+1}\to 0<1\) so the series converges for all x by the ratio test and \(I=(\text{-}\infty ,\infty ).\)

  28. \(\sum _{n=1}^{\infty }{(-1)}^{n}\frac{{x}^{n}}{\text{ln}\ (2n)}\)

  29. \(\sum _{k=1}^{\infty }\frac{{(k\text{!})}^{2}{x}^{k}}{(2k)\text{!}}\)

    Atskleisti atsakymą

    \({a}_{k}=\frac{{(k\text{!})}^{2}}{(2k)\text{!}}\) so \(\frac{{a}_{k+1}}{{a}_{k}}=\frac{{(k+1)}^{2}}{(2k+2)(2k+1)}\to \frac{1}{4}\) so \(R=4\)

  30. \(\sum _{n=1}^{\infty }\frac{(2n)\text{!}{x}^{n}}{{n}^{2n}}\)

  31. \(\sum _{k=1}^{\infty }\frac{k\text{!}}{1\cdot 3\cdot 5\cdots (2k-1)}{x}^{k}\)

    Atskleisti atsakymą

    \({a}_{k}=\frac{k\text{!}}{1\cdot 3\cdot 5\cdots (2k-1)}\) so \(\frac{{a}_{k+1}}{{a}_{k}}=\frac{k+1}{2k+1}\to \frac{1}{2}\) so \(R=2\)

  32. \(\sum _{k=1}^{\infty }\frac{2\cdot 4\cdot 6\cdots 2k}{(2k)\text{!}}{x}^{k}\)

  33. \(\sum _{n=1}^{\infty }\frac{{x}^{n}}{(\begin{array}{l}2n \\ n\end{array})}\) where \((\begin{array}{l}n \\ k\end{array})=\frac{n\text{!}}{k\text{!}(n-k)\text{!}}\)

    Atskleisti atsakymą

    \({a}_{n}=\frac{1}{(\begin{array}{l}2n \\ n\end{array})}\) so \(\frac{{a}_{n+1}}{{a}_{n}}=\frac{{((n+1)\text{!})}^{2}}{(2n+2)\text{!}}\ \frac{2n\text{!}}{{(n\text{!})}^{2}}=\frac{{(n+1)}^{2}}{(2n+2)(2n+1)}\to \frac{1}{4}\) so \(R=4\)

  34. \(\sum _{n=1}^{\infty }{\text{sin}}^{2}n{x}^{n}\)

  35. \(\sum _{n=1}^{\infty }\frac{{(n\text{!})}^{3}}{(3n)\text{!}}{x}^{n}\)

    Atskleisti atsakymą

    \(\frac{{a}_{n+1}}{{a}_{n}}=\frac{{(n+1)}^{3}}{(3n+3)(3n+2)(3n+1)}\to \frac{1}{27}\) so \(R=27\)

  36. \(\sum _{n=1}^{\infty }\frac{{2}^{3n}{(n\text{!})}^{3}}{(3n)\text{!}}{x}^{n}\)

  37. \(\sum _{n=1}^{\infty }\frac{n\text{!}}{{n}^{n}}{x}^{n}\)

    Atskleisti atsakymą

    \({a}_{n}=\frac{n\text{!}}{{n}^{n}}\) so \(\frac{{a}_{n+1}}{{a}_{n}}=\frac{(n+1)\text{!}}{n\text{!}}\ \frac{{n}^{n}}{{(n+1)}^{n+1}}={(\frac{n}{n+1})}^{n}\to \frac{1}{e}\) so \(R=e\)

  38. \(\sum _{n=1}^{\infty }\frac{(2n)\text{!}}{{n}^{2n}}{x}^{n}\)

  39. \(f(x)=\frac{1}{x};a=1\) (Hint: \(\frac{1}{x}=\frac{1}{1-(1-x)})\)

    Atskleisti atsakymą

    \(f(x)=\sum _{n=0}^{\infty }{(1-x)}^{n}\) on \(I=(0,2)\)

  40. \(f(x)=\frac{1}{1-{x}^{2}};a=0\)

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Kaip vartoti: Power Series and Functions

  1. Identify a power series and provide examples of them.
  2. Determine the radius of convergence and interval of convergence of a power series.
  3. Use a power series to represent a function.
  4. The series converges at
  5. The series converges for all real numbers
  6. There exists a real number
  7. To check for convergence, apply the ratio test. We have
  8. Apply the ratio test. For

Klausimai, kuriuos klausia žmonės

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.

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Daugiau informacijos Calculus

LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests